Question

0 b) 0.21 c) 0.42 d) 0.49 e) 0.91 Problem 3: in humans, Rh-positive individuals have the Rh antigen on their red blood cells, while Rh. negative individuals do not. If the Rh-positive phenotype is produced by a dominant gene (A), and the Rh negative phenotype is due to its recessive ailele (a). what is the frequency of the Rh-positive allele if 84% of a population is Rh-positive? a) 0.04 b) 0.16 c) 0.48 d) 0.60 e) 0.84 Problem 4: The Hardy-Weinberg equation is useful for predicting the percent of a human population that may be heterozygous carriers of recessive alleles for certain genetic diseases. Phenylketonuria (PKU) is a human metabolic disorder that results in mental retardation if it is untreated in infancy. In the United States, one out of approximately 10,000 babies is born with the disorder. Approximately what percent of the population are heterozygous carriers of the recessive PKU allele? DI

Answer 1

3) The answer will be 0.60 (Option d).

Explanation: Given, population of Rh positive individual is 84%. So, population of Rh negative individual will be 16% or 0.16.

Now, according to Hardy-Weinberg equilibrium-

                                                    p² + 2pq + q² = 1 & p + q = 1

Where, p = frequency of dominant allele, q = frequency of recessive allele, p² = frequency of homozygous dominant genotype, q² = frequency of homozygous recessive genotype & 2pq = frequency of heterozygous genotype.

So, frequency of homozygous recessive genotype (q²) = 0.16

So, frequency of recessive allele (q) = $\sqrt{q^{2}}$ = = 0.4

V0.16

So, frequency of dominant allele (Rh positive allele) = 1 - q = 1-0.4 = 0.60

4) The answer will be 1.98% (Option b).

Explanation: Now, according to Hardy-Weinberg equilibrium-

                                                    p² + 2pq + q² = 1 & p + q = 1

Where, p = frequency of dominant allele, q = frequency of recessive allele, p² = frequency of homozygous dominant genotype, q² = frequency of homozygous recessive genotype & 2pq = frequency of heterozygous genotype.

According to the question, q² = 1/10000 = 0.0001

So, q = $\sqrt{q^{2}}$ = = 0.01

0.0001

So, p = 1 - q = 1 - 0.01 = 0.99

So, frequency of heterozygous carrier = 2 x p x q = 2 x 0.99 x 0.01 = 0.0198

So, percent of population for heterozygous carrier will be 1.98%.

5) The answer will be 0.16 or 16% (option b) (Actually answer will be 160 as the question asked about the number of individuals. But in option only percent of individuals provided).

Explanation: Now, according to Hardy-Weinberg equilibrium-

                                                    p² + 2pq + q² = 1 & p + q = 1

Where, p = frequency of dominant allele, q = frequency of recessive allele, p² = frequency of homozygous dominant genotype, q² = frequency of homozygous recessive genotype & 2pq = frequency of heterozygous genotype.

According to the question, q² = (1000-640) / 1000 = 0.36

So, q = $\sqrt{q^{2}}$ = = 0.6

V0.36

So, p = 1 - q = 1 - 0.6 = 0.4

So, frequency of homozygous dominant genotype (red eye), p² = (0.4)² = 0.16

6) The answer will be 0.4 (option b).

Explanation: As we know the value of q², we can directly determine the q by square root of q². So, q will be or 0.4.

V0.16

7) The answer will be 0.48 (option d).

Explanation: Now, according to Hardy-Weinberg equilibrium-

                                                    p² + 2pq + q² = 1 & p + q = 1

Where, p = frequency of dominant allele, q = frequency of recessive allele, p² = frequency of homozygous dominant genotype, q² = frequency of homozygous recessive genotype & 2pq = frequency of heterozygous genotype.

According to the question, q² = (200-168) / 200 = 0.16

So, q = $\sqrt{q^{2}}$ = = 0.4

V0.16

So, p = 1 - q = 1 - 0.4 = 0.6

So, frequency of heterozygotes = 2 x p x q = 2 x 0.6 x 0.4 = 0.48