Please answer the following questions on The Hardy-Weinberg Principle
1. A population of 400 individuals, 289 of which are recessive (aa) for a particular trait. Determine the frequency of the dominant and recessive allele in this population.
2. A plane carrying 20 individuals crashes on a deserted island. Although everyone survives, they have no way of communicating with the rest of the world, and help never comes. Eventually, they decide to start a new civilization.
A) If two of the individuals are affected by cystic fibrosis (a recessive genetic condition, aa), what will be the incidence of carriers on the island (Aa) assuming that the frequency of this allele does not change as the population grows?
B) Cystic fibrosis is a known lethal disorder, with affected rarely surviving to reproduce. It is likely that allele frequencies will remain constant throughout subsequent generations, as assumed in part a. Why or why not?
3. The gene pool of a population of fruit flies contains only two eye-colour alleles; the dominant red allele (R), and the recessive purple allele (r). Only 1% of the population has red eyes. Determine the allele and genotype frequencies of this population.
4. Manx cats have no tails and large hind legs. The no-tail traits results from a heterozygous genotype, Tt. Interestingly, TT genotypes are normal cats, while the tt genotype is lethal. In a population of 1000 cats, only 1% are Manx and 99% are normal.
a) What are the allele frequencies in this population?
b) Determine the expected frequency of each genotype in the next generation if there is no selection against any of the alleles.
c) Do you think the expected frequency you calculated in part b will match the observed frequency in the next generation? Explain.
1. To determine the frequency of the dominant and recessive allele, we need to know the total number of alleles in the population. Since each individual has two alleles, the total number of alleles is 2 times the population size (400 individuals x 2 alleles = 800 alleles).
The frequency of the recessive allele (q) can be calculated by dividing the number of recessive alleles (289) by the total number of alleles (800).
q = 289/800 = 0.36125
The frequency of the dominant allele (p) can be determined by subtracting the recessive allele frequency from 1.
p = 1 - q = 1 - 0.36125 = 0.63875
Therefore, the frequency of the dominant allele is approximately 0.63875, and the frequency of the recessive allele is approximately 0.36125.
2.
A) If two individuals are affected by cystic fibrosis (aa), we know that the frequency of aa in the population is 2/20 since there are 2 individuals affected out of 20. Assuming that the frequency of the allele for cystic fibrosis (a) does not change as the population grows, we can calculate the carrier frequency (Aa) using the Hardy-Weinberg equation:
p² + 2pq + q² = 1
where p is the frequency of the dominant allele (A), q is the frequency of the recessive allele (a), and pq represents the frequency of heterozygotes (Aa).
We can assume that the frequency of the dominant allele (A) is 1-q, since only two alleles are considered in this scenario.
p(A) = 1-q = 1 - (2/20) = 0.9
Given that pq = 2pq, the carrier frequency (Aa) can be calculated as:
2pq = 2 x (0.9) x (2/20) = 0.36
Therefore, the incidence of carriers (Aa) on the island will be approximately 0.36 or 36%.
B) Cystic fibrosis is a known lethal disorder, with affected individuals rarely surviving to reproduce. This means that they cannot contribute their alleles to future generations. Without any selection against the alleles, the Hardy-Weinberg principle assumes random mating and no evolutionary forces acting on the population.
Therefore, the allele frequencies will remain constant if the assumptions of the Hardy-Weinberg principle hold true, including no selection, no mutation, no migration, large population size, and random mating. However, in real populations, selection against a lethal disorder like cystic fibrosis will lead to a decrease in the frequency of the recessive allele (a), as affected individuals are unable to reproduce and pass on the allele.
3. If only 1% of the fruit fly population has red eyes, we can assume that this is the frequency of the homozygous dominant genotype (RR) since the red eye allele (R) is dominant. Let's assume that q represents the frequency of the recessive allele (r).
Given that RR = 0.01, we can calculate the frequency of the recessive allele as:
q² = 0.01
q = √0.01 ≈ 0.1
Since p + q = 1, the frequency of the dominant allele can be determined as:
p = 1 - q = 1 - 0.1 = 0.9
Therefore, the allele frequencies in this fruit fly population are approximately 0.9 for the dominant allele (R) and 0.1 for the recessive allele (r).
4.
a) In a population of 1000 cats, since only 1% are Manx and 99% are normal, we can assume that q² represents the frequency of Manx cats (tt) and p² represents the frequency of normal cats (TT). Let's assume that q represents the frequency of the allele for the tailless trait (t).
Given that q² = 0.01, we can calculate the frequency of the recessive allele as:
q = √0.01 ≈ 0.1
Since p + q = 1, the frequency of the dominant allele can be determined as:
p = 1 - q = 1 - 0.1 = 0.9
Therefore, the allele frequencies in this population are approximately 0.9 for the dominant allele (T) and 0.1 for the recessive allele (t).
b) If there is no selection against any of the alleles, we can use the Hardy-Weinberg equation to determine the expected frequency of each genotype in the next generation. The equation is:
p² + 2pq + q² = 1
Given that p = 0.9 and q = 0.1, we can substitute these values into the equation to calculate the expected frequencies.
Expected frequency of TT (normal cats): p² = (0.9)² = 0.81
Expected frequency of Tt (heterozygous Manx carriers): 2pq = 2 x (0.9) x (0.1) = 0.18
Expected frequency of tt (Manx cats): q² = (0.1)² = 0.01
Therefore, in the next generation, we would expect the frequencies of TT, Tt, and tt to be approximately 0.81, 0.18, and 0.01, respectively.
c) The expected frequency calculated in part b may not necessarily match the observed frequency in the next generation. This is because the Hardy-Weinberg principle assumes certain conditions, such as no selection, no mutation, no migration, large population size, and random mating. In reality, these assumptions may not be met, and various evolutionary forces can influence allele frequencies.
If conditions change, such as selection against Manx cats due to their reduced fitness or mating preferences for certain genotypes, the observed frequency can deviate from the expected frequency predicted by the Hardy-Weinberg equation.
1. To determine the frequency of the dominant and recessive allele in a population, we can use the Hardy-Weinberg equation:
p^2 + 2pq + q^2 = 1
Where p is the frequency of the dominant allele and q is the frequency of the recessive allele.
Given that the population has 400 individuals and 289 of them are recessive (aa) for the trait, we can calculate the frequency of the recessive allele:
q^2 = 289 / 400
q^2 = 0.7225
Taking the square root of both sides, we get:
q = 0.85
Since q represents the frequency of the recessive allele, the frequency of the dominant allele can be calculated using:
p = 1 - q
p = 1 - 0.85
p = 0.15
Therefore, the frequency of the dominant allele is 0.15 and the frequency of the recessive allele is 0.85.
2. A) Assuming Hardy-Weinberg equilibrium and that the frequency of the allele does not change, we can calculate the incidence of carriers (Aa) on the island. Since two individuals are affected by cystic fibrosis (aa) and the frequency of aa is q^2, we have:
q^2 = 2 / 20
q^2 = 0.1
Taking the square root of both sides, we get:
q = 0.316
The frequency of the recessive allele (q) is 0.316, so the frequency of the dominant allele (p) can be calculated:
p = 1 - q
p = 1 - 0.316
p = 0.684
The frequency of carriers (Aa) is given by 2pq:
2pq = 2 * 0.684 * 0.316
2pq = 0.432
Therefore, the incidence of carriers (Aa) on the island is 0.432 or 43.2%.
B) In a small population, random events such as genetic drift and the Founder Effect can cause allele frequencies to change over time. Additionally, the lethal nature of cystic fibrosis (aa) means that affected individuals will not survive to reproduce, which will decrease the frequency of the recessive allele in subsequent generations. Therefore, it is unlikely that allele frequencies will remain constant over time.
3. Given that red eye color is the dominant allele (R) and only 1% of the population has red eyes, we can calculate the allele frequencies and genotype frequencies.
Let p be the frequency of the dominant allele (R) and q be the frequency of the recessive allele (r).
Since only 1% of the population has red eyes (RR or Rr), we can denote the frequency of the RR genotype as p^2 and the frequency of the Rr genotype as 2pq.
Given that p^2 represents 1% of the population, we have:
p^2 = 0.01
Taking the square root of both sides, we get:
p = 0.1
Since p represents the frequency of the dominant allele (R), the frequency of the recessive allele (r) can be calculated using:
q = 1 - p
q = 1 - 0.1
q = 0.9
Therefore, the frequency of the dominant allele (R) is 0.1 and the frequency of the recessive allele (r) is 0.9. The genotype frequencies can be calculated as follows:
RR = p^2 = (0.1)^2 = 0.01 or 1%
Rr = 2pq = 2 * 0.1 * 0.9 = 0.18 or 18%
rr = q^2 = (0.9)^2 = 0.81 or 81%
4. a) To determine the allele frequencies in this population, we can use the same approach as in question 3. Let p be the frequency of the dominant allele (T) and q be the frequency of the recessive allele (t).
Since 1% of the population is Manx (tt), we can denote the frequency of the tt genotype as q^2.
Given that q^2 represents 1% of the population, we have:
q^2 = 0.01
Taking the square root of both sides, we get:
q = 0.1
Since q represents the frequency of the recessive allele (t), the frequency of the dominant allele (T) can be calculated using:
p = 1 - q
p = 1 - 0.1
p = 0.9
Therefore, the frequency of the dominant allele (T) is 0.9 and the frequency of the recessive allele (t) is 0.1.
b) If there is no selection against any of the alleles, the expected frequency of each genotype in the next generation can be calculated using the Hardy-Weinberg equation.
TT = p^2 = (0.9)^2 = 0.81 or 81%
Tt = 2pq = 2 * 0.9 * 0.1 = 0.18 or 18%
tt = q^2 = (0.1)^2 = 0.01 or 1%
c) The expected frequency calculated in part b may not match the observed frequency in the next generation for a few reasons. If there are any selective pressures or genetic drift, it could cause changes in allele frequencies that deviate from the expected values. Additionally, the small population size of 1000 cats may lead to random fluctuations in allele frequencies due to genetic drift, which can cause deviations from expectations.