Answer Key Unit 2 Genetic Processes
13. Daughter cells formed during cell division are genetically identical to the parent cell.
Unit Preparation Questions (Assessing Readiness)
14. In plant cells, a cell plate is formed during cytokinesis and divides the cytoplasm into two. Animal cells are divided by a contractile ring rather than a cell plate.
(Student textbook pages 154–7) 1. b 2. In both animal and plant cells: A–cell membrane B–cytosol C–mitochondrion E–endoplasmic reticulum (smooth and rough) G–Golgi apparatus H–nucleus (or nuclear envelope) L–nucleus (or nuclear envelope) In plant cell only: I–central vacuole J–cell wall (or membrane) K–chloroplast In animal cell only: D–ribosomes F–vesicle 3. For efficient movement of materials the ratio of a cell’s surface area to volume must stay within a range that permits diffusion to all parts of the cell within a short amount of time (to support the metabolic activity of the cell). 4. d 5. Nucleus (contains) → chromosomes (composed of) → DNA (which is made up of) → genes (which code for) → proteins 6. d 7. Students should draw a double helix and label any small piece as a gene. 8. a 9. c 10. Mitosis involves the division of the nuclear material in the nucleus and cytokinesis is the process that divides the cytoplasm (including all organelles). 11. During cell division, DNA is tightly coiled in chromosomes and looks “bunched.” The rest of the time, the DNA looks more like long, loose threads. 12. A–centrosome B–chromosomes C–spindle fibres D–centromere
15. A–interphase B–growth and preparation (G1) C–DNA replication D–growth and preparation (G2) E–mitosis F–cytokinesis 16. Any two of: the DNA may be damaged; DNA replication may not have occurred; there were not enough nutrients to support cell growth; no additional cells of that type are needed; cell death may have been signalled 17. Sample answer: Skin cells have a short life span—skin is easily damaged and must often be replaced. Nerve cells have a long life span—these internal cells are rarely damaged and, once an organism is full size, there is no need to divide for growth. 18. Cancer results from uncontrolled cell reproduction and the lack of cell death triggered by errors in DNA. 19. Sample answer: The digestive system is made of organs including the stomach and small intestine, and tissues like muscle (which move food along) or epithelium (which line the tract). Muscle tissue is made of muscle cells and epithelial tissue is made of epithelial cells. Diagrams could look like the one that accompanies question 38 on student textbook page 437. 20. Cell differentiation describes the process by which a cell specializes to perform a specific function. Because they perform different jobs, muscle cells look very different from skin cells and nerve cells, for example. If cells didn’t differentiate, they would all be generalists—they would be able to do all functions, but not very well. 21. Stem cells can specialize to become many types of cell. Other types of adult cells can only divide to produce cells identical to themselves.
Biology 11 Answer Key Unit 2 • MHR TR
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Chapter 4 Cell Division and Reproduction Learning Check Questions (Student textbook page 164) 1. Interphase, mitosis, and cytokinesis 2. Interphase 3. At prophase, the cell’s chromatin condenses into chromosomes. Each chromosome exists as two copies of one chromosome, joined at a centromere. 4. When mitosis is inhibited, healing times increase. 5. Interphase would differ in length between cell types because different cells have different functions that are carried out during interphase. Mitosis and cytokinesis are processes that would likely be consistent for all cell types. 6. The daughter cells would either have twice as much DNA as a healthy cell, or no DNA. Neither cell would be viable. One would lack genetic material and the other would have too much to be able to survive.
(Student textbook page 172) 7. A gamete is a haploid sex cell and a zygote is a diploid cell. 8. Thirty-nine chromosomes; gametes are haploid, meaning they contain half the number of chromosomes that are in a diploid cell 9. See Figure 4.14 on student textbook page 172 for illustration; each homologous chromosome still consists of two sister chromatids. In mitosis, the sister chromatids are separated during anaphase. 10. The gametes would have 46 chromosomes (23 pairs). After fertilization, the zygote would have 92 chromosomes (four sets of 23). 11. The phases in meiosis II (metaphase II, anaphase II, and telophase II) are most like the phases of mitosis because, during these stages, the chromosomes align on the equator of the cell, sister chromatids are separated, and a new nucleus forms for each new cell. 12. During anaphase I and anaphase II, because this is when the homologous chromosomes and sister chromatids, are each separated
(Student textbook page 176) 13. The outcome of mitosis is genetically identical offspring, whereas the outcome of meiosis is one (or four) haploid cells with a genetically diverse
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MHR TR • Biology 11 Answer Key Unit 2
representation of the parent cell. This leads to genetically unique offspring. 14. 27 = 128 different gametes 15. Students may redraw the figure showing chromosomes that are mixes of pieces of blue and yellow. 16. Because of independent assortment and crossing over, all children will have some genetic material from each of their ancestors. 17. Alleles that are found close together on the same chromosome will be inherited together more often than those that are either far apart on a chromosome or on different chromosomes. 18. The ideal donor is an identical twin, because they are genetically identical.
(Student textbook page 185) 19. Artificial insemination allows wider access to highquality male donors. 20. Both of these processes allow for the introduction of genetic variation from different parts of the world. 21. A vector carries the gene of interest into the foreign cell. 22. They are much less expensive to produce in large quantities. 23. A company may use embryo transfer so that they can choose desirable characteristics for their animals, and shipping embryos is much easier than shipping animals. This also allows offspring to grow up in their permanent environment. 24. Sample answer: The bacteria do or do not do something to the protein that would normally occur in the human tissue.
Caption Questions Figure 4.4 (Student textbook page 162): Unequal distribution in anaphase could lead to unequal distribution of chromosomes between new cells. This would lead to chromosomal disorders or cells that were not viable. Figure 4.10 (Student textbook page 167): They are homologous because they are two X chromosomes that are the same length and carry genes for the same traits at the same location. Figure 4.12 (Student textbook page 169): six chromosomes Figure 4.13 (Student textbook page 171): Meiosis I would produce cloned cells, resulting in less genetic diversity introduced into the cell and the organism.
Figure 4.25 (Student textbook page 184): The vector DNA contains DNA only from one source, while recombinant DNA contains the vector DNA and the DNA being cloned. Figure 4.26 (Student textbook page 185): The nucleus is removed from the egg cell so that it will only contain the nucleus with the desired (inserted) genes. Figure 4.28 (Student textbook page 187): Reduced crop loss due to disease will result in larger harvests. This saves the farmer money and allows lower prices. Figure 4.29 (Student textbook page 188): Students may suggest that animal genes should not be manipulated for human benefit, period; that the welfare of the animals should be considered; or that any method available should be used to alleviate human suffering. Any answer supported by an argument is acceptable.
Section 4.1 Review Questions (Student textbook page 168) 1. All living things are composed of one of more cells. Cells are the smallest units of living organisms. New cells come only from pre-existing cells, by cell division. 2. Sample answer: Skin cells undergo mitosis more frequently than nerve cells do. The skin cells will divide and repair the cut before the nerve cells are replaced. 3. In a child, because a child is growing in addition to repairing and replacing tissue 4. G1 is a period of rapid growth and normal cell duties. S is a period of DNA replication so that each cell produced receives a full set of DNA. During G2, cells prepare for division; organelles are made so that new cells have full complement of information. 5. a. anaphase b. prophase c. telophase 6. Sketches should look like Figure 4.4. 7. Sample answer: DNA replicates and I am connected to my twin; we condense to make chromosomes, line up along the equator, are pulled apart to opposite sides, and are then surrounded by a new nuclear membrane. We relax and get to work. 8. Mitosis duplicates and divides the nuclear material and cytokinesis divides the cytoplasm, producing two daughter cells. 9. Daughter cells are genetically identical to parent cells (except in the case of a mutation).
that when cytokinesis occurred, there would be a completely random division of the genetic material or that cell division would stop. 11. The error most likely occurred during anaphase, when the chromosomes are divided for the new daughter cells. The centromere did not divide or spindle fibres only formed from one centrosome, resulting in all of the chromosomes moving to the same pole. When cytokinesis occurred, all of the chromosomes were on one side of the cell. The other daughter cell had no chromosomes. 12. Students should sketch the basic structure of DNA, indicating nucleotides and that a certain section of the DNA contains a gene. Thin threadlike substances should represent chromatin, and something resembling the chromosome structure. Sketches should show that DNA is a component of both chromatin and chromosomes, such as in Figure 4.7 on student textbook page 165. 13. Chromosome pairs are not necessarily identical (due to copying errors, etc.) but the pairs are homologous, meaning their genes are in corresponding locations but may have different coding. 14. Diagrams should combine Figures 4.5 (on student textbook page 163) and 4.9 (on student textbook page 167). 15. Diagram should show an entire set of chromosomes, with three that are identical in both size and banding. 16. The X and Y chromosomes determine the sex of the individual; a female has two Xs and a male has one X and one Y. 17. a. This is a karyotype, prepared by collecting a cell sample that is treated to stop cell division during metaphase of mitosis, then stained to produce a banding pattern on the chromosomes that is clearly visible under a microscope. Chromosomes are then sorted and paired. The autosomes are numbered 1 through 22 and the sex chromosomes are labelled as X or Y. b. Male c. Yes, this individual has the correct number of chromosomes (23 pairs, 46 altogether). 18. By matching homologous chromosomes (which should be the same size and pattern), the doctor would see that one chromosome is shorter than it should be.
10. Since microtubules are responsible for the accurate division of the chromosomes, it is likely that the chromosomes would not move to the poles, and Biology 11 Answer Key Unit 2 • MHR TR
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entanglement of chromatids during prophase 1 that leads to pieces of chromosome changing places.
Section 4.2 Review Questions (Student textbook page 181)
Both lead to variation by creating unique combinations of chromosomes and genes in the sex cells. Sketch should combine information from Figures 4.18 and 4.19 on student textbook page 175.
1. a. meiosis b. fertilization c. mitosis 2. a. 2n = 64 b. n = 32 c. 32 d. 64
12. Sketches should accurately represent the errors in chromosome structure as shown in Table 4.1 on student textbook page 177.
3. fertilization 4. Meiosis produces cells that: • are genetically diverse; and • contain half the number of chromosomes (are haploid) 5. a. Metaphase II b. Sketches should show that the sister chromatids separate and move to opposite poles of the cell. c. Different colours illustrate that genetic material was exchanged during crossing over. d. 2n = 8 6. Meiosis takes place in the reproductive organs (ovaries and testes). 7. Homologous pairs line up next to each other to make a tetrad. This leads to chromosomes becoming tangled (synapse) and DNA trading places. This leads to increased variation in the daughter cells. 8. Mitosis and meiosis II are very similar in that the chromosomes line up on the equator and are separated during anaphase. The significant difference is that mitosis separates the chromosomes of a diploid cell, but in meiosis II the cell is haploid. 9. Sample answer: Spermatogenesis r0001QSPEVDFT0001NBMF0001 TFY0001DFMMT0001 TQFSN
r000100150001DFMMT00010001QSPEVDFE r0001QSPEVDFT0001IBQMPJE0001 r0001PDDVST0001JO0001 HBNFUFT0001GSPN0001 UIF0001UFTUFT EJQMPJE0001DFMMT0001 r0001TUBSUT0001XJUI0001 CZ0001NFJPTJT TQFSNBUPHPOJVN r0001FWFO0001DZUPQMBTNJD0001 EJWJTJPO
Oogenesis r0001QSPEVDFT0001GFNBMF0001 TFY0001DFMMT0001 FHHT
r000100120001DFMM0001BOE000100140001QPMBS0001 CPEJFT0001QSPEVDFE r0001VOFWFO0001DZUPQMBTNJD0001 EJWJTJPO r0001TUBSUT0001XJUI0001 PPHPOJVN r0001PDDVST0001JO0001 UIF0001PWBSJFT
10. 25 = 32 distinct gametes 11. Independent assortment refers to the fact that the orientation of each pair of chromosomes along the equator is independent of the orientation of the other pairs (ensuring a mix of maternal and paternal DNA goes to each pole). Crossing over refers to the 4
MHR TR • Biology 11 Answer Key Unit 2
13. Chromosomes do not separate evenly during nondisjunction. During anaphase II, a centrosome may not divide, leading to sister chromatids both going to the same pole. During anaphase I, a spindle fibre may attach to both homologous chromosomes and pull them to the same pole. This results in one cell having one too many chromosomes (called trisomy) or one too few chromosomes (called monosomy). 14. By looking at the homologous pairings in a karyotype, a clinician can quickly see if one of the chromosomes is missing its homologue (monosomy) or has two homologues/three identical chromosomes (trisomy). 15. Invasive methods of prenatal genetic testing pose a risk to the fetus. If there are no indications of genetic abnormality, there is no reason to put the fetus at risk of miscarriage (i.e., invasive methods can be avoided). 16. Students’ arguments should be supported by statements from either the textbook or individual research. Students will likely find that it was difficult to be entirely in favour of or against this type of testing, illustrating the dilemmas faced by many.
Section 4.3 Review Questions (Student textbook page 190) 1. Selective breeding. Farmers choose the best animals to breed with each other by looking at the traits that they considered favourable (for example, the fastest or those which produced the most milk). 2. Ways to produce these traits should relate to finding parent animals with the same traits. Students may mention selective breeding or artificial insemination. 3. Answers should contain supporting information such as: In favour—anything that makes us better should be encouraged; taking the best characteristics and allowing people to pick and choose their skills will make us more productive Opposing—it is not right to interfere with nature; everyone could end up being highly skilled in the same areas and we wouldn’t have people who would want to or be able to do other jobs
4. Both embryo transfer and IVF involve in vitro fertilization of egg by sperm. In humans, there is usually no genetic basis for choosing the egg and sperm donor. Most human IVF procedures are undertaken as a result of a fertility problem. Usually in humans, the embryo is implanted into the female that donated the egg, whereas in animals, the embryo is usually implanted in an unrelated female 5. Vectors act as carriers of DNA that a scientist wants to clone, enabling that DNA to be copied to a foreign cell. This is important for applications such as gene therapy and making insulin. 6. Gene cloning copies a segment of DNA, usually for the purpose of protein production or study. Therapeutic cloning produces genetically identical cells, usually for medical treatments. Reproductive cloning produces genetically identical individuals. 7. Flowcharts and diagrams should accurately represent the process as shown in Figure 4.25 on student textbook page 184. 8. Binary fission; some cells underwent conjugation or had vectors inserted into them 9. Somatic cell nuclear transfer uses an egg cell (with its nucleus removed) and the nucleus from a somatic cell. The daughter cells are genetically identical to the somatic cell. 10. Producing insulin through transgenic plants is less expensive. 11. The animals produced through reproductive cloning suffer from health problems and reduced lifespan, and many are not even born alive. 12. He can choose to breed his schnauzer with another prize-winning dog (selective breeding) or he could clone it. Cloning would produce the most exact copy. 13. Stem cells are used in regenerative medicine because they are undifferentiated. When they are placed into a patient, they can be stimulated to differentiate and replace the defective cells of the patient. 14. Students may choose examples such as transgenic animals like goats which are designed to produce medical protein products like HGH in their milk, or pigs that can act as organ donors.
15. a. Sample answer: The Canadian government should consider the benefits of the transgenic carrots over regular carrots: are the worms and insects a significant crop risk? Will it cause more economic success? Where did the genes come from? Are the genes naturally-occurring in the area? Will pesticide eliminate desirable bugs as well? Might the pesticide genes cross over to other plants? Are there any possible health risks to those who eat the carrots? b. Sample answer: The biggest advantage is that crops will be larger and of a better quality. Other advantages may be that fewer pesticides are needed, and that pesticides are delivered at the site of concern and in the smallest possible amounts, reducing toxic runoff. Disadvantages might include a higher seed cost, resulting in higher market prices and fewer sales. The public may also be reluctant to buy transgenic food. c. Answers should show an understanding of the arguments mentioned in the subsection entitled “Regulating the Use of Transgenic Organisms” on student textbook page 188. 16. Students may agree or disagree. If they agree, arguments might include that it will save lives (prevent deadly allergic reactions). If they disagree, arguments might include that there are more pressing issues than creating a peanut that doesn’t cause an allergic reaction, or that the causes of allergies are more complex than a simple allergen–reaction relationship, and that other solutions are possible.
Chapter 4 Review Questions (Student textbook pages 195–7) 1. d 2. d 3. e 4. e 5. c 6. a 7. e 8. b 9. growth, repair, and maintenance 10. The bases in DNA connect to each other in a way that looks like the rungs of a ladder. The sugar and phosphate make the rails of the ladder, joining the rungs. The comparison is limited because a real ladder is flat but DNA is twisted in a helix.
Biology 11 Answer Key Unit 2 • MHR TR
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11. Both have 22 homologous pairs of chromosomes (called autosomes). The difference is in the sex chromosome pair: males have one X and one Y chromosome but females have two X chromosomes.
22. a. Students may name any tissue that undergoes growth or repair such as root tips in plants or embryo cells in animals. b. oogonium or spermatogonium
12. Diagram B represents metaphase I of meiosis because the homologous pairs are lined up across from each other on the equator. Diagram A represents metaphase of mitosis because the homologous pairs are lined up along the equator.
23. In females the sex chromosomes are a homologous pair (XX); in males they are not (XY). The X chromosomes contain the same genes in the same locations. X and Y do not contain the same genes in the same locations and are therefore not homologous.
13. haploid cells and cells that are genetically different
24. The micrograph shows crossing over of non-sister chromatids which is significant because it causes genetic variation.
14. Oogenesis produces only one egg because of asymmetrical cytoplasm division, which ensures enough cytoplasm to nourish the fertilized egg. Sperm do not need to have cytoplasm for nourishment, so each of the four cells produced during spermatogenesis can become a sperm cell. 15. When non-disjunction occurs during anaphase I, both sets of homologous chromosomes move to the same side of the cell. If non-disjunction happens in anaphase II, the sister chromatids do not separate. Non-disjunction can result in trisomy disorders such as Down syndrome or monosomy disorders such as Turners syndrome. 16. Answers should show an understanding of the four types of errors described in Table 4.1 on student textbook page 177. 17. If somatic cells can be induced to behave like stem cells, then embryonic stem cells will not be needed to continue stem cell research, thus avoiding a significant ethical dilemma in this type of research. 18. One adult donates an egg, another donates a somatic cell, and a third acts as the surrogate, carrying the clone. The clone is genetically identical to the adult who donates the somatic cell. 19. Applications of transgenic organisms include: drug production, environmental clean-up, providing organs for transplant, and improving the food supply. 20. Coiled chromatin is less prone to damage or tangling while chromosomes move around the cell. During interphase the uncoiled chromosomes are protected inside the nucleus. 21. Because A–T and C–G are complimentary pairs, when the DNA is unzipped for replication, each base acts as a template, attracting its complement. This results in two identical DNA strands. If the strands were not identical, they would not contain the same alleles and therefore not code for the same proteins; the daughter cells would not be identical.
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MHR TR • Biology 11 Answer Key Unit 2
25. Genes would move from one chromosome to another, which could reduce the cell’s function. 26. a. 24 = 16 distinct gametes b. 28 = 256 distinct gametes c. 23 = 8 distinct gametes 27. The major difference between selective breeding techniques and genetic engineering is that genetic engineering deals with the genetics of the organism on a gene-by-gene basis, altering genes if they are not desirable. 28. Chromosomes that are not homologous may not line up properly during metaphase, or may not separate properly during anaphase, which could lead to nonviable gametes. 29. Student answers should include chromosomes, spindle fibres, and centromeres. They will also likely include cell membrane, nuclear membrane, and nucleolus. Props might include balls or balloons, string and pipe cleaners, or modelling clay, or a floor-sized template for human “props” to move on. 30. Mitosis r0001QSPEVDFT0001UXP0001 EJQMPJE0001DFMMT0001 UIBU0001BSF0001 HFOFUJDBMMZ0001 JEFOUJDBM0001UP0001 UIF0001QBSFOU0001DFMM
r0001QSPEVDFT0001 EBVHIUFS0001DFMMT r0001DISPNPTPNFT0001MJOF0001 VQ0001BU0001UIF0001FRVBUPS r0001TFQBSBUFE0001XIFO0001 TQJOEMF0001ñCSFT0001 TIPSUFO 0001QVMMJOH0001 UIF0001DISPNPTPNFT0001 UP0001UIF0001QPMFT0001
Meiosis r0001QSPEVDFT0001GPVS0001 IBQMPJE0001DFMMT0001UIBU0001 IBWF0001IBMG0001UIF0001%/'0001 PG0001UIF0001QBSFOU0001DFMM r0001HFOFT0001SFDPNCJOFE EVSJOH0001DSPTTPWFS
31. Diagrams should look similar to Figure 4.9 on student textbook page 167. 32. Diagrams should look similar to Figure 4.20 on student textbook page 178.
33. Pamphlets should show ultrasound, maternal blood tests, and amniocentesis such as in Figure 4.22 on student textbook page 180. Information should include the value of each type of test, how it is performed, the possible information gathered from the test, and the risks. 34. Sample answer: Method Selective breeding
Description Breeding of individuals selected because of their desirable traits
Pros/Cons Imprecise because of the many genes are involved and their dominant/ recessive nature
Artificial Transfer of semen insemination mechanically/ manually into a female’s reproductive tract
Makes semen from highquality males more widely available; does not require mating
Embryo transfer
Transfer of fertilized egg manually/ mechanically into a female’s reproductive tract
Embryos can be shipped; offspring raised in natural environment have more success
Recombinant DNA that contains a specific gene is inserted into the host
Useful genes can be added to other organisms (e.g., silk-producing goats can be made by inserting spider genes into mammary glands of goats)
Gene cloning
40. a. to c. A list of genetic tests is available at www. scienceontario.com. Arguments could include whether or not the test allows for a change in behaviour (having offspring) or a change in treatment (early diagnosis leads to better treatment). 41. Answers should show consideration for the welfare of the fetus and the ill child. 42. Genetic engineering could allow genes to be inserted into oranges that would make them grow in Ontario’s colder climate. 43. Students might research: somatic gene therapy, AIDS tests, prenatal diagnosis, drugs (e.g., insulin), human growth hormone, rTPA, hepatitis B vaccine, or hemophilia clotting factors. Answers should include the purpose of the drug and its benefits and risks. A list of potential topics is available at www. scienceontario.com. 44. Look for evidence that students have integrated their new learning, synthesizing the information into their opinions. 45. If stem cells can be used to generate tissues and organs for transplantation, enough organs could be grown to meet the demand.
Chapter 4 Self-Assessment Questions (Student textbook pages 198–9) 1. a 2. c
35. Organizers should include benefits such as regenerating nerve cells, treating heart disease, regrowing tissue and limbs, testing new drugs, and growing organs for transplant. Risks of stem cell research applications might include the ethics of human cloning. 36. Diagrams should illustrate crossing over and independent assortment.
3. a 4. c 5. d 6. c 7. e 8. e 9. a
37. Answers should include details about the research (as it applies to this unit), possible benefits to society and the environment, and possible consequences.
10. b
38. The graphic organizer should be in the form of an idea web, a spider map, a fishbone diagram, or a concept map. The information can be related to the Big Ideas found in the unit opener or the section titles. Student answers should reflect the Key Concepts and the Key Terms.
12. One benefit of cuttings is that they produce identical offspring. This is also a risk, as all offspring will be vulnerable to the same diseases and environmental conditions.
39. Artificial chromosomes would have to be able to attach to spindle fibres, so would have to have a functioning centromere.
11. Sketches should be similar to Figure 4.7 on student textbook page 165.
13. The number of chromosomes is reduced by a factor of two; 2n becomes n, diploid becomes haploid.
Biology 11 Answer Key Unit 2 • MHR TR
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14. Student diagrams should compare interphase to metaphase as in Figure 4.4 on student textbook pages 162–3. They should show that DNA is in the form of chromatin during the stage of the cell cycle known as interphase, which includes the G1, S, and G2 phases. During metaphase, however, DNA is highly condensed in the form of chromosomes. 15. Sample answer: chromatid spindle fibre centriole (or centrosome)
centromere
16. Students need not create a graphic organizer; however, one is provided to summarize all possible differences. Characteristic
Meiosis
Mitosis
Number of cells produced
4
2
Number of chromosomes
n (haploid)
2n (diploid)
Number of divisions 2
1
Variation
All different
Identical
Location
Germ cells
Somatic Cells
Product
Gametes
Somatic Cells
Purpose
Sexual reproduction
Repair, growth, replacement, reproduction
17. Sample answer:
blonde
black
brown
blue
18. A horse will produce gametes that have 32 chromosomes. A donkey will produce gametes with 31 chromosomes. When the two combine in meiosis, the diploid number is odd; not every chromosome has
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MHR TR • Biology 11 Answer Key Unit 2
a homologue. This means that meiosis cannot occur, so no sex cells can be produced. 19. a. Karyotyping uses a microscope to observe cells during metaphase of mitosis. The chromosomes were then sorted by length, banding pattern, and position of the centromere. b. This is a male with one too many X chromosomes (XXY). He has Kleinfelter’s syndrome. c. Non-disjunction during either anaphase of meiosis causes Kleinfelter’s syndrome. The chromosomes do not separate properly, resulting in one cell having too many chromosomes and another having too few. 20. a. Trisomy 21 b. Non-disjunction during either anaphase of meiosis; the chromosomes do not separate properly, resulting in one cell having too many and another having too few. 21. Students may suggest that foreknowledge about disorders helps parents prepare for the required lifestyle or that it creates peace of mind if there are no disorders. Students may also suggest that testing may result in pregnancy termination, discrimination, or pose a danger to the fetus (if test is invasive). All answers are acceptable. 22. IVF → mitosis → one cell analyzed by PGD → only healthy embryos are implanted 23. Diagrams should be similar to Figure 4.25 on student textbook page 184. 24. Answers should include some of: Benefits—increased nutritional value, production of medicines, increase food supply Risks—disease transfer, animal rights issues, need to control growth with stronger chemicals that might harm the environment, long-term health effects are not known. Students may suggest limitations such as the inability to fully test organisms or the need to label foods to explain that they are transgenic. Any limitations based on the science are acceptable. 25. In therapeutic cloning, stem cells can be programmed for many different purposes. For example, to be bone marrow cells for cancer treatment, nerve cells to treat neurological diseases, cardiac cells to treat heart disease, or pancreatic cells to treat diabetes. Ethical concerns involve the use of embryos as a source of stem cells, and the question of how long and under what conditions life should be prolonged.
Chapter 5 Patterns of Inheritance
16. All of the plants are tall with purple flowers.
Learning Check Questions
(Student textbook page 223) 17. See Figure 5.11 (student textbook page 219) for sample graphics. a. empty square b. shaded circle c. horizontal line connecting male and female
(Student textbook page 205) 1. A cross of two individuals that differ by one trait 2. 3:1 3. The F1 generation exhibits only the dominant form of the trait, and the F2 generation exhibits both the dominant and recessive forms in a 3:1 ratio. 4. Mendel needed to start with plants with predictable traits so that results were accurate and so that he was able to control his experiments. 5. Student drawings should be similar to Figures 5.3 (student textbook page 204) and 5.4 (student textbook page 205), but showing wrinkled and round seeds. 6. Brown is the dominant eye colour. Each parent has the allele for brown eyes. If a child inherits at least one allele for brown eyes, then the child will have brown eyes. If a child inherits two recessive alleles, the child will have blue eyes.
(Student textbook page 212) 7. a. T and t b. G c. f 8. 3:1 9. 3 round seeds:1 wrinkled seed 10. Sample answer: B
b
B
Bb
Bb
b
Bb
bb
18. The gene for the trait is carried on one of the autosomal chromosomes. 19.
I II
1
1
2
2
3
20. Two unaffected parents having an affected child 21. Because the lethal disorder is an autosomal recessive disease, people with only one recessive allele will carry the gene without getting sick, passing the gene to their offspring. 22. Since affected parents can have an unaffected child, woolly hair must be an autosomal dominant phenotype. Both parents must have one recessive hair gene, if a child has non-woolly hair. The genotypes of the children with non-woolly hair are certain, they must be doubly recessive (ww). You cannot be certain of the genotype of the child with woolly hair as only one dominant allele is necessary to produce woolly hair (W_). Sample pedigree:
I
1
2
11. 210 round-seeded plants:70 wrinkled-seed plants 12. Yes, if both parents are heterozygous for dimples, their child could be non-dimpled.
(Student textbook page 215) 13. A monohybrid cross involves a cross between two individuals that differ in one gene (trait). A dihybrid cross differs by two genes (traits). 14. a. TG, tG, Tg, and tg b. ABc and Abc 15. Sample answer: Plants and animals are made of many genes and many traits, and they work together to provide a healthy animal or plant. Therefore, it makes sense to study the inheritance pattern of groups of genes.
II
1
2
3
Caption Questions Figure 5.3 (Student textbook page 204): The green form seems to have disappeared. The allele for green is still on a chromosome, it is just not expressed. The chromosome doesn’t disappear. Figure 5.7 (Student textbook page 209): Because purple is a dominant allele, the purple plant may still carry the recessive allele. The Punnett square shows the possible offspring. Just because there are four offspring possibilities doesn’t mean that all four will occur from a set of parents.
Biology 11 Answer Key Unit 2 • MHR TR
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Figure 5.8 (Student textbook page 213): Plants of genotype Y_R_ (yellow round) to Y_rr (yellow wrinkled) to yyR_ (green round) to yyrr (green wrinkled) is 9:3:3:1. The ratio of YY:Yy:yy is 1:2:1 and the ratio of RR:Rr:rr is 1:2:1.
8. Sample answer: Parent Phenotypes:
Figure 5.10 (Student textbook page 217): They would likely be inherited together. Figure 5.13 (Student textbook page 221): Individual II-2 must be heterozygous because one offspring is unaffected.
Section 5.1 Review Questions (Student textbook page 207) 1. Dominant refers to an allele that will be expressed when present alone (e.g., freckles). Recessive refers to an allele that will only be expressed in individual who have two alleles for that trait (e.g., no freckles). 2. Traits are determined by pairs of alleles that segregate during meiosis so that each gamete receives one allele. Gametes combine to form a zygote that contains alleles from both the mother and the father. Alleles in the gamete that were not expressed in the parent are the recessive alleles. In the offspring, if the zygote receives two alleles for the recessive trait (e.g., grandmother’s hair), the trait will be expressed. 3. They support the Mendelian ratio because the ratio of second-generation plants is 149 tall to 53 short, or roughly 3:1. 4. a. yellow b. A heterozygous parent may have been used. 5. Cross a true-breeding purple flowered plant with a true-breeding white flowered plant. All of the flowers in the F1 generation will exhibit the dominant allele’s trait. 6. Phenotype is the term applied to the physical expression (traits) of alleles and genotype refers to the combination of all alleles present in an individual. 7. Heterozygous means one of each type of allele for one or more genes and sufficiently describes the genotype. To describe the genotype of a homozygous individual, you need to know whether they have two alleles of the dominant form, such as YY, or two alleles of the recessive form, such as yy.
10
MHR TR • Biology 11 Answer Key Unit 2
tall Parent Genotypes:
homozygous dominant (TT)
short homozygous recessive (tt)
F1 Phenotypes:
all tall F1 Genotypes:
Tt (heterozygous)
F2 Phenotypes:
_3 are tall 4
F2 Genotypes:
_1 are short 4
Tt (heterozygous), TT (homozygous dominant), tt (homozygous recessive)
9. a. Sample answer: F = freckles; f = no freckles b. ff c. Either FF or Ff is possible because F is dominant to f. You can only be sure by knowing the phenotypes of their parents or offspring and looking for a recessive individual. For example, if one parent has a recessive trait, the offspring must be Ff. 10. a. Black coat colour is dominant to white coat colour. b. Each parent must be Bb and the offspring is bb. 11. a. RR b. rr c. Rr 12. a. tall b. tall c. short
1 2 2 4 c. Genotypes: _ TTRR, _ TtRR, _ TTRr, _
Section 5.2 Review Questions (Student textbook page 218)
16
16
1. Sample answer: T
t
T
TT
Tt
t
Tt
tt
BB
Bb
Bb
bb
3. Yes, because the father will pass on the dominant curlyhair gene to some offspring, regardless of whether he is (Cc or CC). cc
Cc
cc
16
16
16
10. The ratio is close to 9:3:3:1, so the parents were likely PpSs. 9 3 11. long and grey: 144 _ ; long and black: 48 _ ; short
( )
16 3 1 _ ; short and black: 16 _ and grey: 48 16 16
( )
c. 25%
Cc
16
16
16
b
c
16
16
B
c
16
3 1 _ short and red, and _ short and yellow
2. a. Both parents are Bb. B b b.
c
16
9 3 tall and red, _ tall and yellow, Phenotypes: _
A Punnett square distributes the genes from each parent in all possible combinations.
C
16
1 1 2 2 1 TTrr, _ ttRR, _ ttRr, _ Ttrr, and _ ttrr TtRr, _
C
C
c
Cc
Cc
c
Cc
Cc
4. A test cross is used to determine the genotype of an individual with the dominant phenotype. 5. By doing a test cross with a white pig, the breeder will be able to determine that the pig is heterozygous if any white piglets are produced. Alternatively, the breeder could examine the pedigree of the unknown pig. 6. A Punnett square must represent all possible combinations of the four gametes from each of the individuals; there are 16 possible combinations. 7. The law of independent assortment states that the two alleles for one gene assort independently of the alleles for other genes during the formation of gametes. If you followed two traits such as plant height and flower colour, the allele for plant height received by a gamete during meiosis would not influence which allele for flower colour was received. 8. a. Tall with purple flowers b. TtPp c. TP, tP, Tp, and tp 9. a. All F1 plants will be TtRr; tall with red fruit. b. TR, Tr, tR, tr
( )
( 16 )
12. Student answers should note the following: genes are carried on chromosomes. The two different alleles of any specific gene are carried on the two homologous chromosomes. When the homologous chromosomes separate during anaphase I of meiosis, the two alleles of any gene on that chromosome move into two different cells. This process corresponds with the concept of Mendel’s law of segregation. When all of the pairs of homologous chromosomes are separating during anaphase I, they do so independently. That is, different chromosomes have no influence on each other with respect to which cell the homologous chromosomes migrate into. Therefore, the combinations of alleles of different genes that migrate into one cell or the other is random. This process corresponds with the concept of Mendel’s law of independent assortment.
Section 5.3 Review Questions (Student textbook page 227) 1. mechanism of inheritance of a trait, to determine the genotype of past generations, and to predict the genotype of future offspring 2. Autosomal recessive traits can remain hidden in a carrier. Because they are caused by a recessive allele, only individuals with two copies will display the trait. This can only happen 25% of the time when two carriers produce offspring. 3. autosomal dominant 4. The CFTR gene that causes cystic fibrosis was identified by Lap-Chee Tsui, a geneticist working at the Hospital for Sick Children, in Toronto. 5. The genetic test to detect the presence of a CFTR allele for cystic fibrosis is only 85–90% accurate because there are so many mutations that cause cystic fibrosis. Her spouse must have had an allele that was not tested for or detected.
Biology 11 Answer Key Unit 2 • MHR TR
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6. Sample answer: Pros
Cons
• plan ahead for healthcare
• increased guilt
• plan whether or not to have children
• discrimination in hiring and for insurance
• prevention or early treatment
• reduced genetic diversity
7. a. autosomal recessive b. I-1, II-3, II-4, and III-3 are aa; I-2, II-1, and II-2 are Aa; III-1, III-2 and III-4 may be Aa or AA because both parents are heterozygous and the gametes sort independently. There is a 50% chance that they are Aa and a 25% chance they are AA. (Students may note that since we already know from the pedigree that III-1, III-2, and III-4 are unaffected, there is no chance of them being aa. Therefore, the probability 1 could be calculated as being _ chance of being AA 3 2 _ and probability of being Aa.) 3
8. The last student is correct; the pedigree could represent either situation. There is not enough information to determine whether the pedigree is recessive (two nonaffected parents having an affected child) or dominant. More generations are required. 9. The parents, I-1 and I-2 are both heterozygous, having normal pigmentation (Aa). The children who are albino have the genotype aa. The other two children could be either AA or Aa. Two-thirds of the non-albino genotypes are heterozygous. 1
Practice Problems (Student textbook page 212) 1. a. A only b. A and a c. a only d. A and a 2. Half the offspring will be Pp (purple) and half will be pp (white). 3. All plants produced will be heterozygous (Gg) with green pods. 4. Long-winged LL offspring will make up 25%, 50% will be Ll (long-winged), and 25% will be ll (curly-winged). 5. 50% 7. a. Both parents are Nn. b. Probabilities are the same: 25% nn and 75% Nn.
2
8. nn and Nn 1
2
3
4
10. Karyotyping is used to identify abnormalities such as additions, deletions, and translocations, at the chromosome level. FISH can be used to look for a smaller abnormality in a chromosome. Genetic testing is used to identify a specific gene mutation. 11. A genetic counsellor may use a pedigree to determine the genotypes of family members, and to explain options for genetic testing, the probability of passing on a disease-causing allele, symptoms, and available treatments. They may also provide emotional support for family members.
12
13. Students may agree, citing that gene therapy could replace a faulty gene that causes a genetic disease. Or, students may disagree, citing that there have been a number of problems (such as immune response and short-lived treatments) that have not yet been solved. Either answer is acceptable, when supported by arguments.
6. The only possible parent combination is Ss and ss.
I II
12. Gene therapy involves inserting a gene into a host cell using a vector, as do all gene cloning techniques. Like other forms of cloning, the hope is that cells that take up the gene will reproduce and produce a normal cell’s product. Unlike many forms of cloning, human cells do not take up the normal gene very effectively, so gene therapy is not as successful as other types of cloning.
MHR TR • Biology 11 Answer Key Unit 2
9. The parents may be either Bb or bb (but not BB) since the white pig must have inherited a recessive b gene from each parent. It is unlikely that the parents are BB and Bb since a ratio of 3:1 black:white would be expected if both parents were heterozygous (Bb). Since black pigs were produced, the parents cannot be homozygous recessive (bb). To be sure, the phenotype of the grandparents would need to be known, or if the pedigree could be re-evaluated after more offspring are born, to see if the ratio gets closer to 3:1 (implying parental genotypes of Bb and Bb) or 1:1 (implying parental genotypes of Bb and bb). 10. GG and gg
(Student textbook page 216) 11. a. Pptt × PpTT, where P = purple, p = white, T = tall, and t = short b. parent 1 can produce Pt and pt gametes and parent 2 can produce PT and pT gametes. 12. Black is dominant to brown, and short hair is dominant to long hair, since brown hair and long hair were not expressed. The parents were most likely BBss and bbSS. 13. 3/16, or 18.75% 14. a. All of the F1 generation will have short black hair. b. One in 16 15. a. 50% curly, dark-haired, and 50% straight, darkhaired b. We can know the genotype of the parents by examining the phenotypes of their ancestors. 16. GGRr, where G = green, R = round, and r = wrinkled 17. 25% 18. Although both sets of parents would produce the set of offspring, the ratios are closer for a mating between one plant heterozygous for both genes and one plant homozygous recessive for both genes. 19. a. Curly hair must be dominant since the straight hair trait was not expressed in the parent generation but reappeared in the children. b. A cleft chin must be dominant since smooth chin was not expressed in the parent generation but reappeared in the children. c. Both parents are HhCc, where H = curly hair and C = cleft chin. 1 , or 6.25% d. _ 16
1 , or 25% 20. _ 4
Chapter 5 Review Questions (Student textbook pages 235–7) 1. d 2. c 3. b 4. d 5. a 6. c 7. c 8. d
9. Reasons to use pea plants for breeding trials include: their rapid growth means many generations are produced in little time; they have seven distinct characteristics that are controlled by a single gene with two alleles each; and it is easy to control their fertilization. 10. The genotype of the P (parent) generation is homozygous for the trait, true-breeding, with each parent displaying the opposite phenotype. Starting with true-breeding individuals ensures that their genotype is known. 11. Yes, if the phenotype is the recessive trait, because an individual displaying the recessive trait must be homozygous recessive. You cannot tell the genotype from the dominant phenotype, because the individual may be either heterozygous or homozygous dominant. 12. A test cross is a cross between a parent unknown genotype and a homozygous recessive parent. This is used to determine the genotype of an individual displaying the dominant trait. 13. 9 (both dominant traits):3 (one dominant trait): 3 (other dominant trait):1 (both recessive traits) 14. a. AB, Ab, aB, and ab b. ABC, ABc, AbC, and Abc 15. Affected individuals have at least one affected parent; two unaffected parents only have unaffected offspring. 16. a. rr b. Rr c. RR 17. Breed two true-breeding individuals with different alleles and look for what phenotype their offspring is. Only the dominant allele will be expressed in the offspring. The other allele must be recessive. 18. a. black b. black-haired (BB) and white-haired (bb) c. black-haired (Bb) 19. a. Dd or DD b. 50% drooping eyelids (Dd) and 50% non-drooping eyelids (dd) c. Probabilities relating to the combination of alleles apply to each offspring individually, not all offspring together. 20. a. P male1 = GG, P female = gg, P male2 = Gg b. 3:1 grey:albino
Biology 11 Answer Key Unit 2 • MHR TR
13
21. One of the parents is heterozygous and the other is homozygous recessive. The dominant allele could be determined by breeding each parent with an individual of the same colour and observing the offspring. For example, the yellow rat should be bred with another yellow rat. If any offspring from this cross are black, you know that yellow is dominant. 22. The diagram displays the results of a dihybrid cross. The F1 generation display only the dominant traits (yellow and round). In the F2 generation, all four genotypes are displayed. This illustrates the principle of independent assortment; the two genes or characteristics sort independently during meiosis. 23. The trait is dominant, because two affected parents have an unaffected child. 24. This shows an autosomal recessive inheritance pattern. All individuals have a normal phenotype except for I-2, III-2, and III-3, who are affected (and therefore must have an ss genotype). Because II-2, II-3, II-4, and II-5 are unaffected offspring of an affected individual, they must have an Ss genotype. The mother (I-1) may be either SS or Ss, having passed on the S gene to all her offspring. II-1 must have the Ss genotype since she is unaffected yet passed on an s to her offspring. III-2 and III-3 are affected, so must have the ss genotype. III-1 and III-4 are unaffected, but may be either SS or Ss. 25. Polling must be an autosomal dominant characteristic, since two individuals with the trait have offspring that do not have the trait. L
l
L
LL
Ll
l
Ll
ll
26. a.
I
Test
What the Test Analyzes
Diagnosed Disorder(s)
Illustration
Karyotyping
Number of chromosomes
Turner syndrome, Down syndrome, Klinefelter’s syndrome
See Figure 4.21 (student textbook page 178)
FISH
Shows chromosome abnormalities
CML
n/a
Gene testing
Mutations
Breast cancer n/a susceptibility
Biochemical testing
Abnormal enzymes
Tay-Sachs disease
n/a
28. Sample answers: Pro—I believe that genetic testing should be a standard medical test paid for by OHIP. There is no danger to the individual and the information gathered from the test could potentially save the person’s life. Knowing that you are at risk for cancer means that you can alter your lifestyle or make sure that you monitor your health and catch any symptoms early. Con—I believe that genetic testing should be optional. Not all people can cope with the stress of a positive test. The presence of these genes does not mean that a person will definitely get cancer. If a person wants to know if they are at risk and are ready to deal with any result, they should be able to have the test paid for by OHIP. 29. Students should include gene therapy benefits and risks discussed in Section 5.3 on student textbook page 226.
1
II
2
1
b. They are both either autosomal dominant or recessive. There is not enough information to be sure. c. The father is DdFf; the mother and son are ddff. 1 d. _ , or 25% 4
14
27. A mind map is well suited to this task. This sample answer uses a table:
MHR TR • Biology 11 Answer Key Unit 2
30. Students can choose any example but should describe how meiosis allows for variation (for example, illustrate the genetic diversity in a litter of kittens). 31. Students may research, for example, sickle cell anemia or Tay-Sachs disease. Students’ presentations should describe the disorder and the study, indicate why that group was chosen, address social and ethical concerns about this type of research, identify who benefits most from the research, and make an argument for who should be allowed access to the genetic information of the people in the study.
32. The graphic organizer should be in the form of an idea web, a spider map, a fishbone diagram, or a concept map. The information can be related to the Big Ideas found in the unit opener or the section titles. Student answers should reflect the Key Concepts and the Key Terms. 33. Does this dog have Von Willebrand’s disease? Do any of his parents, grandparents, or siblings have the disease? 34. Probability is not a certainty, especially in the case of offspring, in which each conception is an individual event. For each child, the probability that he or she will display a specific trait is the same (25%, in this case). The probability that subsequent children will display a trait is not influenced by the presence of that trait in previous offspring. The easiest example is boys and girls. Although the probability of having a boy is 1:2, a son may be born each time. 35. Students may argue that parents do or do not have a responsibility to inform children of recessive disorders they may have inherited. Either answer is acceptable. Sample answer: This information can help children decide whether or not to have their own children. If their spouse has the same gene, being a carrier means that their children could show the genetic disorder. For minor disorders such as colour blindness, this may seem trivial. 36. A list of genetic tests is available at www.scienceontario.com. 37. a.
I II
1
2
Chapter 5 Self-Assessment Questions (Student textbook pages 238–9) 1. a 2. c 3. c 4. c 5. c 6. a 7. a 8. b 9. b 10. d 11. a. Homozygous means having two of the same allele for a gene and heterozygous means having two different alleles for the same gene. b. An allele that is dominant is expressed when there are one to two copies but a recessive allele requires two alleles in order to be expressed. 12. The farmer should breed heterozygous cattle with homozygous dominant cattle. This will produce 50% heterozygous, with the best meat, and the other 50% homozygous dominant, which although will not have the best meat, will not die. 13. 50% 14. Cross the unknown plant with a short plant producing white flowers (a homozygous recessive plant). If the offspring include any plants with the recessive phenotype, then the parent plant must be heterozygous for that characteristic. 15. Both are heterozygous RrSs.
1
2 (Taku)
3 (Sara)
b. Is there a history of cystic fibrosis (CF) in Sara’s family? If not, their children will not have cystic fibrosis since children will not receive the two copies of the defective allele necessary to produce this autosomal recessive disorder. c. Genetic testing to identify whether or not they are carriers of cystic fibrosis d. The test is only 85-90% accurate and they may be concerned by possible use of the test results in the future.
16. Drawings should show chromosomes sorting during meiosis, illustrating that genes sort independently unless they are on the same chromosome. 17. Since recessive traits are only expressed in homozygotes, the trait may not appear in all generations. To identify the presence of an allele in a family, several generations of data must be obtained. 18.
I II
1
1
2
2
3
Biology 11 Answer Key Unit 2 • MHR TR
15
19. There isn’t enough information to determine the mode of inheritance. If two unaffected parents had an affected child, we could tell that the mode of inheritance is autosomal recessive. If two affected parents have an unaffected child, we could tell that the mode of inheritance is autosomal dominant. We need more information from other generations to determine the mode of inheritance. 20. Pedigrees should show both parents affected. Children may or may not be affected despite the 75 percent probability of them having a peaked hairline. Sample answer:
I II
1
1
2
2
3
4
21. a. Cystic fibrosis is an autosomal recessive disorder caused by one of over 1600 different mutations to the CFTR gene on chromosome 7. The genetic tests for this disorder each look for only one specific mutation. b. A child inherits the gene from parents, and must inherit the gene from both parents to be affected. c. A person with cystic fibrosis has a life expectancy in the late 40s. If they have the disorder or are carriers of the disorder, they may not choose to reproduce or may experience difficulty finding life partners. The disorder does not have delayed symptoms so there is no risk of future discrimination because they may get sick. 22. Any personal opinion is acceptable if it shows evidence of incorporating knowledge of genetic diseases, testing, and inheritance. 23. The allele for Huntington disease does lead to a drastically shortened life span, so it is sometimes considered to be a lethal disorder. On the other hand, it may be argued that it is not a lethal disorder since sufferers normally survive long enough to reproduce. 24. Students should discuss skills that include counselling and remaining impartial. Counsellors must be able to explain the alternatives and provide support regardless of the decisions made by a family. 25. Students should discuss that if perfected, gene therapy will be able to replace abnormal genes with normal ones, but that the process is not yet perfected. Problems with the viral vector have led to some deaths, and even if not fatal, there has been no long-term success in human trials. 16
MHR TR • Biology 11 Answer Key Unit 2
Chapter 6 Complex Patterns of Inheritance Learning Check Questions (Student textbook page 244) 1. Incomplete dominance is a condition in which neither allele completely conceals the presence of the other, resulting in an intermediate phenotype. Codominance is a condition in which both alleles in a heterozygote are fully expressed. 2. Multi-letter notations represent the combination of allele dominance. Lower case letters are only used to represent alleles that are recessive to another allele. Capital letters show that neither trait is recessive to the other. 3. a. Light purple b. White and purple 4. Having some cells that are sickle-shaped provides the carrier with immunity to malaria, a deadly disease prevalent in some parts of Africa. 5. Sample answer: Red and white flowers that produce pink flowers suggest an intermediate phenotype, and codominance is found in roan cows. 6. When scientists were able to take a closer look at the blood cells in heterozygous individuals, they saw that both normal and sickle cell blood cells were present, supporting the idea of codominance.
(Student textbook page 253) 7. Linked genes do not assort independently. 8. Linked genes are found on the same chromosome and are inherited together; phenotypic ratios that they produce do not match the Mendelian ratio. 9. The more frequently linked genes get separated, the farther apart they are on a chromosome. This provides an idea of the relative positions of genes on a chromosome, which can be visualized in a gene map. 10. Yes, otherwise the ratio would be 9:3:3:1, which is expected for unlinked traits resulting from a dihybrid cross. 11. Sample answer: The genes for the traits are likely to be located on the X and Y sex chromosomes. 12. Because linked genes can be separated from each other during meiosis, the gene that is being tested for may not always indicate the presence of the diseasecausing allele.
(Student textbook page 262) 13. Determined the DNA sequence of the human genome; identified all of the genes; made the genes accessible to all, address social, ethical, and legal issues that arise; and sequenced other organism’s genomes 14. about 2% 15. biological molecule sequencing technology, computer software, and communication technology 16. Bioinformatics allowed for organization and analysis of large amounts of sequence data that was generated from labs all over the world. 17. The Internet allowed for rapid and easy transfer of data (and communication) between labs all over the world. 18. Sample answer: A new DNA sequence of the genome for an organism has been identified and they want to compare that sequence to that of the human genome.
Because this disorder shows intermediate levels of LDL in heterozygotes, it is incomplete dominance. If it was a case of codominance, you might see that levels are high one day and low the next. 3. a. Students should use an uppercase letter for the gene, for example C, and then use two uppercase letters for the superscripts, for example C R (red) and C W (white). The genotypes for the three radishes would then be C RC R (red), C WC W (white), and C RC W (purple). b. 1:2:1 red:purple:white 4. a. Feather colour must be a case of codominance since each allele is expressed independent of the other. b. black (C BC B) and speckled black and white (C BC W)
Caption Questions
5. a. Students should draw a blue organism (Bb). b. Students should draw a green organism (C BC Y). c. The student should draw an organism, parts of which are yellow and parts of which are blue (C BC Y).
Figure 6.6 (Student textbook page 245): Agouti: CC, Cc ch, Cc h, Cc; chinchilla: c chc ch, c chc h, c chc; Himalayan: c hc h, c hc; albino: cc
6. a. I-1, I-2, II-2, II-3, and II-4-Hb AHb S, II-1 and III-1 are Hb SHb S. All genotypes are determinable. b. 25%
Figure 6.9 (Student textbook page 248): As the number of gene pairs involved increases, so does the range of possible phenotypes.
7. a. Both parents are heterozygoous c chc and c hc. b. 2:1 chinchilla:Himalayan
Figure 6.10 (Student textbook page 251): PPLL, PpLL, PPLl, ppll Figure 6.11 (Student textbook page 252): Alleles that are close together will likely be transferred on the same bit during crossing over. The farther apart two alleles are, the more likely that they will get split up during crossing over. Figure 6.13 (Student textbook page 254): 50% of females will be X RX r (red-eyed); 50% of females will be X rX r (white-eyed); 50% of males will be X RY (red-eyed); 50% of males will be X rY (white-eyed)
Section 6.1 Review Questions (Student textbook page 250) 1. a. incomplete dominance b. codominance 2. Hypercholesteremia leads to a build up of LDL (bad) cholesterol in the blood. Homozygotes for the normal allele have the lowest LDL levels, heterozygotes have an intermediate level, and homozygotes for the disease allele have the highest levels, demonstrating the intermediate phenotype for the heterozygote. In incomplete dominance there is an intermediate phenotype that is a blend of the extreme phenotypes.
8. 0%; To get an agouti rabbit, you must have at least one agouti parent since it is the most dominant allele. 9. Sample answer: ABO Blood Groups
Example
Multiple alleles
I A, I B, and i
Codominance
I A and I B alleles, both are expressed, producing blood type AB
Dominant/recessive
ii results in O blood type I Ai and I Bi result in A and B blood types
10. The gene that controls coat colour in Siamese cats may be influenced by environmental factors including temperature. Areas of colder temperature (such as extremities: ears and tail) may lead to a darker coat colour. 11. The continuous variation expressed in human skin colour (from very light to very dark) suggests that skin colour is a polygenic trait.
Biology 11 Answer Key Unit 2 • MHR TR
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Section 6.2 Review Questions (Student textbook page 259) 1. Sample answer: Start with true-breeding plants for both traits. Cross to obtain the F1 generation which should all display the dominant traits. Cross two plants from the F1 generation. If the two genes are not linked, the ratio of phenotypes in the F2 will be 9:3:3:1. If the genes are linked, the ratio will be closer to 3:1. 2. Genes that are on the same chromosome are linked. When crossing over occurs between non-sister chromatids, the alleles of the linked genes can segregate, or become “unlinked.” Alleles that were once on the same chromosome are then located on two different chromosomes. 3. Linked genes would not produce the expected Mendelian ratio for a dihybrid cross (9:3:3:1). 4. Alleles P and Q are closest on the chromosome, since the crossover frequency is the lowest. 5. Diagrams should look similar to Figure 6.11 (student textbook page 252). 6. They reproduce rapidly, produce many offspring, and sex determination is controlled by XY chromosomes. 7. a. The woman must be a carrier, X CX c, and the man must be X CY. b. The colour vision deficient child must be a boy, since a girl would inherit a normal allele from her father and cannot have CVD. 8. X hY (father) and X ?X h_ (mother) The father must have hemophilia since the woman inherited one of her X h from him and he only has one to give. The mother must have an X h also, but her other allele could be X H or X h. 9. A girl with Turner syndrome has only one X chromosome, so if she inherits an allele for CVD from her mother and no X chromosome from her father, she could have CVD. 10. Students will likely disagree. Although there are more males affected than females, the pedigree does not show the standard pattern of affected male through daughter to her son (Individual II-1/II-2). Students could argue X-linked recessive if they say that II-2 is a carrier. 11. An autosomal trait will have a comparable number of males and females affected. For X-linked recessive traits, there should be more males affected. Both will show traits skipping generations, but in an X-linked recessive inheritance pattern, it will skip generations (from an affected grandfather through the mother to an affected son). 18
MHR TR • Biology 11 Answer Key Unit 2
12.
I
1
II
1
2
2
3
3
The chance that his sisters are carriers is 50%, since they inherit a normal allele from their father, and have a 50% chance of inheriting the disease allele from their mother. 13. X-inactivation occurs in females, so a female with one allele for the disease trait will likely have only about 50% of her cells affected. 14. See Figure 6.14 (student textbook page 255) for the format of the diagram. 15. Because different X chromosomes are deactivated in patches of cells, some sweat glands will not contain the disorder and will therefore produce sweat. Other sweat glands will have an active, recessive X so they will not be able to produce sweat.
Section 6.3 Review Questions (Student textbook page 267) 1. Sequencing an organism’s genome makes it useful in genetic research. Also, knowing other sequences provides a basis for comparison with the human genome. 2. Sample answer: Book Pages
Genome Genome
Paragraphs
Chromosomes
Sentences
Genes
Words
Codons
Letters
Nucleotides
3. Facts learned through the Human Genome Project include: only 2 percent of nucleotides code for genes; there are 25 000 genes; 50 percent of our DNA is made of repeating sequences; and about 99.9% of the DNA sequence is almost exactly the same in all people. 4. Ongoing research includes determining the function of genes that were discovered and determining the function of the non-coding repeating regions of DNA. 5. Sample answer: Yes, if it was not for the Human Genome Project, we would not know the location of disorders that can be identified through genetic tests and we would not be able to treat disorders using gene therapy, or to create insulin through transgenic organisms.
6. The difference is that with the advent of bioinformatics, much of the genetic research that takes place involves the use of computers “(“dry lab”) rather than just “wet lab” equipment. 7. Sample answer: The components of the term bioinformatics are “bio” in reference to biology, “info” in reference to information, and “matics” in reference to mathematics. Alternative names include genometrics, biomancy, macrogenetistry, computational biology, and bionomics. 8. Genomics is the study of genomes and how genes work together to control phenotype. Multiple genes are considered at one time and the entire organism is studied. This compares to Mendel looking at one trait/ gene at a time. This will be helpful in understanding the inheritance of diseases like cancer and to help develop treatments for these diseases. 9. The International HapMap Project is an international sequencing project that is attempting to map variations in the human genome. The main goal is to provide researchers with enough information to allow them to connect specific variations to specific genetic diseases. 10. Epigenetics suggests that gene function may change without the actual DNA sequence of the gene changing. In other words, it looks at the mechanisms that control whether our genes are on or off. 11. By looking at thousands of genes at one time, scientists can see which genes are active under certain conditions. This helps them to understand how gene expression is coordinated. 12. Sample answer: Benefits
Risks
Medical treatments can be designed to match a specific individual’s genome
Can be used to discriminate against an individual in the case of insurance
13. Sample answer: No, if the profile is not kept private, the person could be discriminated against by insurance companies and employers. Just having a particular gene does not guarantee that you will get a particular disorder or disease and the knowledge could cause undue stress or discrimination.
14. a. Some issues that might be considered are: Who owns the genetic information? Should companies have the right to sell DNA information to other companies without the permission of the people who provided the samples? Should companies that use DNA in medical research be required to share the results of their work with the individuals or communities whose genetic information was used? Where is the boundary between public and private genetic property rights? Should the legal system have the right to the information? b. Sample answer: Providing genetic information should be voluntary. Genetic information can provide medical benefits to many. Research should be funded so as to maximize benefit, for as many as possible. It is reasonable for companies to expect a return on their investment in genetic research. The outcome of the research should be widely available. c. Sample answer: To keep research going, information could be gathered but kept private through coding that only the individual will recognize. This will prevent discrimination while still providing information to researchers and the individual. 15. Sample answer: I think that people should be responsible for themselves. If they want to work in an environment that may potentially prove harmful to them due to their genetic profile, that should be the individual’s decision to make. To expect that an employer use information in the profile to create an idea work environment yet not discriminate against an employee for a potential genetic issue is unfair to the employer and has too great a potential to be misused.
Practice Problems (Student textbook page 247) 1. A, AB, or B (if the woman is heterozygous) 2. AB or A 3. The baby cannot be theirs because each of her type O blood parents gave her an i allele. This woman’s baby would have received either an I A or I B allele from her. 4. Yes, parents with the genotypes I Ai and I Bi could produce offspring with any of these blood types. 5. c chc and c hc 6. 1:1 chinchilla:Himalayan 7. A mating between a chinchilla rabbit with a c chc h genotype, and an albino rabbit (cc) has a 50 percent chance of producing a Himalayan rabbit (c hc).
Biology 11 Answer Key Unit 2 • MHR TR
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8. a. Sample answer: Mother IB
IB
father i
I Bi
I Bi
i
I Bi
I Bi
Mother IB
IB
father I B
I BI B
I BI B
IB
I BI B
I BI B
b. Students should choose whichever cross produces a higher percentage of offspring with the genotype I B_, those with blood type B. In this sample, that would be either of the sets of parents. An alternative cross could have been two heterozygous parents producing 50% of offspring with blood type B. Students would then choose one of the samples shown, both of which have 100% of offspring with blood type B. 9. a. Parent I-1 is A Sa t and parent I-2 is a ya t. b. 50% c. A Sa y or A Sa t
17. The inheritance is most likely X-linked dominant since the daughters of affected males are affected, as are some daughters and sons of affected females. Affected males did not pass the trait to their sons. 18. a. All of the female offspring will be normal, the male offspring will all be hearing impaired. Since this is carried on the Y chromosome, the information about the female dog is irrelevant. b. No offspring will be hearing impaired. 19. Cross the red-eyed female with the male to obtain carrier females. Cross the F1 female (carrier) with the original white-eyed male to get half the female offspring with white eyes. 20. His genotype is LlX cY. Her genotype could be any that has X C and l. (For example, X CX CLl.) Because none of their children have CVD, she is not likely a carrier of CVD but best guess is that she is homozygous dominant since there are no offspring with long fingers.
10. Sample answer: dark coloured dog AS
at
sandy coloured a y
A Sa y
a ta y
t
S t
t t
dog a
Aa
aa
(Student textbook page 258) 11. a. The mother is X CX c and the father is X cY. b. Out of four possible, one would be a carrier female (X CX c), one a colourblind female (X cX c), one a normal male (X CY), and one a colourblind male (X cY). Mother 12. Xc
Xc c c
c
c c
XX
XX
Y
X cY
X cY
Father X
13. 100% 14. a. X NY (father) and X NX n (mother) b. The boy is X nY because he is affected by the disease. The girl must be X NX n or X NX N because she could get either the dominant or the recessive gene from her mother and had to get the dominant gene from her father.
Quirks and Quarks Feature Questions (Student textbook page 249) 1. autosomal recessive; for an individual to be deaf due to the mutation they must have two copies of the mutated gene 2. heterozygous advantage refers to an advantage that a heterozygous individual has compared to individuals who are homozygous dominant or homozygous recessive for the trait. People with one copy of the mutated gene are not deaf and also gain an advantage—a thicker layer of skin that could offer better protection and defence against infection. 3. Sample answer: Knowledge in the following fields would be an advantage: molecular biology (study of biology at the level of the molecules in the cell); genetics; human diseases that are caused by genetic mutations; and research skills for how to study cells and the genetic material in cells.
Chapter 6 Review Questions (Student textbook pages 273–5) 1. b 2. d
15. 50%
3. c
16. a. The pattern of inheritance is likely X-linked recessive, since the yellow disappeared in the F1, and reappeared only in F2 generation males. b. The females are X TX T and X TX t, and the males are X TY and X tY. c. 50%
4. d
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MHR TR • Biology 11 Answer Key Unit 2
5. a 6. b 7. c
8. c 9. a. incomplete dominance b. codominance c. red as dominant 10. 1:2:1
19. No, you could not establish true-breeding platinum foxes, as the genotype for the form is heterozygous. A cross between C PC p and C PC p (platinum-coloured fur) parents will produce 25% C PC P (lethal), 50% C PC p (platinum-coloured fur), and 25% C pC p (silvercoloured fur).
11. Heterozygous advantage describes a situation in which an individual who is heterozygous for a trait has a reproductive advantage over individuals who are homozygous for the trait. An example is sickle cell anemia in areas of the world that have malaria. Homozygous dominants are susceptible to malaria, and homozygous recessives have sickle cell disease, which reduces their oxygen–carrying capacity. Being heterozygous with the sickle cell trait increases resistance to malaria but does not greatly compromise oxygen capacity as only some of their cells are sickleshaped.
22. The rabbit could have a gene that is activated and turns its hair black in response to cold.
12. Sample answer: Sometimes there are variations in the expression of a gene. This can be shown through blood type inheritance where there is an allele that makes the A protein on blood cells, a different allele that makes a B protein and a third allele that makes no protein. Due to the multiple alleles, there are more than two possible phenotypes for the trait.
25. You may conclude that the gene for wing shape is Y-linked. Since the X chromosome is inherited from the female parent, some of the males would have normal wings if the gene for wing shape were X-linked.
13. The continuous variation in height suggests that it is a polygenic trait. 14. Linked genes are genes located close together on a chromosome. They are not usually separated during crossing over and therefore are always sorted together (not independently) during meiosis. 15. Duchenne muscular dystrophy is linked to the X chromosome, which a boy receives only from his mother. Any normal mother of affected offspring must be a carrier. 16. A person’s genetic profile is that person’s complete genotype (including mutations). It can help doctors make generalized risk assessments about disease but could also be used as a tool for discrimination if it is accessed by insurance companies or employers (for example). 17. This is likely incomplete dominance since one of their children has an intermediate phenotype. 18. a. Individuals I-4 and I-6 both have blood type A. b. No. Since female III-2 has blood type O, her genotype is ii. The father with AB blood has the genotype I AI B. Their children can only have either type A blood (genotype I Ai) or type B (genotype I Bi).
20. Blood types among the children could be A, B, and AB. If any child is blood type A then the man must be heterozygous type B. 21. I AI B, I Ai, I AI A, I BI B, or I Bi
23. The distance between the genes may mean that crossovers occur often, resulting in frequent recombination between these genes. 24. None, because female offspring have XX sex chromosomes and therefore cannot pass a Y chromosome to their sons.
26. a. Mendel would have seen combinations of traits not present in the parents and different phenotypic ratios such as 0.42:0.41:0.09:0.8. b. Sample answer: Mendel might have hypothesized that if the traits were carried on the same chromosome, then a cross between two truebreeding plants for dominant and recessive traits would produce offspring that only display the dominant phenotype. c. Mendel could have crossed the two parental plants, e.g., maternal pure breeding dominant for both traits, and paternal pure breeding recessive for both traits. He may have observed recombinants in the F1 phenotypes, indicating that there was crossing over of the maternal and paternal genes. 27. a.
I II
1
2
1
3
4
2
b. There is no chance of having a girl with CVD. A daughter would either be a carrier, or not have the gene at all. c. There is a 50% probability of their son having normal vision.
Biology 11 Answer Key Unit 2 • MHR TR
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28. Students should draw or construct a homologous pair of chromosomes, containing a number of genes. They should illustrate crossing over, and show that genes that are close together on a chromosome tend to be transferred together during crossing over.
Description
Benefit
Concern
Proteomics and
Topic
Studying threedimensional shapes of proteins and determining their functions
Identifying proteins can lead to the identification of genes responsible for certain diseases
Very complex to get the DNA sequence, as many variations in the order of nucleotides can result in the production of the same protein
Epigenetics
Studies how changes in the inheritance of certain traits or phenotypes are based on gene functions rather than gene sequences
Identifying the causal factors (triggers) of genetic diseases that are latent
29. See Figure 6.11 (student textbook page 252). The left side of that figure shows the contradiction. The right side shows independent assortment. 30. Answer could show consequences of discrimination or positive results of research curing a disease. 31. Sample answer: Nucleotides are like notes, DNA is like a chord, genes are like phrases, and chromosomes are like the song. 32. A mind map or wheel organizer is well-suited to this task. Uses of bioinformatics should be at the centre. The “spokes” should detail: the analyzing of genomic data, identifying genes associated with human health, preventative treatments, and cures for diseases. 33. Sample answer: Read the first column from top to bottom to see connections between the topics, then grow out along each row to build concept map. Topic
Description
Benefit
Concern
Determined the sequence of DNA in the entire human genome
Showed how many genes we have, how similar we are to other organisms and each other
What the information could lead to
Bioinformatics Applies which leads to computer technology to create and maintain databases of information (chemist Margaret Dayhoff )
Allows everyone easy access to information
Genetic profiles could be improperly used
Genomics and
Identifying complex diseases (such as cancer)
Human Genome Project is a result of
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The study of genomes and the complex interactions of genes that result in phenotypes
MHR TR • Biology 11 Answer Key Unit 2
Many diseases involve a complex array of factors. Results could be misleading
34. Organizers should include all of the Key Terms and Key Concepts listed in the chapter summary. 35. a. The farmer could selectively breed tall corn plants and harvest the seeds from the offspring and then repeat the process again and again. b. The farmer’s work will be most effective if the height of corn plants were determined by multiple alleles or codominance, because then only the tall plants could be selected to breed instead of working with varying heights produced by polygenic inheritance. c. The farmer could cross various parent plants to see if the offspring of particular crosses show evidence of the disease resistance. d. The farmer would avoid selecting any plant for breeding that showed the disease and then select only the tall plants for breeding from the diseasefree offspring. 36. a. to c. Answers should show evidence of independent research. 37. a. to c. Sample answer: The goal of Enhancing Canola Through Genomics was to improve canola seed and apply that knowledge to other related crops. It is important to make sure that the research done does not have huge consequences.
18. The gene for it occurs on the X chromosome. Males only have one X and therefore if they inherit the recessive allele it will always be expressed. For a pedigree, see the diagram in question 17 on student textbook page 277.
Chapter 6 Self-Assessment Questions (Student textbook pages 276–7) 1. d 2. b 3. e
19. The gene for it occurs on the X chromosome. Males only have one X and therefore if they inherit the recessive allele it will always be expressed. A carrier must be heterozygous.
4. b 5. d 6. b
20. Barr bodies are inactive X chromosomes. Fur colour is X-linked. Because only one X is active at a time, a heterozygous cat will have some patches of fur with one X active (and a fur colour associated with that X) and other patches with the other X active, causing a different colour.
7. c 8. b 9. c 10. e 11. 1:2:1 of red:purple:white 12. a. All offspring will be green b. Offspring will be green and yellow; Could be in patches, could be that some seeds are green and others are yellow c. All offspring will be intermediate in colour between green and yellow––yellowish green or greenish yellow. 13.
B
Alleles
O
A
A
A
A
O
B
B
B
O
Type A
O
O
21. So that we can compare the human genome to them and see which sequences are shared, and therefore which proteins are shared 22. Sample answer: The individual nucleotides are like the letters in the language. Groups of nucleotides make words and groups of words make sentences that have a purpose (this is what a gene is). A chromosome is composed of many genes, just as a paragraph is composed of many sentences. If you don’t understand the structure of the language, you can not decode it. AC TG AATCG G T TAAC TATCG T TACG
Type O
word/base pair sentence/gene
Type B
paragraph/chromosome
14. The production of chlorophyll is controlled by the light turning the genes on or off. Chlorophyll is essential for the process of photosynthesis. By researching this, scientists can better understand how to control the production of chlorophyll and therefore increase productivity.
23. Genomics research focuses on the interaction of genes in an organism. By identifying disease- or disordercausing genes, we can look at methods that could cure the disorders or screen for them.
15. See Figure 6.11 on student textbook page 252. A and b are linked. They can become unlinked if crossing over occurs.
25. Variation in non-coding regions can be used in DNA fingerprints to help identify forensic evidence.
16. 50% red-eyed fruit flies X RX r and 50% red-eyed male fruit flies X RY.
Unit 2 Review Questions (Student textbook pages 281–5)
17. a. It is sex-linked because only males are affected. It is X-linked recessive because two unaffected parents have an affected child. b. X HX h c. 25%
24. Bioinformatics is the production of a research database. The Human Genome Project uses bioinformatics.
1. d 2. c 3. c 4. c 5. a 6. a
Biology 11 Answer Key Unit 2 • MHR TR
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7. b 8. e 9. c 10. e 11. G1: rapid growth and cell activity; S: DNA synthesis and replication; G2: cell prepares for division 12. A karyotype is a picture of the set of chromosomes that an individual has. It is used to determine sex, diagnose monosomy and trisomy disorders or chromosome abnormalities. 13. same size, position of centromere, and banding patterns; they are not identical because each chromosome could carry different alleles for the same genes 14. Haploid cells contain half the set of chromosomes. They are also called gametes. In humans they are found in the testes and ovaries. Diploid cells contain a full set of chromosomes, and are called somatic cells. They are found in all tissues except the germ tissues. 15. One essential outcome of meiosis is variation in the cells produced. This occurs during prophase when crossing over happens and in anaphase when chromosomes sort to either pole. The other essential outcome is that the cells produced are haploid. This occurs during anaphase I when half of the chromosomes go to one pole and the other half go to the opposite pole. 16. a. 2 pairs b. 2 chromosomes c. 4 17. Pea plants are self-fertilizing (they have male and female organs), are true-breeding, grow quickly, and have only two alleles for each trait studied. Each allele shows simple dominant/recessiveness. 18. Dominant and recessive refer to alleles. Dominant alleles make proteins and are always expressed in phenotype. Recessive alleles do not make proteins and are only expressed when the individual has the homozygous recessive genotype. 19. Monohybrid crosses examine the inheritance of one trait. Dihybrid crosses involve two different traits. Punnett squares show the fertilization possibilities for each cross. The male gametes are placed along the top and the female along the side. The intersections have genotypes entered which represent zygotes.
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MHR TR • Biology 11 Answer Key Unit 2
20. Autosomal recessive inheritance refers to the inheritance of a recessive trait that is found on one of the autosomes (non-sex chromosomes). Cystic fibrosis is an autosomal recessive disorder. A person will only have CF if they are homozygous recessive for the trait. 21. Genes are located on chromosomes, which provide the basis for the segregation and independent assortment of alleles. Walter Sutton studied the process of synapsis and migration of chromosomes during meiosis, and related his observations to the behavior of the factors described by Mendel. 22. Students should be able to describe any three of the genetic tests summarized in Table 5.4 on student textbook page 224. 23. Because there are three phenotypes (one for each genotype) and the intermediate phenotype (heterozygous) has some cells that are normal and some that are sickle shaped 24. When a single gene has multiple alleles, there are more than two alleles for the trait. Sometimes these alleles have a dominance continuum and sometimes there is codominance between some of the alleles. Blood type is an example. When both dominant alleles are present in the genotype, the blood is type AB, when just one dominant allele is present, the blood type is either A or B, and when the genotype is homozygous recessive, the blood type is O. 25. Because males have one X chromosome, if they have the recessive allele it will always be expressed. 26. By collating the data in a central bank, information can be used by several scientists at one time. 27. Mitosis and meiosis both involve cytokinesis and interphase (growth and DNA replication) and have phases called prophase, metaphase, anaphase, and telophase. Mitosis occurs in somatic cells; meiosis occurs in germ cells. Mitosis makes two identical cells that are diploid. Meiosis makes four haploid cells that are different. 28. A gene is a segment on a chromosome that codes for a trait. An allele is the form of the gene (each gene will have different ways of being expressed). 29. a. Mutations (see Table 4.1 on student textbook page 177) and non-disjunction (see Figure 4.20 on student textbook page 178). b. Karyotyping and biochemical analysis 30. They don’t; the only difference is that fertilization occurs internally in artificial insemination and externally in embryo transfer.
31. See the answer to question 9 in the Section 4.2 Review. 32. a. BB b. bb c. Bb d. B; b; B, b 33. a. Round is dominant since all F1 offspring are round. b. The data is very close to the Mendelian ratio. I would expect 75 percent of the offspring to be round and 75.8 percent were round. The percent may not be the same due to the small number of offspring observed. 34. Yes, if both parents are heterozygous for hairline 35. Gamete from Gamete from Genotype Male Parent Female Parent of Offspring TY
tY
TtYY
Phenotype of Offspring tall plant, yellow seed colour
Gt
gt
Ggtt
green pod, short plant
Yg
yg
Yygg
yellow seed and pod colour
36. a. Both parents are PpTt. b. PT, Pt, pT, pt c. 17 37. a. Dominant. If shaded was recessive, all offspring would have to be recessive. b. If A = affected, I-1, I-2 are Aa, II-1 is AA or Aa (since both parents are heterozygous and the disorder-causing allele is dominant), and II-2 and II-3 are aa. 38. a. C RC W (pink); each parent gives one allele. The red parent can only give red alleles, the white only white alleles. This makes the offspring heterozygous and flower colour is incompletely dominant. b. pink C RC W, white C WC W, 1:1 39. Baby 1 belongs to Mr. and Mrs. Jones. Baby 2 belongs to the Guttierez family. Mr. and Mrs. Guttierez cannot have a type O child as neither carries the recessive allele. 40. If the father is not hemophiliac there is no chance of a daughter having the disease. There is a 25 percent chance that the son is a carrier since the mother is a carrier.
41. Epigenetics studies the changes of inheritance of certain traits based upon gene function, not DNA sequence. Genetics focuses on DNA sequence. The interaction between the environment and the genotype can turn on or off certain phenotypes (fur colour in Siamese cats, for example). 42. See Figures 4.7 (student textbook page 165) and 4.9 (student textbook page 167). 43. See Figure 4.13 on student textbook page 171. 44. Diagrams should show detail from the right side of Figure 4.14 on student textbook page 172, with the addition of details on anaphase I/telophase I, and meiosis II. The first should indicate that chromosomes look different as a result of crossing over. For meiosis II, a note should indicate that chromosomes go to daughter cells randomly as a result of independent assortment. 45. Sample answer: Cons—To increase fungal resistance, genes from other plants or bacteria can be added to a crop to allow it to encode enzymes which break down chitin or glucan, respectively, which are essential components of fungal cell walls. Unless the crop is clearly labelled as being genetically modified, there could be a danger of allergic reaction in some people when they ingest food from the crop. (For example, potatoes containing peanut genes may cause a reaction in a person allergic to nuts.) If the crop is near another crop, cross pollination of the genes may also occur. Pros—Fungi are responsible for a range of serious plant diseases such as blight, grey mould, bunts, powdery mildew, and downy mildew. Crops of all kinds often suffer heavy losses. By genetically modifying these crops, they will be more productive and reduce or eliminate the need to use chemical fungicides or heavy metals which are very damaging to the environment. 46. See Figure 4.29 on student textbook page 188. 47. Let T be tall, t be short. Each cross in the Punnett square represents a 25 percent probability. Parent genotypes: Tt Punnett square: F1 T
t
T
TT
Tt
t
Tt
tt
Genotypic ratio of TT:Tt:tt is 1:2:1. Phenotypic ratio of tall:short is 3:1.
Biology 11 Answer Key Unit 2 • MHR TR
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48. Perform a test cross by crossing the unknown parent with a homozygous recessive parent (tt). One short offspring tells us that the parent is heterozygous: T
?
t
TT
Tt
t
Tt
tt
No short offspring tells us the parent might not be heterozygous: T
T
t
TT
Tt
t
Tt
Tt
49. a. Answers should include: autosomal dominant; brain deterioration over 15 years; decreased muscle coordination; there is no cure; medications can lessen symptoms. Diagnosis is done by looking for CAG repeats in the gene called huntingtin. b. Supporting answers might include: Accuracy—35 or fewer repeats will not be affected but 36–39 are at risk and over 39 will definitely get the disease. Uncertainty— the number of repeats may increase or decrease between generations Odds—due to the autosomal dominant nature, if one parent has Huntington’s, there is at least a 50 percent chance that children will inherit the gene. Knowing how severe the impact on the huntingtin gene is will help plan the future. 50. See Figure 6.11 (student textbook page 252). 51. Dominance—Mendel’s peas exhibited dominance/ recessiveness. Tall is dominant over short, therefore heterozygous plants were tall. Incomplete dominance—Snapdragon flower colour exhibits incomplete dominance: white and red are both expressed and the heterozygote shows a blend of the two alleles (it is pink). See Figure 6.1 (student textbook page 242). Codominance—both alleles are expressed. Roan cattle show codominance in their hairs: some patches of fur are white and other patches are red. The phenotype is not a blend but is an intermediate. See Figure 6.2 (student textbook page 243). Sex-linked inheritance—the allele is carried on the portion of the X chromosome that does not have a homolog on the Y chromosome, this results in more males having characteristics such as colour deficient vision. See Figure 6.13 (student textbook page 254).
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MHR TR • Biology 11 Answer Key Unit 2
52. See Figure 6.14 (student textbook page 255) as a reference. In this case, every letter B or b should be an H or h. The key would be X HX H = normal female; X HX h = carrier female; X hX h = hemophiliac female; X HY = normal male; and X hY = hemophiliac male. This pedigree has more affected males than females. The affected characteristic can skip a generation (as shown in generation II) but the next generation shows only affected males. 53. Look for evidence that students have used knowledge from this unit. 54. To try to make it so that the person can make their own insulin. By gene therapy, it is hoped to correct the defective gene. 55. Reduced genetic variability. Studies indicate that animals born and raised in their native environment do better than those that are imported. 56. a. Embryos used to be the only source of stem cells. Now scientists can make specialized adult stem cells into undifferentiated stem cells that can differentiate into any of the three germ layers. b. Sample answer: Dr. Lynn Megeney is focusing on adult cardiac muscle stem cells. He is developing small molecule compounds to stimulate the regeneration of cardiac muscle tissue. This will lead to being able to repair hearts disease damaged hearts. 57. Sample answer: Ethically, we cannot force people to breed just to study the genotypic ratios of their offspring. Practically, we cannot control the environmental factors that influence gene expression. Scientifically, many genes are linked, some phenotypes are polygenic and some phenotypes are not activated unless certain environmental conditions are present. 58. a. As certain traits were bred into the breed, natural variation was decreased. Perhaps by increasing the jaw strength of a dog, you also decrease the hip stability because the genes are linked. Some breeders breed to close relatives, which increases the chance for disorders to be exhibited. b. Breed with dogs from different areas of the country or world. Make sure that you are not breeding close relatives.
59. a. CF is an autosomal recessive disorder. A defective protein in the cell membrane leads to increased mucus production. b. By using a pedigree, a genetic counsellor can predict whether either parent is a carrier and therefore calculate the chance of having a child with CF. c. Gene therapy could replace the gene (CFTR) on chromosome 7. The challenge is that any one of 1600 mutations to the gene causes CF. 60. Answers should include use of a vector-like virus to remove and replace the gene. It will be more accurate if it is done in early development and only acts on a simple mutation (unlike CF). A possible plot twists is that the newly inserted gene could be from another animal and carry with it other characteristics. 61. a. Sample answer: Thermal imaging or blood tests to monitor protein production. b. It reproduces asexually through budding and therefore all offspring are identical (unless mutation occurs). It can be transformed easily. It also reproduces very quickly under optimum conditions and is eukaryotic (therefore has large, non-coding regions of DNA like human DNA does). 62. a. As data is collected from cancer patients, it can be analyzed for similarities. DNA sequences can be identified as markers to help predict cancer. b. Students may name the Ontario Institute of Cancer Research. c. Refer students to BLM A-10 Presentation Checklist or BLM A-32 Presentation Rubric (on the accompanying CD) for assistance in preparing presentations or assessing them. 63. Knowing your genetic profile may enable you to seek treatment, change your lifestyle to reduce risk, or make an informed decision on the risk of having children. However, this knowledge could reduced your freedom, create discrimination (in hiring, getting insurance, or finding a life partner), and increase stress. These implications show that society needs to consider how the information can and should be used. 64. Sample answer: The ability to use adult stem cells to create stem cells that can differentiate into any germ layer has eliminated the need to use embryonic stem cells (an ethical problem) for stem cell research that could repair damaged heart and nerve tissues.
Unit 2 Self-Assessment Questions (Student textbook pages 286–7) 1. d 2. d 3. e 4. d 5. b 6. c 7. b 8. c 9. b 10. c 11. Haploid cells that are genetically unique; reproductive cells (sperm or egg) 12. The cells that result from meiosis merge during fertilization to produce the first cell (zygote) of an organism. This cell will carry all errors that occurred during meiosis and is the foundation for the entire organism. Mitosis begins and continues for the life of the organism. Only the cells that divide from a cell that has a mistake in mitosis will carry that error. 13. See the right-hand side of Figure 4.14 (student textbook page 172). 14. See Table 4.2 (student textbook page 179). 15. Sample answer: Artificial insemination is used to increase the availability of high-quality male semen to breeders all over the world. This can be combined with embryo transfer and genetic screening to ensure that genetic disorders are not enhanced. 16. Rr and rr; R is round; The ratio of the offspring is close to 1:1. 17. Ss and ss. The long-haired cat must be heterozygous to have homozygous recessive offspring. 18. 25% 19. The pedigree indicates whether there is any chance of a person or their child having a disorder. The genetic testing can verify the findings of the counsellor. The tests are expensive and should only be done to verify the genotype of an individual.
Biology 11 Answer Key Unit 2 • MHR TR
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20. Sample answer: Pro • improved quality of life • lower stress on health–care system because people are healthier • longer life • cure for disease • ability to repair tissues, including missing limbs • potential for designer organisms
Con • transgenics—should a gene from one organism be introduced into a another? • increased stress on healthcare system because people live longer • virus may not find the correct point in the DNA, leading to more damage • potential for designer babies • “playing god” • strong immune response to the virus is possible
21. In the case of incomplete dominance, the offspring will be grey (a blend of the two parental phenotypes). In the case of codominance, the offspring will have black and white patches. 22. The Rousseaus are parents of Baby 1; Baby 2 belongs to the Sakic family. 23. a. Let X m = Duchenne muscular dystrophy allele and let X M = normal allele; female carrier X M X m, normal male X MY, affected male X mY b. 25% c. No. The father must have the recessive allele on his X chromosome for him to father a daughter with the disorder (she must be homozygous for the allele). The father is normal and therefore has the dominant allele. 24. Sample answer: By documenting the variations in the human genome, scientists will have more data to analyze and may find that some diseases are caused by multiple genes or only certain environments. 25. The Human Genome Project determined the sequence of the human genome (as well as other organisms). This has lead to advancements in gene therapy (for example, being able to identify a sequence that is faulty and replace it), genetic testing (looking for faulty sequences), and increased understanding of protein action and treatments.
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MHR TR • Biology 11 Answer Key Unit 2