In an electrophoretic study of enzyme variation in a species of bird, you find 6

Question

In an electrophoretic study of enzyme variation in a species of bird, you find 64 A1A1, 47 A1A2, and 9 A2A2 individuals in a sample of 120. A) What are the starting genotype and allele frequencies? B) What genotypic frequencies are expected in the next generation if the population reaches Hardy Weinberg equilibrium? C) What allele and genotype frequencies would you predict for the next generation if the A2A2 individuals face a coefficient of selection of 0.3? D) Beginning with the initial allele frequencies that you found in part A, what allele and genotypic frequencies would you predict for the next generation if A2A2 individuals face a coefficient of selection of 0.3 and the heterozygote experiences just 1/3 of the selective disadvantage of the A2A2 individuals.

Explanation / Answer

given from the question ;

A1A1 = 64 , HOMOZYGOUS

A1A2 = 47 HETEROZYGOUS

A2A2 =9 HOMOZYGOUS

TOTAL NUMBER OF SAMPLES ARE 120

What are the starting genotype and allele frequencies /

ANSWER =   GENOTYPES = A1A2   CROSS A1A2 = 1/4 A1A1 ,1/4 A 2 A2 , 1/2 =A1 A2

ALLELIC FREQUENCIES ;

HARDY WEINBURG EQUATION

P2+ 2PQ+Q2 =[ 0.25 ]2 +2* 0.25*0.25 +[ 0.25]2 = 0.5 +0.125 +0.5 = 1.1   , DEVIATION FORM THE HARDY WEINBURG EQUATION

What genotypic frequencies are expected in the next generation if the population reaches Hardy Weinberg equilibrium?

THE FRACTION OF THE GENOTYPIC FREQUENCIES ARE CALUCULATED NY THE FREQUENCY OF AELLES IN THE PREVIOUS GENERATION

What allele and genotype frequencies would you predict for the next generation if the A2A2 individuals face a coefficient of selection of 0.3

IF WE TAKE THE Q = 0.3 , THEN THE P VALUVE WILL BE 0.7 , IF WE CONSIDER IT IN THE HADY WEINBURG EQUATION WE GET THE GENOTPIC FEQUENICES

]0.3 ] +2.03.*0.7+ [0.7 ]2   = 0.54+2.0.42+0.83 =1.72 IS THE GENOYPIC ALLELIC FRQUNCIES FREQUNCIES

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