There are 100 students in a class Ninetysix did well in the

There are 100 students in a class. Ninety-six did well in the course whereas four blew it totally and received a grade of F. Sorry. In the highly unlikely event that these traits are genetic rather than environmental, if these traits involve dominant and recessive alleles, and if the four (4%) represent the frequency of the homozygous recessive condition, please calculate the following: a. The frequency of the recessive allele. b. The frequency of the dominant allele. c. The frequency of heterozygous individuals. Within a population of butterflies, the color brown (B) is dominant over the color white (b). And 40% of all butterflies are white. Given this simple information, which is something that is very likely to be on an exam, calculate the following: a. The percentage of butterflies in the population that are heterozygous. b. The frequency of homozygous dominant individuals. A rather large population of Chipucabras have 396 red-sided individuals and 557 tan-sided individuals. Assume that red is totally recessive. Please calculate the following: a. The allele frequencies of each allele. b. The expected genotype frequencies. c. The number of heterozygous individuals that you would predict to be in this population. d. The expected phenotype frequencies. e. Conditions happen to be really good this year for breeding and next year there are 1, 245 young Chipucabras. Assuming that all of the Hardy-Weinberg conditions are met, how many of these would you expect to be red-sided and how many tan-sided? A very large population of randomly-mating laboratory mice contains 35% white mice. White coloring is caused by the double recessive genotype, \"aa\" a. Calculate allelic and genotypic frequencies for this population. After graduation from Chupacabra academy, 20 of the beautiful creatures (lets say 10 males and 10 females) charter a plane to go on a round-the-world tour. Unfortunately, they all crash land (safely) on a deserted island. No one finds them and they have to start a new population totally isolated from the rest of the world. Two of them carry (i.e. are heterozygous for) the recessive cystic fibrosis allele (c). a. Assuming that the frequency of this allele does not change as the population grows, what will be the incidence of cystic fibrosis on the island?

Solution

You can solve the given problems using Hardy-Weinberg equaltion.

Hardy -Weinberg equilibrium states that in a population the frequency of alleles and genotype will remian constant from generation to generation until there is no other evolutionary influence.

Hardy - Weinberg euqation is a mathematical expression introduced by Hardy and Weinberg, that describes the basic principle of population genetics.

Hardy- Weinberg equation is

p² + 2pq + q² = 1

where p is the frequency of dominant allele,

q is the frequency of recessive allele.

  pq refers to the heterozygous alleles.

If you consider a single locus with two alleles denoted by A for dominant allele and a for recessive allele, then

In the above equation, p² represents the percentage of homozygous dominant alleles (AA)

q² represents the percentage of homozygous recessive alleles (aa)

2pq represents the percentage of heterozygous genotype (Aa)

In any population, the sum of allele frequencies of all the alleles at a locus must be equal to 1.

That means p + q = 1

Thus, if the allele frequencies of p and q are known, then the allele frequencies of all the genotypes can be calculated using the above equation.

If the above concept is clearly understood, it is easy to calculate the given problems.

Problem 1:

a. It is given that the homozygous recessive allele (aa) is 4%.

That is q² is 4% which is equal to 0.04

  q² = 0.04. Therefore q = 0.2

Frequency of the recessive allele = 0.2

b. Calculate the frequency of homozygous dominant allele (AA) using the above formula p + q = 1

p + 0.2 = 1

p = 1 - 0.2

= 0.8

Frequency of the dominant allele = 0.8

c. The frequency of heterozygous individuals can be calculated by incorporating obtained values in 2pq.

The allele frequency of dominant allele is 0.8 and frequency of the recessive allele is 0.2   

Substitute the values of p and q

2pq = 2 ( 0.2 X 0.8)

= 0.32

the heterozygous allele frequency is 0.32.

Thus the frequency of heterozygous individuals is 32 %.

Problem 2:

It is given that white color is recessive and thus bb is 40%.

bb is nothing but q²

Calculate q value from this.

  q² = 40%

= 0.4

When q² is 0.4 q is equal to 0.632 (since 0.632 X 0.632 = 0.4)

q = 0.632

We know the formula p + q = 1

Therefore p + 0.632 = 1

p = 1 - 0.632

= 0.368

Now using p and q values , you can calculate answers for a and b of the question.

a. The % of butterflies that are heterozygous = 2pq

2 ( 0.368) ( 0.632 ) = 0.4651

b. The frequency of homozygous dominant individuals = p²

   0.368 X 0.368 = 0.1354

Problem 3:

It is given that in the population tan sided are dominant and red sided are recessive.

Given that individuals with dominant character = 557 and individuals with recessive character = 396

Total population is 557 + 396 = 953

a. The recessive individuas = 396 / 953 = 0.4155

That means q² = 0.416

Therefore q = 0.645 ( since 0.645 X 0.645 = 0.416 )

As we know, p + q = 1

p + 0.645 = 1

p = 1 - 0.645

= 0.355

Thus allele frequency of each allele is 0.355 and 0.645

b. Expected genotype frequencies

dominant homozygous (AA) = p²

(0.355 )² = 0.126

recessive homozygous (aa) = q²

   ( 0. 645 ) ² = 0.416

heterozygous (Aa) frequency = 2 (pq )

= 2 (0.355 ) ( 0.645 )

= 0.458

c. The number of heterozygous individuals = 0.458 X 953

436.47

That means about 436 individuals.

d. The expected phenotype frequencies:

the frequency with dominant phenotype = 0.126 + 0.458 which is equal to 0.584

the frequency with recessive phenotype = 0.416 which was calculated already in section a of this question.

e. Under the above given conditions, breeding results in 1,245 individuals.

If the Hardy- Weinberg conitions were met, then the dominant tan sided individuals are 1245 X 0.584 = 727.08

That means 727 individuals are tan sided.

Remaining are the recessive red sided individuals . 1245 - 727 = 518.

Problem 4:

Double recessive (aa) mice are 35 % which means q² is 0.35

For knowing the allele frequencies, calculate q and p values;

q² = 0.35 Thus q = 0.59.

using the formula p + q = 1, calculate p value

p = 1 - 0.59 = 0.41

Now calculate the genotype frequencies.

p² = (0.41 X 0.41)= 0.17

q² = (0.59 X 0.59) = 0.35

2pq = 2 (0.41 ) ( 0.59 ) = 0.48

You can cross check the Hardy- Weinberg equation

p² + 2pq + q² = 1

The some of the obtained values must be 1. 0.17 + 0.48 + 0.35 = 1 . Hence the equation is checked.

Problem 5:

It is given that conditions, there are 20 individuals in the island, which means 40 alleles.

out of the 40 alleles, 2 alleles carry cystic fibrosis allele

you need to calculate the frequency of cystic fibrosis allele first.

frequency of cystic fibrosis allele =  2/ 40 = 0.05

That means 5% of alleles are of cystic fibrosis which is a recessive allele.

Hence the incidence of cystic fibrosis in the island will be (0.05)²

which is equal to 0.0025. This indicates that 0.25 % individuals will be born with cystic fibrosis.

The incidence of cystic fibrosis is 0.25% in the given island.