A recessive mutation causes short sightedness (ss) in cats. The frequency of homzygous wild type (+/+), heterozygous (+/s) and homzygous recessive (ss/ss) individuals was assessed in two populations of cats. The data is shown in the chart below. a. Are the two populations in Hardy-Weinberg equilibrium? Explain. b. If one population is not, what might cause this deviation?

b. The Population of City cats is not in Hardy-Weinberg equilibrium because the proportion of dominant and recessive genes have been forced to change by outside force of migration, selection and mutation

Explanation:

By the Hardy-Weinberg Law, we have;

p² + 2·p·q + q² = 1

p + q = 1

p = Dominant allele frequency within the population

q = Recessive allele frequency within the population

p² = Percentage of individuals that are h o m o zygous dominant

q² = Percentage of individuals that are h o m o zygous recessive

2· p · q = Heterozygous individuals percentage

Therefore, for the country cats, we have;

p² = 0.49

p = √0.49 = 0.7

q² = 0.09

q = √0.09 = 0.3

2×p×q = 2×0.7×0.3 = 0.42 = The value given in the table

Which shows that the country cats are in Hardy-Weinberg equilibrium

For the city cats, we have;

p² = 0.52

p = √0.52 = √(13)/5

q² = 0.03

q = √0.03 = √3/10

2×p×q = 2×√(13)/5×√3/10 = √(39)/25 ≠ 0.45 which is the value given in the table

Which shows that the City cats is not in Hardy-Weinberg equilibrium

Therefore, only one of the two populations is in Hardy-Weinberg equilibrium

b. The Population of City cats is not in Hardy-Weinberg equilibrium because the proportion of dominant and recessive genes have been forced to change by outside force of migration, selection and mutation.

marianaegarciaperedo marianaegarciaperedo · Answer 2 · 2022-01-25T15:50:44+01:00

Following the Hardy-Weinberg theory, we can say that the country cats population is in equilibrium, while the City cats population is not. This is probably due to the action of evolutive forces.

Hardy-Weinberg equilibrium

The Hardy-Weinberg equilibrium theory states that the allelic frequencies in a locus are represented as p and q. Assuming a diallelic gene,

p is the frequency of the dominant allele,

q is the frequency of the recessive allele.

The genotypic frequencies after one generation are

p² (H0m0zyg0us dominant genotypic frequency),

2pq (Heter0zyg0us genotypic frequency),

q² (H0m0zyg0us recessive genotypic frequency).

If a population is in H-W equilibrium, it gets the same allelic and genotypic frequencies generation after generation.

The addition of the allelic frequencies equals 1

p + q = 1.

The sum of genotypic frequencies equals 1

p² + 2pq + q² = 1

Solving the problem

We are given the genotypic frequencies of each population.

Now, we need to say if these populations are in H-W equilibrium, and if not, why?

So let us analyze each population separately.

Country Cats

We have the following genotypic frequencies, F(xx).

F(+/+) = p² = 0.49
F(+/ss) = 2pq = 0.42
F (ss/ss) = q² = 0.09

Now, we need to get the allelic frequencies, p and q.

To do it, we will add half the heter0zyg0us frequency to each of the h0m0zyg0us frequencies.

p = 0.49 + (0.42/2) = 0.49 + 0.21 = 0.7

p = 0.7 ⇒ Frequency of the dominant allele

q = 0.09 + (0.42/2) = 0.09 + 0.21 = 0.3

q = 0.3 ⇒ Frequency of the recessive allele

Now, if this population is in H-W equilibrium, it will get the same allelic and genotypic frequencies generations after generation.

So now, we will get the genotypic frequencies (p², 2pq, and q²) of the following generation.

If p = 0.7 ⇒ p² = 0.7² = 0.49

If q = 0.3 ⇒ q² = 0.3² = 0.09

2pq = 2 x 0.7 x 0.3 = 0.42

The given genotypic frequencies in one generation are equal to the calculated genotypic frequencies in the next generation.

We can conclude this country's cats population is in Hardy-Weinberg equilibrium. No evolutive events are going on.

City Cats

We have the following genotypic frequencies, F(xx).

F(+/+) = p² = 0.52
F(+/ss) = 2pq = 0.45
F (ss/ss) = q² = 0.03

Now, we need to get the allelic frequencies by adding half the heter0zyg0us frequency to each of the h0m0zyg0us frequencies.

p = 0.52 + (0.45/2) = 0.52 + 0.225 = 0.745

p = 0.745 ⇒ Frequency of the dominant allele

q = 0.03 + (0.45/2) = 0.03 + 0.225 = 0.255

q = 0.255 ⇒ Frequency of the recessive allele

Now, if this population is in H-W equilibrium, it will get the same allelic and genotypic frequencies generations after generation.

So now, we will get the genotypic frequencies (p², 2pq, and q²) of the following generation.

If p = 0.745 ⇒ p² = 0.745² = 0.555

If q = 0.255 ⇒ q² = 0.255² = 0.065

2pq = 2 x 0.745 x 0.255 = 0.379

The given genotypic frequencies in one generation differ from the calculated genotypic frequencies in the next generation.

Generation 1 Generation 2

p² 0.52 0.555

2pq 0.45 0.379

q² 0.03 0.065

We can conclude that this city's cats population is not in Hardy-Weinberg equilibrium.

Probably some evolutive events are going on.

The evolutive forces affecting the H-W equilibrium are natural selection, genetic drift, and genetic flow. The mating system might also affect the equilibrium in this population.

You can learn more about H-W equilibrium at

https://brainly.com/question/3406634

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