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A hypothetical population of 200 cats has two alleles, TL and TS, for a locus that codes for tail length. The table below describes the phenotypes of cats with each possible genotype, as well as the number of individuals in the population with each genotype. Which statements about the population are true?

Genotype Phenotype(tail length) Number of individuals
TLTL long 60
TLTS medium 40
TSTS short 100

Select the five statements that are true.
Heterozygotes make up 20% of the population.
Homozygotes make up 80% of the population.
Homozygotes make up 30% of the population.
In the entire cat population, 60% of the alleles are TS.
In the entire cat population, the frequency of the TS allele is 0.5.
In the entire cat population, the frequency of the TL allele is 0.4.
Assuming random mating, each gamete has a 50% chance of having a TL allele and a 50% chance of having a TS allele.
Assuming random mating, each gamete has a 40% chance of having a TL allele and a 60% chance of having a TS allele.
Heterozygotes make up 20% of the population.
Homozygotes make up 80% of the population.
In the entire cat population, 60% of the alleles are TS.
In the entire cat population, the frequency of the TL allele is 0.4.
Assuming random mating, each gamete has a 40% chance of having a TL allele and a 60% chance of having a TS allele.

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Answer:

Heterozygotes make up 20% of the population.

Homozygotes make up 80% of the population.

Homozygotes make up 30% of the population.

In the entire cat population, the frequency of the TS allele is 0.5.

Assuming random mating, each gamete has a 50% chance of having a TL allele and a 50% chance of having a TS allele.

Explanation:

The heterozygote species are forty in number. If a percentage is evaluated for heterozygote species out of the total species it would be

[tex]\frac{40}{200} * 100\\= 20[/tex]%

Hence statement A is correct

Homozygous species expressed in terms of percentage with respect to total number of species is equal to

[tex]\frac{160}{200} * 100\\= 80[/tex]%

Hence, statement B is correct

Statement C is incorrect

Frequency of recessive allele is equal to

[tex]\sqrt{\frac{100}{200} } \\\sqrt{0.5} \\= 70.7[/tex] %

Hence option D is incorrect

Frequency of TS allele population is equal to 0.5

Assuming independent assortment, there are 50% chances of expression of each allele.

Hence statement "Assuming random mating, each gamete has a 50% chance of having a TL allele and a 50% chance of having a TS allele." is also correct

TLTL homozygotes make up 30% of the population .

[tex]\frac{60}{200} * 100\\= 30[/tex]%

Histrionicus

The statements about the population are true -

Assuming random mating, each gamete has a 40% chance of having a TL allele and a 60% chance of having a TS allele.

In the entire cat population, 60% of the alleles are TS.

In the entire cat population, the frequency of the TL allele is 0.4.

Heterozygotes make up 20% of the population  

- Hozygotes make up 80% of the population

Given:

Genotype     Phenotype  (tail length)      Number of individuals

TLTL               long                                           60

TLTS             medium                                      40

TSTS               short                                        100

From hardy weinberg equilibrium -

Genotype frequency:

long tail (TLTL) - [tex]\frac{60}{200}[/tex] = 0.3

short tail - [tex]\frac{100}{200}[/tex] = 0.5

medium tail - [tex]\frac{40}{200}[/tex] = 0.2

Allele frequency

dominant: [tex]\frac{120+40}{400}[/tex]= [tex]\frac{160}{400}[/tex] = 0.4

recessive: [tex]\frac{200+40}{400}[/tex] = [tex]\frac{240}{400}[/tex] = 0.6

so we can say the following on the given above allele frequency and genotype:

Assuming random mating, each gamete has a 40% chance of having a TL allele and a 60% chance of having a TS allele.

In the entire cat population, 60% of the alleles are TS.

In the entire cat population, the frequency of the TL allele is 0.4.

Heterozygotes make up 20% of the population  

- Hozygotes make up 80% of the population

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