MENDEL S PRINCIPLES CHAPTER SUMMARY QUESTIONS. Hyde Chapter 2 Solutions 3

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1 Hyde Chapter 2 Solutions 3 2 MENDEL S PRINCIPLES CHAPTER SUMMARY QUESTIONS 2. Perform a testcross by crossing the tall pea plant of unknown genotype with a homozygous recessive dwarf plant (dd). If the offspring consist only of tall pea plants, then the plant in question would be homozygous (cross: DD dd Dd). If the offspring consists of both tall and dwarf pea plants, then the plant in question would be heterozygous (cross: Dd dd 1/2 Dd:1/2 dd). 4. Cross AABB AABBCC AABBCCDD (P generation) aabb aabbcc aabbccdd Number of different F 1 gametes Number of different F 2 genotypes Number of different F 2 phenotypes Degrees of freedom in chi-square test 6. a. The offspring are in an approximate 3:1 phenotypic ratio of red eyes to brown eyes. The 3:1 ratio suggests that this is a monohybrid cross and that red eye is dominant and brown eye is recessive. The cross would be Bb Bb. b. There are two phenotypes in approximately a 1:1 ratio in the offspring. We don t know which allele is dominant from this cross. The two phenotypes indicate one gene, and the 1:1 ratio indicates a mating of a heterozygote and homozygous recessive individuals (the testcrossing of a heterozygote). Therefore, the cross can be depicted as Aa aa.

2 4 Hyde Chapter 2 Solutions 8. In the table, D = dominant; R = recessive; M = mutant; + = wild type. Dominant Mutant Alternative Dominant Allele or Recessive or Wild Type Allele or Recessive y + D + y R Hw D M Hw + R Ax + R + Ax D Co D M Co + R rv + D + rv R dow R M dow + D M(2)e + R + M(2)e D J D M J + R tuf + D + tuf R bur R M bur + D b, 2. a, 3. b, 4. c (one can pick the ace of diamonds). EXERCISES AND PROBLEMS 12. a. AA Aa, aa Aa, AA aa, and Aa Aa. b. AA AA and aa aa. 14. The 3:1 phenotypic ratio of the F2 generation is composed of a 1:2:1 genotypic ratio that corresponds to 1/4 DD, 1/2 Dd, and 1/4 dd. The 1/4 dwarf F2 (dd), when selfed, produce all dwarf progeny (dd). The tall F2 (3/4 of total F2), when selfed, fall into two categories: 1/4 (DD, 1/3 of the tall F2) produces all tall, and 1/2 (Dd, 2/3 of the tall F2) produces tall and dwarf progeny in a 3:1 ratio. We can compute the relative proportions of genotypes and phenotypes in the F3 generation using the following table. Fraction of Genotype in F 2 F 2 Genotype Generation F 3 Genotype DD Dd dd Dd (dwarf) 1/4 1 1/4 = 1/4 DD (tall) 1/4 1 1/4 = 1/4 Dd (tall) 1/2 1/4 1/2 = 1/2 1/2 = 1/4 1/2 = 1/8 1/4 1/8 F 3 genotypic 3/8 DD 1/4 Dd 3/8 dd ratio F 3 phenotypic 5/8 tall 3/8 dwarf ratio

3 Hyde Chapter 2 Solutions 5 Overall, the F 3 are 3/8 DD (tall), 2/8 Dd (tall), and 3/8 dd dwarf. 16. When a decahybrid is selfed, it would produce 2 10 = 1024 different gametes; 1/(2 10 ) 2 or approximately of the F 2 would be homozygous recessive; 3 10 = 59,049 different genotypes yielding 2 10 = 1024 different phenotypes would appear. If the decahybrid were testcrossed, it would produce 2 10 different gametes; 1/2 10 = of the F 2 would be homozygous recessive; 2 10 different genotypes and phenotypes would appear. 18. The round, yellow F 2 plants are made up of four genotypes; the round, green of two genotypes; the wrinkled, yellow of two genotypes; and the wrinkled, green of one genotype. Testcrossing all these genotypes produces the following results: W G : 1/16 WWGG wwgg all WwGg (round, yellow) 2/16 WwGG wwgg 1/2 WwGg (round, yellow), 1/2 wwgg (wrinkled, yellow) 2/16 WWGg wwgg 1/2 WwGg (round, yellow), 1/2 Wwgg (round, green) 4/16 WwGg wwgg 1/4 WwGg (round, yellow), 1/4 Wwgg (round, green), 1/4 wwgg (wrinkled, yellow), 1/4 wwgg (wrinkled, green) W gg: 1/16 WWgg wwgg all Wwgg (round, green) 2/16 Wwgg wwgg 1/2 Wwgg (round, green), 1/2 wwgg (wrinkled, green) wwg : 1/16 wwgg wwgg all wwgg (wrinkled, yellow) 2/16 wwgg wwgg 1/2 wwgg (wrinkled, yellow), 1/2 wwgg (wrinkled, green) wwgg: 1/16 wwgg wwgg all wwgg (wrinkled, green) 20. Examine each trait separately. There are 104 long:34 short; and 69 brown:69 red. Length appears in a 3:1 ratio; therefore, long is dominant and the 3:1 ratio in the progeny tells us that each parent is heterozygous. Eye color appears in a 1:1 ratio. We can t conclude which allele is dominant; all we can conclude is that one parent is a recessive homozygote and that one parent is a heterozygote. (We will assume that red is the wild type.) If bw + = red, bw = brown, s + = long, and s = short, one possible way to indicate the cross is bw + bw s + s bwbw s + s. 22. You could set up a Punnett square and count the boxes; but unfortunately, this is an 8 8 matrix that yields 64 squares to count. A very tedious proposition! Set up the cross: DdGgWw DdGgWw and look at one gene at a time. a. The chance of getting a dominant trait from a monohybrid cross is 3/4. Therefore, the chance of all three dominant traits together is 3/4 3/4 3/4 = 27/64. b. The chance of getting a recessive trait from a monohybrid cross is 1/4. Therefore, the chance for all three recessives is 1/4 1/4 1/4 = 1/64. c. The chance of dwarf is 1/4, the chance of green is 1/4, and the chance of round

4 6 Hyde Chapter 2 Solutions is 3/4. Therefore the total chance producing the dwarf, green, round plant is 1/4 1/4 3/4 = 3/ a. Let e = ebony, e + = wild type, b = black, and b + = wild type. The first cross is ee b + b + e + e + bb. This will yield only e + e b + b. Thus, all F 1 flies will exhibit a wild type phenotype. b. Selfing of e + e b + b results in the following 9/16 e + b + wild type 3/16 e + bb black 3/16 ee b + ebony 1/16 ee bb ebony, black Since it is difficult to distinguish black and ebony, 7/16 will be dark-bodied and 9/16 will be normal. c. The first backcross is e + e b + b ee b + b +. Progeny = 1/2 ebony:1/2 normal. The second backcross is e + e b + b e + e + bb. Progeny = 1/2 black:1/2 normal. In each case we have a testcross situation for only one gene. 26. The F 1 progeny are AaBbCcDdEe. The chance of getting any individual with a particular homozygous genotype is (1/4) 5. Since we are looking for two different possibilities, we have 2(1/4) 5 = 2/1024 = 1/ This cross can be set up as a dihybrid cross: AaPp AaPp, where a is the recessive albinism allele and p is the recessive PKU allele. a. The chance of a recessive genotype (aa) from two heterozygotes is 1/4. b. The probability of having albinism (aa) but not PKU (P ) is 1/4 3/4 = 3/16. The probability of having PKU (pp) but not albinism (A ) is 1/4 3/4 = 3/16. For an either/or situation, we add the probabilities, 3/16 + 3/16 = 6/16 = 3/8. c. For two independent events to occur simultaneously, we multiply their individual probabilities, 1/4 1/4 = 1/ We calculate the frequency of deaths from cancer as 300/900 = 1/3, and the deaths due to heart disease as 200/900 = 2/9. a. 1/3 b. For death by cancer or heart disease, we add the probabilities, 1/3 + 2/9 = 5/ The cross is Aa aa. Therefore, there is a one-half chance of either a taster or nontaster offspring. We can imagine five different birth order scenarios that give one taster child (symbolized as T) and four nontasters (symbolized as N) children. T, N, N, N, N (where the taster child is the first-born); N, T, N, N, N (where the taster child is the second-born); N, N, T, N, N; N, N, N, T, N; and N, N, N, N, T. The probability of each scenario is 1/2 1/2 1/2 1/2 1/2 = 1/32. The probability

5 Hyde Chapter 2 Solutions 7 that one of five children will be a taster is given by the sum rule, 1/32 + 1/32 + 1/32 + 1/32 + 1/32 = 5/ Hypothesis: WwGg WwGg produces W G :W gg:wwg :wwgg in a 9:3:3:1 ratio. Critical chi-square at 0.05, 3 df, = Offspring W G W gg wwg wwgg Total Observed numbers (O) Expected ratio 9/16 3/16 3/16 1/16 Expected numbers (E) O E (O E) (O E) 2 /E = 2 Since this chi-square, 0.470, is less than the critical chi-square, we fail to reject (that is, accept) our hypothesis of two-locus genetic control with dominant alleles at each locus. 36. a. This is a dihybrid cross. From the F 1 progeny, we can determine that longwinged and tan-bodied must be dominant to short-winged and dark-bodied, respectively. Because this is a dihybrid cross, we expect the F 2 progeny should be in a 9:3:3:1 ratio. b. Our null hypothesis: This is a dihybrid cross with dominant alleles at each locus. Offspring Long, Long, Short, Short, Tan Dark Tan Dark Total Observed numbers (O) Expected ratio 9/16 3/16 3/16 1/16 Expected numbers (E) O E (O E) (O E) 2 /E = 2 Critical chi-square at 0.05, 3 df, = The chi-square for our null hypothesis, 3.133, is less than the critical chi-square. Therefore, we fail to reject (that is, accept) our hypothesis. 38. a. 1/2 1/2 1/2 1/2 1/2 = 1/32. b. (1/2 1/2 1/2 1/2 1/2) + (1/2 1/2 1/2 1/2 1/2) = 2/32 or 1/16. c. P(A bb C D E ) = 3/4 1/2 1/2 3/4 1/1 = 9/64. d. The probability of an individual with a parental phenotype is 9/64 + 9/64 = 9/32.

6 8 Hyde Chapter 2 Solutions Therefore, the remaining 1 9/32 = 23/32 will be phenotypically unlike either parent. e. P = 0, because all offspring of this cross will have the dominant E trait. 40. You should change your choice because the box you chose originally has a 1/3 chance of containing the prize, whereas the remaining box has a 2/3 chance of containing the prize. The 1/3 chance of your choice is set by the fact that there were three equally likely choices at the beginning. When your friend eliminated an empty box, she left two choices: your original box and the third box. Since the probability of your original choice has not changed, the probability that the remaining box contains the prize must be 2/3 to give a combined probability of 1.0.