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The next generation of lizards has 1092 individuals with green scales and 108 individuals with blue scales. is the population in hardy-weinberg equilibrium? solve for p and q.

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saadhussain514
There are two equations necessary to solve a Hardy-Weinberg Equilibrium question:               p + q = 1

P is the frequency of the dominant allele.q is the frequency of the recessive allele.
P² is the frequency of individuals with the homozygous dominant genotype.
2pq is the frequency of individuals with the heterozygous genotype.
The q² is the frequency of individuals with the homozygous recessive genotype.

Given Data: 
Green individuals = 1092
Blue individuals    = 108

The frequency of individuals = individuals/total population

The frequency of Green individuals = 1092/1200

The frequency of Green individuals = 0.91

                                                        q² = 0.91
                                                     √ q² = √0.91
                                                          q = 0.953

Use the first Hardy-Weinberg equation:
                                                                 p + q        = 1
                                                                 p + 0.953 = 1
                                                                 p               = 1 - 0.953
                                                                 p               = 0.046
obasola1

Answer:

p=0.7

q=0.3

the recessive allele frequency () has changed, the population is not

Hardy-Weinberg Equilibrium.

Explanation:

+ = 1

² + 2 + ² = 1

Two relations involved in hardy-weinberg equilibrium

is the frequency of the dominant allele.

is the frequency of the recessive allele.

² is the frequency of individuals with the homozygous dominant genotype.

2 is the frequency of individuals with the heterozygous genotype.

² is the frequency of individuals with the homozygous recessive genotype.

Frequency of individuals with the homozygous dominant genotype

the frequency of individuals with the homozygous recessive genotype

²=108/1200

²=0.09

q=0.3

+ = 1 , recall

p+0.3=1

p=0.7

the recessive allele frequency () has changed, the population is not

Hardy-Weinberg Equilibrium.

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