Question

You observe a population of three-toed frogs that lives in the cloud forests of Panama. In this population, you observe the following genotype counts at a locus controlling coloration: 411 (AA), 800 (Aa), and 911 (aa). This population is not at Hardy-Weinberg Equilibrium (HWE). For this population to be in HWE, what would the frequency of homozygous AA frogs have to be?

		0.000
		0.194
		0.411
		0.146

Answer 1

FIRST CALCULATE ALLE LE CALCULATE GENOTYPE FREQUENCY THEN WE CAN FREQUENCY BY HARDY-WEINBERG Y BY HARDY-EINBERG EQUATION ALLELE FREQUENCY = NUMBER OF COPIES OF ALLELE NUMBER OF COPIES OF ALL ALLELES. NUMBER Of ALLELES, 41 'AA' = 440182 - 822 'A' ALLELES, 800 Aa 800 A AND 800 'a ALELBS. 911 aa' = 91X2 = 1822 a' ALLELES . TOTAL A ALLELES - 822+ 800 = 1622 TOTAL ALLELES = 42 44 (PFREQUENCY OF A ALVELE. = 1622 = 0.38 4244 NOW, HARDY WEINBERG EQUATION (P+Q) = 1 (P+)2 = p²+299 +92 at P= FREQUENCY OF DOMINANT ALLELE (A) p2_ FREQUENCY of te Homozy GOUS DOMINANT GENOTYPE. tok a FOR RECESSIVE ALLELE AND 92 FOR HOMOZYGOUS RECESSIVE GENOTYPE FREQUIBNCY, 'FRECON ENCY OF "AR? pz (0.38)2 0.1452 = 0.146 OP TLON Cd) 0.146 Is CORRECT ANSWER.