Question

Suppose when you count the actual number of offspring produced and determine their genotypes, you get the following: 469 AA, 448 Aa and 83 aa. What are the allele frequencies for these individuals in this generation?

frequency of dominant allele _____.

frequency of recessive allele ______.

YES OR NO - Is this population in Hardy-Weinberg equilibrium (pick one)

Simply and briefly - explain why ________

Answer 1

frequency of dominant allele 0.69.

frequency of recessive allele 0.31.

YES OR NO - Is this population in Hardy-Weinberg equilibrium (pick one): YES

Simply and briefly - explain why FIND THE BELOW CALCULATION

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The steps involved here

Step 1: Calculation number of alleles

Genotype	Freequency	Allele A	Allele a	Total
AA	469	938	0	938
Aa	448	448	448	896
aa	83	0	166	166
Total	1000	1386	614	2000

Step 2: Calculation of allele frequencies

Frequency = No of a alleles/Total no of alleles

Allele A (Dominant allele)	= 1386/2000	= 0.69
Allele a (Recessive allele)	= 614/2000	= 0.31

Step 3: Expected genotype frequencies estimation

Genotype	Genotype frequencies	For 1000 population the freqeunecies
AA	= 0.69*0.69= 0.480	*1000	=480
Aa	2*0.69*0.31= 0.426	*1000	=426
aa	0.31*0.31= 0.094	*1000	=94

Step 4: Chi-square test

Null hypothesis: The observed values are not deviating from the expected values.

Category	AA	Aa	aa
Observed values	469	448	83
Exprected Values	480	426	94
Deviation	-11	22	-11
D^2	126.54	506.16	126.54
D^2/E	0.26	1.19	1.34	2.80
X^2	2.80
Degrees of freedom	1

Inference: The calculated chisquar value i.e. 2.80 is less than the table value i.e.3.84 at 1 DF and 0.05 probability, hence the null hypothesis is accepted, which means the population is in HW equilibrium.