24.31 Fifty tiger salamanders from one pond in west Texas were examined for gene

Question

24.31 Fifty tiger salamanders from one pond in west Texas were examined for genetic variation by using the technique of protein electrophoresis. The genotype of each salamander was determined for five loci (AmPep, ADH, PGM, MDH, and LDH-1). No variation was found at AmPep, ADH, and LDH-1; in other words, all individ- uals were homozygous for the same allele at these loci. The following numbers of genotypes were observed at the MDH and PGM loci. MDH Genotypes Number of Individuals PGM Genotypes Number of Individuals 35 10 AB 35 DE Calculate the proportion of polymorphic loci and the het erozygosity for this population

Explanation / Answer

A genetic locus with two or more alleles, at which the most common form has a frequency not exceeding 0.95 in a given population is called polymorphic locus.

The proportion of Polymorphic Loci s calculated by determining the number of polymorphic loci and dividing by the total number of loci examined. In this question, 5 loci are evaluated and only 2 are polymorphic.

So, Polymorphic ratio: 2/5: 0.4=40%

The Hardy Weinberg equilibrium statement is (p + q)2=1

or

p2+2pq+q2=1

p2 represents the frequency of the homozygous genotype AA, q2 represents the frequency of the homozygous genotype BB and 2pq represent the heterozygosity.

Let us calculate the frequency:

total number of organisms: 50.

Number of AA organisms: 11.

Frequency of AA = 11/50 = 0.22

Similarly,

Frequency of BB= 4/50= 0.08

substituting these values in the equation: p2+2pq+q2=1

0.22 +2pq + 0.08 = 1

0.30 + 2pq = 1

2pq = 1 - 0.30 = 0.70.

This means the heterozygosity at the MDH locus is 0.70.

We will follow the same procedure to determine the frequency of the population at PGM locus.

Frequency of DD=35/50= 0.70

Frequency of EE = 5/50= 0.10

p2+2pq+q2=1

0.70 + 2pq + 0.10 = 1

0.80 + 2pq = 1

2pq = 1 - 0.80 = 0.20.

The mean heterozygosity of the population will be a sum of both heterozygosities divided by the number of heterozygosities.

(0.70 + 0.20)/2 = 0.90/2 = 0.45.

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.