Download SBI 3U Genetics Test Review Sheet

Genetic Processes – Textbook Review Solutions Ch. 4 -­‐ pg. 176-­‐179 #3, 6-­‐8, 19, 20, 22, 25, 29, 33, 46, 49, 50 and 64 3. (c) 6. (c) 7. (d) 8. (c) 19. (a) (ii) (b) (i) (c) (v) (d) (iii) (e) (iv) 20. (a) (iv) (b) (iii) (c) (i) (d) (ii) 22. The offspring will each have six chromosomes in each somatic cell. 25. If I was viewing cells with a microscope, I would identify cells that are in metaphase I by looking for cells in which duplicated chromosomes are arranged as tetrads and aligned by their centomeres across the middle of the cell. 29. A pair of chromosomes that are similar in size and shape and carry genetic information for the same genes are called homologous chromosomes. 33. Answers may vary. Either monosomy and trisomy (which are types of disorders), and also any two specific non-­‐disjunction disorders, such as Down syndrome, Turner syndrome, Klinefelter syndrome, etc… 46. During meiosis, crossing over increases genetic diversity when the sister chromatids exchange segments of DNA so that they are no longer genetically identical. Random assortment also increases genetic diversity, because it results the set of homologous chromosomes in daughter cells being different than the set in the parent cells. 49. The chance of a human offspring inheriting the genetic characteristic of Down syndrome increases with the age of the mother and with exposure to X-­‐rays. 50. Answers do not need to be presented as a table. 64. (a) The individual who provided the cell sample for this karyotype was female. (b) Human sex is determined by inheritance of X and Y chromosomes. This individual has two X chromosomes and lacks a Y chromosome. (c) A karyotype can be used to test for non-­‐disjunction disorders, because it groups each copy of every chromosome in an individual. A non-­‐disjunction disorder is caused by loss of one copy of a chromosome in a pair (monosomy) or more than two copies of a chromosome, such as three copies (trisomy). These changes in the number of chromosomes in a pair would appear on the karyotype. Genetic Processes – Textbook Review Solutions Ch. 5 – pg. 220-­‐222 #1-­‐6, 8, 21-­‐23, 25, 26, 28, 31 and 33 1. (a) 2. (c) 3. (b) 4. (c) 5. (a) 6. (a) 8. (c) 21. An allele is a form of a gene. 22. Homologous chromosomes separate during anaphase of meiosis I. 23. The possible genotypes of the offspring are HH, Hh, and hh 25. Your diagram should be similar to the following: 26. The thick mucus produced by individuals with cystic fibrosis clogs the lungs, leads to infections, and blocks the release of enzymes from the pancreas. 28. In a dihybrid cross of two heterozygous individuals, there are nine genotypes possible in the offspring. 31. Answers may vary and may include any two of the following: pea plants reproduce quickly, it is easy to control which parents produce offspring, garden peas have a number of readily visible characteristics that have two distinct forms, and that true-­‐breeding varieties that had one or the other of these traits were available to Mendel. 33. No, the individual is just as likely to pass on the recessive allele as she is to pass on the dominant allele. Whether an allele is dominant or recessive affects only its expression, not its probability of inheritance. Genetic Processes – Textbook Review Solutions Ch. 6 – pg. 258-­‐261 #1, 18, 44 1. (b) 18. (a) (vi) (b) (v) (c) (iv) (d) (ii) (e) (i) (f) (iii) 44. The damaged section must have had a base sequence of TACGCGATA. We know this because the structure of DNA involves two complementary strands. In complementary DNA strands, adenine is always paired with thymine, and cytosine is always paired with guanine. Genetic Processes – Textbook Review Solutions Unit 2 -­‐ pg. 268-­‐275 #1, 3-­‐14, 16-­‐18, 21, 32, 34, 37, 39, 51, 60, 69 1. (a) 3. (c) 4. (c) 5. (b) 6. (d) 7. (a) 8. (c) 9. (a) 10. (a) (iv) (b) (ii) (c) (iii) (d) (i) 11. (a) (iii) (b) (i) (c) (ii) 12. Genetic information is stored in DNA molecules. 13. Crossing over occurs in prophase I of meiosis. 14. Trisomy is the meiosis error of non-­‐disjunction that leads to three copies of homologous chromosomes. In Down syndrome, there are three copies of chromosome 21, so the condition is also called trisomy 21. 16. A dog that has codominant black and white alleles for coat colour will likely have a grey coat. 17. In X-­‐linked inheritance, the male is more likely to possess the recessive phenotype and the female is more likely to be a carrier of the recessive allele. 18. In autosomal inheritance, the gene for a characteristic is located on an autosome (a non-­‐sex chromosome), and its inheritance is unaffected by the sex of the organism, so males and females have an equal probability of inheriting any of the gene’s alleles. 21. see diagram of DNA given in class. 32. The genetic material of offspring produced by sexual reproduction differs from either of their parents. The genetic material of offspring produced by asexual reproduction is identical to that of their parent. 34. Crossing over is the exchange of genetic information between homologous chromosomes, which adds to the genetic diversity that arises from the process of independent assortment of homologous chromosomes. 37. Answers will vary. Dominant blood types may be either type A or type B. Only type AB is codominant. 39. The two laws of heredity that Mendel discovered as a result of his experiments on peas plants include: (1) the law of segregation, which states that for each characteristic, an organism carries two factors, one from each parent and that organisms donate only one copy of each gene in their gametes and; (2) the law of independent assortment (alleles of different genes separate into gametes independently of each other). 51. (a) Since some of the offspring have yellow seeds, the genotypes of the green-­‐seeded both P generation plants must be Gg. These plants are heterozygous for seed colour. (b) The genotypes of the F1 generation are GG, Gg, or gg. Half of the F1 generation plants are heterozygous and half are homozygous. (c) The phenotypes of the F1 generation are green-­‐seeded or yellow-­‐seeded. (d) Yellow-­‐seeded plants can only have the genotype gg. (e) All member of the F2 generation would have genotype gg and phenotype yellow-­‐seeded. (f) This problem is an example of a monohybrid cross, because only one characteristic is being tracked. (g) No. A test crosses will only provide information about the genotype of organisms exhibiting dominant traits, and yellow seeds (part (d)) are a recessive trait in this plant. 60. Since the woman’s father is colour-­‐blind, she e would have had to inherit the colour-­‐blindness allele on the X chromosomes that was inherited from her father. Since she has normal vision, the X chromosome she inherited from her mother did not carry the colour-­‐blindness allele, and she is therefore heterozygous. Her husband also has normal vision, so his X chromosome does not carry the colour-­‐blindness allele. The Punnett square analysis of their mating is: Genotypic ratio: ¼ XCXC : ¼ XCXc : ¼ XCY : ¼ XcY Phenotypic ratio: 50% female with normal vision: 25% male with normal vision: 25% male with colour-­‐
blindness. 69. (a) The major disadvantage of test crossing is the time it takes for the animals to produce offspring. (b) The typical reproductive characteristics of organisms used in test crosses are that they reproduce quickly and produce many offspring. (c) No. Horses take too long to produce offspring and they usually only produce only one at a time. (d) Advances in molecular biotechnology has reduced the need for test crosses because geneticists can now test for specific alleles within the genotype of an organism directly and get answers very quickly.