homework 4 ans
4.01 The French biologist Cuenot crossed wild, gray-colored mice with white (albino) mice. In the first generation, all were gray. He then crossed the F1 mice among themselves. From many litters, he obtained in the F2 generation 198 gray mice and 72 white mice.
4.01 The French biologist Cuenot crossed wild, gray-colored mice with white (albino) mice. In the first generation, all were gray. He then crossed the F1 mice among themselves. From many litters, he obtained in the F2 generation 198 gray mice and 72 white mice.
(a) Propose a hypothesis to explain these results. Diagram both generations (phenotype and genotype) based on your hypothesis.
(a) Propose a hypothesis to explain these results. Diagram both generations (phenotype and genotype) based on your hypothesis.
Answer: The hypothesis is that gray (G) is dominant to (g)
Answer: The hypothesis is that gray (G) is dominant to (g)
(b) Test your hypothesis using a chi-square test.
(b) Test your hypothesis using a chi-square test.
Answer:
Answer:
For df=1, at the 0.05 probability level, the critical value from the table is 3.84. 0.4 < 3.84, so the data are consistent with the hypothesis
For df=1, at the 0.05 probability level, the critical value from the table is 3.84. 0.4 < 3.84, so the data are consistent with the hypothesis
4.02 In peas, axillary flowers (A) is dominant to terminal flowers (a) and colored flowers (R) is dominant to white (r). A true-breeding plant with white, terminal flowers is crossed with a true-breeding plant having colored, axillary flowers. The F1 plants are then allowed to self-pollinate.
4.02 In peas, axillary flowers (A) is dominant to terminal flowers (a) and colored flowers (R) is dominant to white (r). A true-breeding plant with white, terminal flowers is crossed with a true-breeding plant having colored, axillary flowers. The F1 plants are then allowed to self-pollinate.
(a) Using the forked line method, list all possible F2 phenotypes and calculate the probability of each.
(a) Using the forked line method, list all possible F2 phenotypes and calculate the probability of each.
Answer:
Answer:
(b) The F2 generation consisted of 308 plants with axillary, colored flowers; 34 plants with terminal, white flowers; 118 plants with axillary, white flowers; and 132 plants with terminal, colored flowers. Using the chi-square test, determine if this outcome is consistent with the predicted ratios for this cross.
(b) The F2 generation consisted of 308 plants with axillary, colored flowers; 34 plants with terminal, white flowers; 118 plants with axillary, white flowers; and 132 plants with terminal, colored flowers. Using the chi-square test, determine if this outcome is consistent with the predicted ratios for this cross.
Answer:
Answer:
The calculated chi-square value is 6.53 which is less than the table value of 7.815 for df=3. Data are consistent with the expected ratio.
The calculated chi-square value is 6.53 which is less than the table value of 7.815 for df=3. Data are consistent with the expected ratio.
4.03 You are doing a cross with Drosophila using the following two traits. Curly wings is dominant over straight wings, and round eyes is dominant over elliptical eyes. You cross a female fly that is known to be heterozygous for both genes with a male that is heterozygous for the wing gene but has elliptical eyes. This cross produces 74 flies with curly wings and round eyes, 61 with curly wings and elliptical eyes, 24 with straight wings and round eyes, and 21 with straight wing and elliptical eyes. Calculate the expected phenotype ratios for this cross, then use the chi-square test to see if the observed data are consistent with the expected numbers.
4.03 You are doing a cross with Drosophila using the following two traits. Curly wings is dominant over straight wings, and round eyes is dominant over elliptical eyes. You cross a female fly that is known to be heterozygous for both genes with a male that is heterozygous for the wing gene but has elliptical eyes. This cross produces 74 flies with curly wings and round eyes, 61 with curly wings and elliptical eyes, 24 with straight wings and round eyes, and 21 with straight wing and elliptical eyes. Calculate the expected phenotype ratios for this cross, then use the chi-square test to see if the observed data are consistent with the expected numbers.
Answer: C- = curly wings, cc = straight wings; R- = round eyes, rr = elliptical eyes
Answer: C- = curly wings, cc = straight wings; R- = round eyes, rr = elliptical eyes
The cross is CcRr x Ccrr. Calculate the expected probabilities with a forked line.
The cross is CcRr x Ccrr. Calculate the expected probabilities with a forked line.
4.04 Calculate the probability of obtaining each of the following outcomes when flipping two coins simultaneously: (a) both heads, (b) both tails, and (c) one head and one tail. Now generate some actual data by flipping two coins 30 times are recording the number of each kind of outcome. Then use a chi-square test to see if your data are consistent with your predicted probabilities.
4.04 Calculate the probability of obtaining each of the following outcomes when flipping two coins simultaneously: (a) both heads, (b) both tails, and (c) one head and one tail. Now generate some actual data by flipping two coins 30 times are recording the number of each kind of outcome. Then use a chi-square test to see if your data are consistent with your predicted probabilities.
Answer: P(both heads) = 1/4, P(head, tail) = 1/2, P(both tails) = 1/4
Answer: P(both heads) = 1/4, P(head, tail) = 1/2, P(both tails) = 1/4
Chi-square test should indicated that your actual data is consistent with these probabilities (assuming you use a fair coin, of course)
Chi-square test should indicated that your actual data is consistent with these probabilities (assuming you use a fair coin, of course)
4.05 An ordinary 6-sided die is rolled repeatedly with the following outcome: 63 ones, 31 twos, 54 threes, 46 fours, 51 fives, and 55 sixes. Would you conclude from this data that it is a fair die? Give statistical support (chi-square test) for your conclusion.
4.05 An ordinary 6-sided die is rolled repeatedly with the following outcome: 63 ones, 31 twos, 54 threes, 46 fours, 51 fives, and 55 sixes. Would you conclude from this data that it is a fair die? Give statistical support (chi-square test) for your conclusion.
Answer:
Answer:
 You would expect an equal number of each if this is a fair die. The calculated chi-square value is 11.76 which is slightly greater than the table value of 11.07 (for df=5). The data are not consistent, suggesting that the die is not fair.
 You would expect an equal number of each if this is a fair die. The calculated chi-square value is 11.76 which is slightly greater than the table value of 11.07 (for df=5). The data are not consistent, suggesting that the die is not fair.
4.06 The pedigree shows the pattern of inheritance for a rare human disease. Filled symbols indicate individuals with the disease; open symbols indicate normal individuals.
4.06 The pedigree shows the pattern of inheritance for a rare human disease. Filled symbols indicate individuals with the disease; open symbols indicate normal individuals.
(a) What is the most probable mode of inheritance (dominant or recessive) for this trait?
(a) What is the most probable mode of inheritance (dominant or recessive) for this trait?
Answer: Recessive
Answer: Recessive
(b) What is the most convincing evidence you can give for your answer to (a)?
(b) What is the most convincing evidence you can give for your answer to (a)?
Answer: The best evidence is that it is possible for two normal parents to have a diseased child. This is seen in the mating of II-6 and II-7. If the disease was dominant, every person with the disease would have at least one diseased parent.
Answer: The best evidence is that it is possible for two normal parents to have a diseased child. This is seen in the mating of II-6 and II-7. If the disease was dominant, every person with the disease would have at least one diseased parent.
(c) What are the most probable genotypes of individuals II-I and II-2?
(c) What are the most probable genotypes of individuals II-I and II-2?
Answer: Using N of normal and n for diseased, II-1 must be nn and II-2 must be Nn. The father has the disease, so he must be homozygous recessive. The mother is normal, but the couple has two diseased children, so she must be a carrier (heterozygous).
Answer: Using N of normal and n for diseased, II-1 must be nn and II-2 must be Nn. The father has the disease, so he must be homozygous recessive. The mother is normal, but the couple has two diseased children, so she must be a carrier (heterozygous).
(d) If individuals II-6 and II-7 have another child, what is the probability that it will be normal?
(d) If individuals II-6 and II-7 have another child, what is the probability that it will be normal?
Answer: The probability is 3/4. This is a simple 4-box Punnett square problem. Both parents are heterozygous.
Answer: The probability is 3/4. This is a simple 4-box Punnett square problem. Both parents are heterozygous.
4.07 The following pedigree shows the pattern of inheritance for a rare human disease. Filled symbols indicate individuals with the disease: open symbols indicate normal individuals.
4.07 The following pedigree shows the pattern of inheritance for a rare human disease. Filled symbols indicate individuals with the disease: open symbols indicate normal individuals.
(a) What is the most probable mode of inheritance (dominant or recessive) for this trait? On what do you base your answer?
(a) What is the most probable mode of inheritance (dominant or recessive) for this trait? On what do you base your answer?
Answer: The trait is probably dominant because it occurs in every generation and every person that has it had a parent that had it. It is actually possible that the trait is recessive because there is no case where both parents have it but kids no not. However, this is very, very unlikely since the trait is rare. If it was recessive, I-2, II-1, and II-5 would all have to be carriers. The chance of that would be exceedingly small.
Answer: The trait is probably dominant because it occurs in every generation and every person that has it had a parent that had it. It is actually possible that the trait is recessive because there is no case where both parents have it but kids no not. However, this is very, very unlikely since the trait is rare. If it was recessive, I-2, II-1, and II-5 would all have to be carriers. The chance of that would be exceedingly small.
(b) What are the most probable genotypes of individuals I-1 and I-2?
(b) What are the most probable genotypes of individuals I-1 and I-2?
Answer: I-1 is Aa, I-2 is aa.
Answer: I-1 is Aa, I-2 is aa.
(c) What are the most probable genotypes of individuals II-9 and II-10?
(c) What are the most probable genotypes of individuals II-9 and II-10?
Answer: II-9 is aa, II-10 is aa
Answer: II-9 is aa, II-10 is aa
(d) If individuals II-1 and II-2 have another child, what is the probability that it will have the disease?
(d) If individuals II-1 and II-2 have another child, what is the probability that it will have the disease?
Answer: II-1 is aa and II-2 is Aa. Using a Punnett square, you find that there is a 1/2 chance of a child being Aa and 1/2 being aa, so the chance that their next child has the trait is 1/2.
Answer: II-1 is aa and II-2 is Aa. Using a Punnett square, you find that there is a 1/2 chance of a child being Aa and 1/2 being aa, so the chance that their next child has the trait is 1/2.
4.08 In the following hypothetical pedigree, there are two traits considered. One gene determines if a person is a vampire or not; the other gene determines if a person is a werewolf or not. The two genes sort independently. Study the diagram then answer the questions below. (HINT: Examine each trait separately when working out the genotypes.)
4.08 In the following hypothetical pedigree, there are two traits considered. One gene determines if a person is a vampire or not; the other gene determines if a person is a werewolf or not. The two genes sort independently. Study the diagram then answer the questions below. (HINT: Examine each trait separately when working out the genotypes.)
(a) Is being a vampire due to a dominant or recessive allele? How can you tell?
(a) Is being a vampire due to a dominant or recessive allele? How can you tell?
Answer: When examining this pedigree, you should consider the two genes separately. First, just see if an individual is a vampire or not (don't worry about whether they might be a werewolf). Vampire is dominant. Look at the mating of II-3 and II-4. Both are vampires, but they have a normal (non-vampire) son.
Answer: When examining this pedigree, you should consider the two genes separately. First, just see if an individual is a vampire or not (don't worry about whether they might be a werewolf). Vampire is dominant. Look at the mating of II-3 and II-4. Both are vampires, but they have a normal (non-vampire) son.
(b) Is being a werewolf due to a dominant or recessive allele? How can you tell?
(b) Is being a werewolf due to a dominant or recessive allele? How can you tell?
Answer: Now consider the werewolf trait, ignoring the vampire trait. Werewolf is recessive. Look at the mating of II-6 and II-7. Both are normal (non-werewolves), but they have a werewolf child.
Answer: Now consider the werewolf trait, ignoring the vampire trait. Werewolf is recessive. Look at the mating of II-6 and II-7. Both are normal (non-werewolves), but they have a werewolf child.
(c) Give the most likely genotypes for individuals I-2, II-3, and II-6. Use D and d for the alleles of the vampire gene and R and rfor the alleles of the werewolf gene.
(c) Give the most likely genotypes for individuals I-2, II-3, and II-6. Use D and d for the alleles of the vampire gene and R and rfor the alleles of the werewolf gene.
Answer: I-2 is DdRr, II-3 is DdRr, II-6 is ddRr
Answer: I-2 is DdRr, II-3 is DdRr, II-6 is ddRr