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- 1. Given the karyotype shown here, is this a male or a female? What would the genotype/phenotype of this individual be?8. The autosomal (not X-linked) gene for brachydactyly, short fingers, is dominant to normal finger length. Assume that a female with brachydactyly in the heterozygous condition is married to a man with normal fingers. What is the probability that(a) their first child will have brachydactyly?(b) their first two children will have brachydactyly?(c) their first child will be a brachydactylous girl?3. A man who has polydactyly and normal vision marries a color-blind woman with a normal number of hand and foot digits. The man's father has normal hand and foot digitation, and both the woman's parents have normal hand and foot digitation. A) Polydactyly is autosomal dominant, which means [ Select ] the gene is carried on the X chromosome b. the gene is carried on a non-sex chromosome a. c. the gene is carried on the Y chromosome d. the dominant allele is actually recessive B) Red-green color blindness is X-linked recessive, which means that in females, [ Select ] a. the recessive allele is actually dominant b. the gene is carried on a non-sex chromosome c. the gene must have both recessive alleles to be expressed in the phenotype d. the gene is carried on the Y chromosome C) What proportion of their sons would be color blind and polydactyly? [ Select ] a. half b. none C. one out of four d. all D) What proportion of their daughters would be color blind and polydactyly? [ Select ] а.…
- 6. An X-linked ichthyosis is a recessive form of a family of skin diseases caused by a hereditary deficiency of the steroid sulfatase (STS) enzyme. A woman heterozygous for this mutation mates with the phenotypically normal man and produces an XXY son who suffers from the disorder. Which of the answer choices can explain this result? Briefly justify your answer. A) nondisjunction in meiosis I in the father B) nondisjunction in meiosis I in the mother C) nondisjunction in meiosis II in the father D) nondisjunction in meiosis II in the mother8. Huntington’s disease is a degenerative disease of the nervous system that strikes in middle age. The allele that causes the disease (H) is dominant to the allele that results in the normal condition (h). Answer the following questions about the inheritance of this disease. A. What is the genotype of a man who is normal but whose father had Huntington’s disease? B. What is the genotype of a woman who has Huntington’s disease if both of her parents had Huntington’s disease? C. If a man who is heterozygous for Huntington’s disease marries a woman who is normal, what would you expect for the genotypes and phenotypes of their children? D. If a normal man marries a woman who is homozygous for Huntington’s disease, what do you expect for the genotypes and phenotypes of their children?1. If an A a Bb C cdd male mates with an A a Bb CC DD female. (a) What is the minimum number of ridge-producing genes possible for a child of this couple? (Show your work. Hint: solve gene by gene) (b) what would be the TRC for this child if it is male? (Show your work: give the formula, place appropriate number in formula and then solve equation). (c) what would be the TRC for this child if it is female? (Show your work: give the formula, place appropriate number in formula and then solve equation).
- 2. Uniparental disomy is a rare phenomenon in whichonly one of the parents of a child with a recessivedisorder is a carrier for that trait; the other parent ishomozygous normal. By analyzing DNA polymorphisms, it is clear that the child received both mutantalleles from the carrier parent but did not receive anycopy of the gene from the other parent.a. Diagram at least two ways in which uniparentaldisomy could arise. (Hint: These mechanismsall require more than one error in cell division,explaining why uniparental disomy is so rare.)Is there any way to distinguish between thesemechanisms to explain any particular case ofuniparental disomy?2. Uniparental disomy is a rare phenomenon in whichonly one of the parents of a child with a recessivedisorder is a carrier for that trait; the other parent ishomozygous normal. By analyzing DNA polymorphisms, it is clear that the child received both mutantalleles from the carrier parent but did not receive anycopy of the gene from the other parent.a. Diagram at least two ways in which uniparentaldisomy could arise. (Hint: These mechanismsall require more than one error in cell division,explaining why uniparental disomy is so rare.)Is there any way to distinguish between thesemechanisms to explain any particular case ofuniparental disomy?b. How might the phenomenon of uniparental disomyexplain rare cases in which girls are affected withrare X-linked recessive disorders but have unaffectedfathers, or other cases in which an X-linked recessive disorder is transmitted from father to son?c. If you were a human geneticist and believed oneof your patients had a disease syndrome caused…20. In epistasis, one gene masks or interferes with the expression of another. True False
- 9. Explain, giving reasons, whether the following pedigrees are compatible with autosomal dominant, autosomal recessive or X-linked dominant and X-linked recessive inheritance. (Note that a pedigree may be compatible with more than one type of inheritance.) 1:1 12 1:1 12 1:1 1:2 I12 I:4 IIS II:1 I12 II:1 I1:2 Il:3 I1:4 II2 II:1 III3 II:1 II:2 III:3 III:4 a. b. c.6. A person is simultaneously heterozygous for two autosomal genetic traits. One is a recessive condition foralbinism (alleles A and a); this albinism gene is foundnear the centromere on the long arm of an acrocentricautosome. The other trait is the dominantly inheritedHuntington disease (alleles HD and HD+). TheHuntington gene is located near the telomere of oneof the arms of a metacentric autosome. Draw all copies of the two relevant chromosomes in this person asthey would appear during metaphase of (a) mitosis,(b) meiosis I, and (c) meiosis II. In each figure, labelthe location on every chromatid of the alleles forthese two genes, assuming that no recombinationtakes place.4. In fruit flies, opaque wing color is due to an X-linked dominant version of a gene. A female with opaque wings mates with a male with opaque wings. Based on this information, which of the following statements is accurate and can be stated with 100% confidence iregarding their offspring? A) female offspring have a 50% chance of having opaque wings B) males are guaranteed to have a 100% chance of having opaque wings C) female offspring cannot have opaque wings D) males are guaranteed to have a 50% chance of having opaque wings E) none of the above statements are completely accurate due to inadequate information provided