Question

Researchers interested in the population structure of the Eastern red-spotted newt, collected many juvenile red efts from Rothrock State Forest. Three alleles are found at a locus coding for phosphoglycerate mutase (PGM) in the red efts. Each red efts's genotype at the PGM locus was determined. The following numbers of genotypes were found: Genotype Numbers P1P1 P1P2 35 P2P2 20 P1P3 53 P2P3 76 P3P3 62 What are the frequencies of the P1, P2, and P3 alleles in the population and what is the expected number of individuals of the P1P2 genotype? O 0.2 P1,0.3 P2,0.5 P3; 30.48 individuals O 0.1 P1,0.4 P2,0.5 P3; 10.16 individuals O 0.4 P1,0.3 P2,0.3 P3; 60.96 individuals O 0.2 P1,0.3 P2,0.5 P3; 15.24 individuals O 0.1 P1,0.4 P2,0.5 P3; 20.32 individuals

Answer 1

Ans:: This question can be solved by the Hardy Weinberg equilibrium.

According to the principle of this equilibrium, a population consisting of different alleles responsible for genetic variation in that population, the sum of the frequencies of these alleles remain constant from generation to generation considering that population is very large and there occurs random mating in between the individuals of the population, no mutation occurring.

• In this population of eastern red-spotted newt, three alleles: P1, P2, P3 are found at a locus.
• Numbers of different genotypes are given in this table. we have to calculate the total number of the population by adding these given numbers.
• After adding all the numbers, Total individuals in the population comes out to be 254.
• To calculate the frequency of each genotype that is given, we have to divide the number of particular genotype with the total population. For example: number of organisms with P1P1 genotype is 8. Hence, the frequency of P1P1 genotype = 8/254 = 0.03
• Allelic frequency of P1 = square-root of 0.03 = 0.17, approximately 0.2.

Similarly, in this given table. all the frequencies of given genotype are solved.

Allelic frequency squareroot of frequency obtained Allelic frequency (approximate) 0.177471302 0.2 = P1 Genotypic frequency Genotype Numbers Frequency (Numbers/254) P1P1 | 8 0.031496063 P1P2 350.137795276 P2P2 20 0 .078740157 P1P3 53 0.208661417 P2P3 76 0.299212598 РЗp3 62 0.244094488 Total 254 0.280606767 0.3 = P2 0.494059195 1 0 .5 = P3

• As shown in column 3, total of all these genotypic frequencies is 1. Also, frequencies of P1, P2, P3 obtained are 0.2,0.3,0.5 respectively.
• It is following the hardy-weinberg equilibrium, as for three alleles, i.e., (P1 + P2+ P3)² = 1
• Values obtained by calculation that is (0.2 + 0.3 + 0.5)² =1 for P1, P2, P3 respectively are following the Hardy- weinberg equilibrium.
• Expected population of P1P2 will be 2 X P1 frequency X P2 frequency X total population = 2.X 0.2 X 0.3 X 254 = 30.48
• Hence, expected individuals of P1P2 genotype = 30.48

Correct option is 1 = 0.2P1, 0.3P2, 0.5P3, 30.48

Hope this will help you.