If the purple allele (p) is dominant, then the red phenotype must be associated with the recessive allele, which we can represent as q. Given that 76 out of 100 plants are red, this means that 24 are purple. Assuming Hardy-Weinberg equilibrium, the frequency of the red phenotype (q^2) is 0.76, so q is the square root of 0.76, which is approximately 0.87. Therefore, the frequency of the purple allele (p) is 1 - q, which is roughly 0.13.
In the population of 100 rabbits, 10 have white fur, indicating that they are homozygous recessive (aa). This means that the frequency of the recessive allele (a) can be calculated using the formula ( q^2 = \frac{10}{100} = 0.1 ), so ( q = \sqrt{0.1} \approx 0.316 ). The frequency of the dominant allele (A) is then ( p = 1 - q \approx 1 - 0.316 = 0.684 ). Thus, the allele frequency for the dominant allele is approximately 0.684.
When a homozygous dominant female (genotype AA) is crossed with a homozygous recessive male (genotype aa), all offspring will inherit one dominant allele from the mother and one recessive allele from the father, resulting in a genotype of Aa for all offspring. The phenotype ratio will show all offspring displaying the dominant trait. Thus, the genotype ratio is 100% Aa, and the phenotype ratio is 100% expressing the dominant trait.
If one parent is homozygous dominant (AA) and the other parent is homozygous recessive (aa), all offspring will inherit one dominant allele and display the dominant phenotype. Therefore, the probability of their offspring exhibiting the dominant phenotype is 100%.
The frequency of dominant which is smooth seed early is 0.556 or 6%.
In this population, 20 out of 100 rabbits have white fur, indicating they are homozygous recessive (bb). The frequency of the recessive genotype (bb) is 0.20. Using the Hardy-Weinberg principle (p² + 2pq + q² = 1), we know q² = 0.20, so q (the frequency of the recessive allele) is the square root of 0.20, which is approximately 0.447. Thus, the allele frequency for the recessive allele is about 0.447.
, 0.34 Apex
.77
A. 0.45 Apex
100% of the offspring will display the dominant trait because the homozygous dominant parent can only pass on the dominant allele. The offspring will inherit one dominant allele from the dominant parent and one recessive allele from the recessive parent, resulting in a heterozygous genotype expressing the dominant trait.
In the population of 100 rabbits, 10 have white fur, indicating that they are homozygous recessive (aa). This means that the frequency of the recessive allele (a) can be calculated using the formula ( q^2 = \frac{10}{100} = 0.1 ), so ( q = \sqrt{0.1} \approx 0.316 ). The frequency of the dominant allele (A) is then ( p = 1 - q \approx 1 - 0.316 = 0.684 ). Thus, the allele frequency for the dominant allele is approximately 0.684.
0.16
An example of allele frequency is when in a population of 100 individuals, 60 individuals have the dominant allele (A) for a specific gene, while 40 individuals have the recessive allele (a). The frequency of the dominant allele (A) would be 0.6, and the frequency of the recessive allele (a) would be 0.4.
To find q (the frequency of the green allele), use the Hardy-Weinberg equation: p^2 + 2pq + q^2 = 1. Given that 11% are green, q^2 = 0.11. Therefore, q = √0.11 ≈ 0.33.
When a homozygous dominant female (genotype AA) is crossed with a homozygous recessive male (genotype aa), all offspring will inherit one dominant allele from the mother and one recessive allele from the father, resulting in a genotype of Aa for all offspring. The phenotype ratio will show all offspring displaying the dominant trait. Thus, the genotype ratio is 100% Aa, and the phenotype ratio is 100% expressing the dominant trait.
If all of the children have freckles, that means that both parents had dominant genotypes. (Mother; FF and Father; FF). Or, one parent could have a hybrid genotype. (For example, Mother; Ff and Father; FF). Based on the outcome of a Punnett Square, either one parent must have a hybrid and the other dominant, or both must have dominant genotypes.
If one parent is homozygous dominant (AA) and the other parent is homozygous recessive (aa), all offspring will inherit one dominant allele and display the dominant phenotype. Therefore, the probability of their offspring exhibiting the dominant phenotype is 100%.
The frequency of dominant which is smooth seed early is 0.556 or 6%.