ER 10 TABLE 10.2 Collared lizards in the New Mexico region. Blue Yellow A Az 988

Question

ER 10 TABLE 10.2 Collared lizards in the New Mexico region. Blue Yellow A Az 988 Generation 1 Total Orange AIA 551 Phenotype Genotype Number of individuals Observed genotype frequencies Allele frequencies Expected genotype frequencies Are these lizards in Hardy-Weinberg equilibrium (yes/no)? Generation 2 Phenotype Genotype Number of individuals Observed genotype frequencies Allele frequencies Expected genotype frequencies Are these lizards in Hardy-Weinberg equilibrium (yes/no)? Generation 3 Phenotype Genotype 361 = 1900 = 1.0 =1.0 Yellow A A2 901 Blue Total Orange AIA 510 289 -1700 = 1.0 Orange A Ai 546 Yellow A A2 1053 Blue A,A 351 Total Number of individuals = 1950 = 1.0 = 1.0 Observed genotype frequencies Allele frequencies Expected genotype frequencies Are these lizards in Hardy-Weinberg equilibrium (yes/no)? disequilibrium forces more likely in the Ozark population than the New Mexico nonula-

Explanation / Answer

Collared lizards in New Mexico region ; Generation 1

Allele frequency of A1 = Number of individuals with allele A1/ Total number of individuals

= 551+494/ 1900

=1045/1900

=0.55 (p)

A1A2 togetherly constitute 998. so, among them number of A1 allele is 988/2, which is equal to 494.

Similarly allele frequency of A2 = Number of individuals with allele A2/ Total number of individuals

=361+494 /1900

=0.45 (q)

Observed genotype frequencies of A1A1 = 551/1900

=0.29

Observed genotype frequencies of A2A2 = 361/1900

= 0.19

Observed genotype frequency of A1A2 is 988/1900

=0.52

Expected genotype frequency of A1A1 = p2 = 0.55 * 0.55

=0.3025

Expected genotype frequency of A2A2 = q2 =0.45 *0.45

= 0.2025

Expected genotype frequency of A1A2 = 2* p* q

=2*0.55* 0.45

=0.495

Multiplying each of the genotype frequency with the total number of individuals in a population, we find that there should be,

574 orange ( 0.3025 * 1900)( in actual 551)

940 yellow (0.495 * 1900) (in actual 988)

384 blue (0.2025 * 1900) ( in actual 361)

Here, in this example the population is not in Hardy- weinberg equilibrium as the Observed and Expected values doesn't match.

Generation 2

Allele frequency of A1= Total number of individuals with allele A1/ Total number of individulas in a population

=510+450.5/ 1700

=960.5/1700

= 0.565 (p)

Allele frequency of A2= Total number of individuals with allele A2/ Total number of individuas in the population

=289+450.5/1700

=739.5/1700

=0.435 (q)

Observerd genotype frequency of A1A1= 510/1700 =0.3

Observerd genotype frequency of A1A2=901/1700=0.53

Observerd genotype freuency of A2A2=289/1700=0.17

Expected genotype frequency of A1A1=p2 = 0.565*0.565 =0.3192

Expected genotype frequency of A1A2=2pq = 2* 0.565* 0.435 =0.4915

Expected genotype frequency of A2A2=q2 = 0.435*0.435 =0.1892

Multiplying each of the expected genotype frequency with ttal number of individuls in a population we get,

542 orange ( 0.3192 *1700) ( but in actual 510)

835 yellow (0.4915*1700) ( but in actual 901)

321 blue (0.1892*1700) ( but in actual 289)

Here also, the population is not in Hardy- Weinberg equilibrium as expected and observed doesnot match.

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