A human female who is heterozygous for the recessive, sex-linked trait causing red-green color blindness, marries a normal male. What proportion of their male progeny will have red-green color blindness?
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A human female who is heterozygous for the recessive, sex-linked trait causing red-green color blindness, marries a normal male. What proportion of their male progeny will have red-green color blindness?
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- A WOMAN IS HETEROZYGOUS FOR TWO HARMFUL RECESSIVE ALLELES IN DIFFERENT CHROMOSOMES, ONE FOR PHENYLKETONURIA (PKU) AND THE OTHER FOR CYSTIC FIBROSIS (CF). SHE MARRIES AN UNAFFECTED MAN WHO IS A CARRIER FOR NEITHER DISEASE. IF SHE HAS A DAUGHTER, WHAT IS THE PROBABILITY THAT THE CHILD WILL CARRY NEITHER OF THE RECESSIVE ALLELES? EXACTLY ONE? BOTH?Red-green colorblindness in humans is caused by a sex-linked recessive allele. A color-blind man marries a woman with normal vision whose father was color-blind. Complete the Punnett square to answer the following questions: What is the probability that they will have a color-blind daughter? (X = wild type allele, Xc = color blind allele) Xc Y XcXc XY XCXY 1/2 3/4 None X Yc XcY XXc YYC 1/4 100% XC Y Probability of having a colorblind daughter? AA 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?
- Imagine that a couple is planning to have children. The male is heterozygous for Huntington’s disease and homozygous dominant for Tay-Sachs. The female is homozygous recessive for Huntington’s disease and heterozygous for Tay-Sachs. The couple is curious about the possibility and probability of their offspring inheriting Tay-Sachs and/or Huntington’s. For humans, Huntington’s disease is dominant (H) over the “normal” condition (h), and the “normal” condition is dominant (T) over Tay-Sachs (t). Complete a Punnett square for this cross and record the probabilities for genotypes and phenotypes of the offspring as ratios.A woman with keratosis, a skin condition caused by a rare dominant allele, marries a man with normal skin. If they have a son, what is the probability that he will have normal skin?A woman with blonde hair and blues eyes, which are recessive genes, marries a heterozygous man who has brown hair and eyes. What are the possible genotypes and phenotypes of their children?
- A 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. They have a daughter who is a dwarf with normal vision. What is the probability that she is heterozygous for both genes? а. 0% b. 25% c. 50% d. 75% е. 100%Duchenne muscular dystrophy (DMD), marked by muscular degeneration, results from an X- linked recessive gene. Thus, a female who is heterozygous for this gene and does not have the disease can be a carrier. What kind of offspring can you expect from a DMD-affected male and a carrier female? Can there be a carrier male?Cystic fibrosis is a recessive human condition. A male with Cystic fibrosis and a woman with a dominant phenotype have sevral children, in which one displays Cystic fibrosis. What can you conclude about the genotype of the maternal parent and what is the probability that a child who does not display Cystic fibrosis is heterozygous?
- One form of the bleeding disorder known as von Willebrand disease is an autosomal recessive disease. A man who is a carrier marries a woman who is also a carrier of the disease. (a) What percentage of their children are likely to have a disease phenotype? (b) What percentage of their children are likely to have a normal phenotype? (c) What percentage of their children are likely to be carriers of the disease?Imagine a cross between a man and a woman having the following genotypes: Genotype of the Husband Genotype of the Wife Aa bb Dd Gg Hh RR Tt X Aa Bb Dd gg Hh Rr Tt Assuming that the dominant allele in each gene pair exhibits complete dominance over the recessive allele, what is the probability that this couple would have a daughter who exhibits the recessive phenotype with respect to all seven of the gene pairs?In humans, the genes for colorblindness and hemophilia are both located on the X chromosome with no corresponding gene in the Y. These are both recessive alleles. If a man and a woman, both with normal vision, marry and have a colorblind son, draw the Punnet square that illustrates this. If the man dies and the woman remarries a colorblind man, draw a Punnet Square showing the type of children that could be expected from the second marriage. How many/what percentages of each could be expected.