please help me with this problem 2! Problem 2. In humans, the genes for color blindness and hemophilia are both on the X chromosome. Let A be the wild-type allele for color blindness and a be the mutant allele, and let B be the wild-type allele and b be the mutant allele for hemophilia Male sex chromosomes are XY, female sex chromosomes are XX, and the combination of alleles of each gene on the X chromosome is indicated by XAB,XAb, XaB, and Xab The family tree shown below is part of one larger family tree. Answer questions I to III. Note that the mutant allele was inherited from the ancestor and was not newly generated in the generation shown in the family tree. Males Females Normal hemophilia color blindness . 10 both diseases 1) What are the allelic combinations of the 3 and 6 male sex chromosomes, respectively? 11) Assume that no homologous recombination occurs between the gene for color blindness (A or a) and the gene for hemophilia (B or b) during meiosis. Who in the figure would be considered to have the variant allele b that causes hemophilia? Please answer with a number. 111) Crossover occurs during meiosis, and 10% of the gametes produced undergo a single homologous recombination between the gene for color blindness (A or a) and the gene for hemophilia (B or b). What is the probability that 9 and 10 women each have the variant allele b? Also, what is the percentage chance that their son will have hemophilia?

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