Albinism, an autosomal recessive trait characterized by an absenc...

Question

Albinism, an autosomal recessive trait characterized by an absence of skin pigmentation, is found in 1 in 4000 people in populations at equilibrium. Brachydactyly, an autosomal dominant trait producing shortened fingers and toes, is found in 1 in 6000 people in populations at equilibrium. For each of these traits, calculate the frequency of

For albinism only, what is the frequency of mating between heterozygotes?

Accepted Answer

Hey, everyone. Let's take a look at this question together. Albinism is an artisanal recessive trait that affects one in 4000 individuals in a population that is in equilibrium. What is the frequency of carriers? Is it answer choice? A one in 256 answer choice B one in 128 answer choice C one in or answer choice D one in 32. Let's work this problem out together to try to figure out which of the following answer choices is the frequency of carriers in this population. So we know that albinism is an autosomal recessive trait and it affects one in 4000 individuals, which means that since it is autosomal recessive, the individuals that experience albinism are the homozygous recessive individuals. So the homozygous recessive genotype is found in one out of 4000 individuals. And since the population is in equilibrium, we know that we are following Hardy Weinberg equilibrium, which means that the frequency of the recessive allele A is represented as Q and that the frequency of the homozygous recessive genotype is represented as Q squared. So that means that Q is equal to the square root of one divided by 4000, which means that Q is equal to 0.158. And since we want to know the frequency of carriers, we have to determine the frequency of the dominant allele, which we know that the dominant allele is represented as P, the P is equal to that dominant A allele. But those carrier individuals are the individuals with the heterozygous genotype. And we know that the value for the heterozygous genotype is calculated as two multiplied by P multiplied by Q. And we also know that P plus Q is equal to one which we already have the value for Q. So that means that P is equal to one minus Q which is 0.158. So the value for P is equal to 0.9842. And we have the value for Q and we have the value four P. So the frequency of carriers is represented as two multiplied by the value of P which is 0.98 or two, multiplied by the value of Q which is 0.158. And we end up with 0.3 which calculated into a fraction is one divided by 32. So answer choice D one in 32 is the correct answer for the frequency of carriers. Since the frequency of carriers is equal to the value of the frequency of the heterozygous genotypes which is calculated as two multiplied by the value four P multiplied by the value for Q which gives us 0.3 and one divided by 32 equals 0.3. I hope you found this video to be helpful. Thank you and goodbye.

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