The TAS2R38 gene encodes a receptor protein that influences the ability to taste bitterness. The gene has two alleles: a dominant, wild-type allele that enables an individual (taster) to taste bitterness and a recessive, mutant allele that interferes with the ability of an individual (nontaster) to taste bitterness. In a sample of 2,400 people, 1,482 were found to have the dominant (taster) phenotype. Assuming that the population is in Hardy-Weinberg equilibrium, approximately how many individuals in the sample are expected to be heterozygous for TAS2R38

Answer:

The correct answer is - 1133.

Explanation:

According to H-W equilibrium;

p2 +2pq + q2 = 1 ........... eq.1   and

(p+q)2 = 1 ..............eq.2

Here q = frequency of the recessive allele,

and p = frequency of the dominant allele in the population

The genotype of p2 + 2pq shows the dominant(taster) phenotype.

The genotype of q2 shows the recessive(non-taster) phenotype.

q^2 = (2400 - 1482)/2400

= 0.3825  

= 0.618

using eq. 2 we get p= 1-q

=0.382

2pq = No. of heterozygotes/2400

No. of heterozygotes = 1133.1648

During which cell division phase does the nuclear membrane disappear?
A) telophase
B) metaphase
C) anaphase
D) prophase

A prospective groom, who is not affected by cystic fibrosis (CF), has a sister with cystic fibrosis, an autosomal recessive disease. Their parents are also not affected. The brother plans to marry a woman who has no history of CF in her family. What is the probability that they will have a child with CF? They are both Caucasian and the overall frequency of CF in the Caucasian population is 1/2500 - that is, 1 affected child per 2500 (assume the population meets the Hardy-Weinberg assumptions).

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