Assuming the requirements of HardyWeinberg equilibrium are m

Assuming the requirements of Hardy-Weinberg equilibrium are maintained for 1 generation, what genotypic and phenotypic frequencies would you expect?

allel frequencies are .78 and .22

expected genotype freq: P^2= .61 q^2=.05 2pq=.34

According to my chi square table, the populaiton isn\'t even in HW equilibrium

Solution

Assuming the requirements of Hardy-Weinberg equilibrium are maintained for 1 generation, what genotypic and phenotypic frequencies would you expect?

Allele frequencies are

p=0.78

and

q=0 .22

expected genotype freq:  P^2= .61 q^2=.05 2pq=.34

OBSERVED

EXPECTED

(O-E)^2/E

p^2

HOMOZYGOUS DOMINANT

.78 x .78

0.6084

0.61

(0.61-0.61)^2/61 =0

2pq

HETEROZY

GOTES

2 x0.78 x0.22

0.3432

0.34

(0.34-34)^2/34 =0

q^2

HOMOZYGOUS RECESSIVE

0.22 x0.22

0.0484

0.05

(0.5-0.5)^2/0.5 =0

1

0

By using gene count method

Frequency of p allele= D+ H/2

= 0.6084 + 0.3432/2 =0.78

Frequency of Q allele= R+ H/2

= 0.0484+ 0.3432/2 =0.22

Calculate t value is less than table value at 5% level of significance

NULL HYPOTHESIS : POPULATION IN HARDY WEINBERG EQUILIBRIUM

ALTERNATIVE HYPOTHESIS : POPULATION IS NOT IN HARDY WEINBERG EQUILIBRIUM

ACEPT NULL HYPOTHEIS AND REJECT ALTERNATIVE HYPOTHEIS

THERE IS NO SIGNIFICANT DIFFERENCE BETWEEN OBSERVED AND EXPECTED FREQUENCIES