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In 1927 , the ophthalmologist George Waaler tested 9049 schoolboys in Oslo, Norway, for red-green color blindness and found 8324 of them to be normal and 725 to be color blind. He also tested 9072 schoolgirls and found 9032 that had normal color vision while 40 were color blind.

a. Assuming that the same sex-linked recessive allele cc causes all forms of red-green color blindness, calculate the allele frequencies of cc and CC (the allele for normal vision) from the data for the schoolboys. (Hint: Refer to your answer to Problem 12a.)

b. Does Waaler's sample demonstrate Hardy-Weinberg equilibrium for alleles of this gene? Explain your answer by describing observations that are either consistent or inconsistent with this hypothesis. On closer analysis of these schoolchildren, Waaler found that there was actually more than one cc allele causing color blindness in his sample: one kind for the prot type (cp)\left(c^p\right) and one for the deuter type (cd)\left(c^d\right). (Protanopia and deuteranopia are slightly different forms of red-green color blindness.) Importantly, some of the apparently normal females in Waaler's studies were probably of genotype cp/cdc^p / c^d. Through further analysis of the 40 color-blind females, he found that 3 were prot (cp/cp)\left(c^p / c^p\right), and 37 were deuter (cd/cd)\left(c^d / c^d\right).

c. Based on this new information, what are the frequencies of the cp,cdc^p, c^d, and CC alleles in the population examined by Waaler? Calculate these values as if the frequencies obey the Hardy-Weinberg equilibrium. (Note: Again, refer to your answer to Problem 12a.)

d. Calculate the frequencies of all genotypes expected among men and women if the population is at equilibrium.

e. Do these results make it more likely or less likely that the population in Oslo is indeed at equilibrium for red-green color blindness? Explain your reasoning.

Solution

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Answered 4 months ago
Answered 4 months ago

a) From 12a we know that the allelic frequencies for sex linked genes in males correspond to the genotype frequencies of the phenotype. So based on that:

C=NormalboysAllboys=83249049=0.92C=\dfrac{Normal\,boys}{All\,boys}= \dfrac{8324}{9049}=0.92

c=ColorblindboysAllboys=7259049=0.08c=\dfrac{Color-blind\,boys}{All\,boys}= \dfrac{725}{9049}=0.08

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