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# Genetics: pattern baldness calculation using Hardy-Weinberg

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Pattern baldness in humans is controlled by a single autosomal locus with two alleles (B and B'), but pattern baldness is a sex-influenced phenotype: baldness is dominant in males (BB and BB' males become bald), but recessive in females (only BB females become bald). if a population is in Hardy-Weinberg equilibrium for this locus and 4% of the women in the population become bald,

1) what percentage of the men are bald?
2) what percentage of the nonbald women could give a gene for baldness to their sons?
3) what percentage of all possible marriages are expected to produce daughters who develop baldness? (assume that the marriages are not influenced by the baldness of the individuals)

https://brainmass.com/biology/dna-chromosomes-and-genomes/genetics-pattern-baldness-calculation-hardy-weinberg-55605

#### Solution Preview

Hardy-Weinberg equilibrium is mathematically expressed by the following:

p^2 + 2pq + q^2 = 1, where p + q = 1

p stands for one of the alleles and q stands for the other allele. For this question, let's assign B for p and B' for q.

That is to say that p^2 represents the frequency of BB, 2pq represents the frequency of BB' and q^2 represents the frequency of B'B'.

What do we know?

We know that 4% of women are bald (BB). Therefore, p^2 = 0.04, or if we take the square root, we get: p = 0.2.

Now, if p = 0.2, then q = 0.8 since p + q = 1.

Therefore, we can ...

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