Question

PKU (Phenylketonuria) is an autosomal recessive disease, in which the synthesis of amino acid Tyrosine from Phenylalanine is blocked. As a result, an excess of Phenylalanine gets converted into phenylketones, which appear in the urine. In severe conditions it may also result in damage to the brain. The gene responsible for this is p, whereas the gene P is responsible for normal synthesis of Tyrosine. In a small population of Brazilian natives, the frequency of gene p, responsible for this disease, is 0.3. What must be the frequency of people who are heterozygous for this disease? ( p + q = 1, p2 + 2pq + q2 = 1)

Answer 1

Answer:

Thus, the frequency of people who are heterozygous for this disease is 0.21

Explanation:

Given,

The frequency of gene p is $0.3$

As per Hardy- Weinberg equation,

Sum of allele frequencies for all allele at locus $= 1$

Which means sum of allele frequencies of recessive and dominant allele is 1

$p + q = 1$

Substituting the value of p in above equation, we get -

$0.3 + q = 1\\q = 1 - 0.3\\q = 0.7$

Another equation of Hardy- Weinberg is

$p^2 + 2pq + q^2 = 1$

Substituting the value of p and q in above equation, we get -

$(0.3)^2 + (0.7)^2 + 2pq = 1\\2pq = 1 - 0.09 - 0.49\\pq = \frac{0.42}{2} \\pq = 0.21$

Thus, the frequency of people who are heterozygous for this disease is 0.21

Answer 2

Answer: Frequency of heterozygous people is 0,42

Explanation: taking into account Hardy-Weinberg equations for allele frequencies it is known that the sum of the fequencies of the alleles is 1

p + q = 1

If we know that p allele has  a frequency of 0,3, the frequency for P allele will be 0,7 since

1- q = p    or   1 - 0,3 = 0,7

Now, the other equation p2 +2pq+q2 = 1 express that sum of  genotype frequencies is 1, where p2 is homozygous dominant frequency, 2pq is heterozygous frequency and q2 is homozygous recesive frequency.

As we know the values for p and q we can calculate fequency for heterozygous:

2pq = 2 (0,7) (0,3) = 0,42