dharjma
Forum Senior
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Posts: 182
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In a certain population the prevalence of colour blind males is 1 in 100. assuming the population is in hardy weinberg equilibrium at this locus, the frequency of colour blind females would be:
0.0001
0.0005
0.01
0.02
0.025
explanation:
Correct answer is A. colour blindness is an x-linked recessive trait. a male is hemizygous for the x chromosome, and this has only one copy of each trait. the frequency of an x-linked recessive in males is thus equal to the frequency of the allele in this population. from this, we know that q = 0.01 and p = 0.99. a female has two copies of each gene on the x chromosome so the equation for hardy weinberg equlibrium is the same as for the autosomal traits. in this case, a homozygous recessive female would occur at a frequency of q2 or 0.0001
whenever i encounter these types of questions i run and hide. i dont know what are the basic assumptions that i must have or where i should even begin. how do i begin answering this? what does "a female has two copies of each gene on the x chromosome so the equation for hardy weinberg equlibrium is the same as for the autosomal traits" mean? anyone? please! thanks..
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| sturge_weber
Forum Guru
Topics: 77
Posts: 1,042
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well when u have an x linked recessive disease in a male, he has only one x and hence, when the q says prevalence is 1/100. it means that q=1/100=0.01
now he says what is the gene frequency in females, who re colour blind, it would be , q2. this is because they have two x chromosomes, and hence in them the disease to be present should be on both chr. hence they will be q2= [1/100]2.. hence it would be 1/10000=0.0001
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| dharjma
Forum Senior
Topics: 11
Posts: 182
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my gosy it makes sense. it makes so much sense it almostmakes me feel i was dyslexic yesterday!  thanks dude 
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| sturge_weber
Forum Guru
Topics: 77
Posts: 1,042
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