The gene for polydactyly (P) is autosomal and dominant to normal fingers (p). Hemophilia is sex-linked and recessive (X h ). A man and his wife both of whom are polydactylous and have normal red blood cells have a child who is non-polydactylous and with hemophilia. What are their genotypes?
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The gene for polydactyly (P) is autosomal and dominant to normal fingers (p). Hemophilia is sex-linked
and recessive (X h ). A man and his wife both of whom are polydactylous and have normal red blood
cells have a child who is non-polydactylous and with hemophilia. What are their genotypes?
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- Clara has polydactyly, an autosomal dominant that is 80% penetrant. Clara inherited her polydactyly from her mother, her father had no polydactyly in his family. She has an extra toe on her left foot. Her husband Ralph has no polydactyly in his family. A) What is the chance that Clara will have a child that will inherit polydactyly? B) What is the chance that the child will express polydactyly. You can leave these as un-calculated equations.Sickle cell anemia is inherited as an autosomal recessive condition. It also exhibits incomplete dominance in that the heterozygous genotype displays a mild form of the disease known as sickle cell trait while individuals with the homozygous recessive genotype have a severe form of SCA. A man who has severe sickle cell anemia marries a woman who suffers from a mild trait. What is the probabilitu they will have a child with severe SCA?What is the probability they will have a child with mild SCA? What is the probability they will have a normal child? Show ALL work using punnett squares.Hemophilia is a sex-linked recessive trait. A male hemophiliac and phenotypically normal female have a son with hemophilia. They would like to have one more child. What is the probability of having a child without hemophilia? Explain using a Punnett square. Is it possible for a girl to be born with hemophilia? Explain.
- A mother is heterozygous for the X-linked gene for colorblindness and also heterozygous for the autosomal inherited sickle cell anemia. She is married to a man who can see color normally and who is heterozygous for sickle cell trait. Using b (colorblind), B (normal color), S (normal hemoglobin), s (sickle cell), answer the following: a. What are the genotypes of the parents? b: What is the probability of having a child who is both color blind and has sickle cell anemia?Albinism, lack of pigmentation in humans, results from an autosomal recessive gene designated a. Two parents with normal pigmentation have an albino child. What is the probability that their next child will be albino? What is the probability that the next child will be an albino girl? If the child is normal, what is the probability that it will be a carrier (heterozygous) for the albino gene?Neurofibromatosis-1 (NF1) is an autosomal dominant disorder where tumours form in the base layer of the skin or in nerve tissues. What is the probability that individuals II-1 and II-2 will have a genetic son with NF1? Find the image attached.
- 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 their first child will be a brachydactylous girl? ¼ 1/2 1/8 3/4 2/3A form of hemophilia is caused by a sex-linked (X-linked) recessive gene. A phenotypically normal woman whose father had hemophilia marries a man who suffers with hemophilia. What is the probability that their first daughter will have hemophilia?Hemophilia is an X-linked recessive disorder and blood type is autosomal. If two healthy parents, one of which is blood type A and the other blood type B, produce a son who has hemophilia and is type O, what is the probability that their next child will be a son with hemophilia and blood type B? 1/2 3/16 1/4 1/8 1/16
- Both red/green color blindness (R=normal, r=colorblind) and Duchenne-type muscular dystrophy (D=normal, d=muscular dystrophy) are X-linked recessive traits that map close to each other. A woman has a father who is red/green color blind. Her mother's family has a history of Duchenne's muscular dystrophy. This woman is apparently healthy with neither color blindness nor muscular dystrophy. She marries a healthy man and they have four sons and two daughters. Half the sons are healthy but color blind, the other half have normal color vision, but have Duchenne's muscular dystrophy. The daughters are both normal. What is the genotype of the woman? ** The notation is written as X/X (so the / separates the two X chromosomes in a female) Group of answer choices RD/rd rd/rd Rd/rD RD/RDA man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man's father was 6 feet tall, and both the woman's parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive. a) How many of their daughters might be expected to be color-blind dwarfs? b)What proportion of their sons would be color-blind and of normal height? c)They have a daughter who is a dwarf with normal color vision. What is the probability that she is heterozygous for both genes?1. A) Apply the concept of sex linkage to explain why color blindness is more prevalent in men than in women. B) Mary is concerned that she may be a carrier for hemophilia, a sex-linked condition located on the X chromosome. Mary is married to John, who doesn't have hemophilia. Assuming Mary is a carrier, what are the genotype ratios expected for Mary and John's kids (specify for boys and for girls)? C) Mary and John have 2 boys, none of them has hemophilia. Can we use this fact as proof that Mary does not carry the allele for hemophilia? Explain your argument.