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)

a. 0.56

b. 0.35

c. 0.42

Question

Jamiya Duke

Biology

11 March, 08:55

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)

a. 0.56

b. 0.35

c. 0.42

+4

Answers (1)

Know the Answer?

Maryjane Ashley · Answer 1

The answer is c. 0.42.

If:

p - the frequency of dominant allele P,

q - the frequency of recessive allele p,

the frequencies of the genotypes are:

p² - for PP genotype (dominant homozygote without the disease),

2pq - for Pp genotype (heterozygote for disease),

q² - for pp genotype (recessive homozygote with the disease).

It is given:

p = 0.3

2pq = ?

Since p + q = 1 ⇒ q = 1 - p

q = 1 - 0.3 = 0.7

Knowing p and q, we can calculate the frequency of people heterozygous for this disease (2pq):

2pq = 2 · 0.3 · 0.7 = 0.42

Therefore, the frequency of people heterozygous for this disease is 0.42