In a population where 8% of the men are red-green colour blind, what are the expected proportions of genotypes for the males and females? (Assume the trait does not affect survival or fertility.)
can anyone help me with this? Im stuck :(
Without knowing the proportion of female carriers in the population (heterozygous for colorblindness), I wouldn't know how to calculate the proportions either.
Sorry that I can't help more. Thanks for asking.
Sure, I can help you with this. To determine the expected proportions of genotypes for males and females in a population where 8% of the men are red-green color blind, we first need to understand the inheritance pattern of this trait.
Assuming the trait follows a simple recessive pattern, where the color blindness allele (c) is recessive to the normal allele (C), we can use the Hardy-Weinberg equation to calculate the expected proportions.
The Hardy-Weinberg equation is:
p^2 + 2pq + q^2 = 1
Where:
p represents the frequency of the dominant allele (C)
q represents the frequency of the recessive allele (c)
p^2 represents the expected proportion of individuals with the CC genotype (normal vision)
2pq represents the expected proportion of individuals with the Cc genotype (carriers of color blindness)
q^2 represents the expected proportion of individuals with the cc genotype (color blind)
Since 8% of the men are color blind (cc), we know that q^2 = 0.08. Therefore, q (the frequency of the recessive allele) can be calculated as the square root of 0.08.
q = √(0.08) ≈ 0.28
Now, to calculate the frequency of the dominant allele (p), we can subtract q from 1, since p + q = 1.
p = 1 - q
p = 1 - 0.28
p ≈ 0.72
Using p and q, we can calculate the expected proportions of genotypes for males and females separately.
For males:
- The proportion of CC genotype (normal vision) is p^2 = (0.72)^2
- The proportion of Cc genotype (carriers of color blindness) is 2pq = 2(0.72)(0.28)
- The proportion of cc genotype (color blind) is q^2 = (0.28)^2
For females, assuming random mating:
- The proportion of CC genotype (normal vision) is p^2 = (0.72)^2
- The proportion of Cc genotype (carriers of color blindness) is also 2pq = 2(0.72)(0.28)
- The proportion of cc genotype (color blind) remains q^2 = (0.28)^2
Keep in mind that these proportions are expected values in a large population and may not perfectly reflect the actual observations in individual families or small samples.
Of course, I can help you with that!
To determine the expected proportions of genotypes for males and females with respect to red-green color blindness, we need to consider the inheritance pattern of the trait. Color blindness is generally an X-linked recessive trait, meaning it is carried on the X chromosome.
Let's break down this problem step by step:
Step 1: Determine the genotypes for males:
- Since males have one X chromosome and one Y chromosome, they can only inherit one allele for each X-linked trait.
- Assuming the trait doesn't affect survival or fertility, red-green color blindness is caused by a mutation in the gene located on the X chromosome. Let's represent the normal allele as "X" and the mutant allele as "x".
- The possible genotypes for males are:
- Normal: X^mY (not color blind)
- Color blind: X^mX^m (homozygous for the mutant allele, where X^m represents the normal X chromosome)
Step 2: Determine the genotypes for females:
- Females have two X chromosomes, allowing for multiple combinations of alleles.
- Since the color blindness trait is recessive, females need to inherit two copies of the mutant allele (xx) to be color blind. If they only receive one copy, they will be carriers (X^mX).
- The possible genotypes for females are:
- Normal: X^mX^m (homozygous for the normal allele)
- Carrier: X^mX (heterozygous, carrying one copy of the mutant allele)
- Color blind: XX (homozygous for the mutant allele)
Step 3: Calculate the expected proportions:
- We are given that 8% of the men are red-green color blind. This means that 8% of the men have the genotype X^mX^m and the remaining 92% have the genotype X^mY.
- Since the trait does not affect survival or fertility, we assume Hardy-Weinberg equilibrium. In this case, the frequency of the mutant allele (x) can be calculated using the formula: q = sqrt(p), where p is the frequency of the normal allele (X).
- Given that X^m represents the normal allele, and assuming X^M represents the mutant allele, the frequency of the normal allele (p) is 1 - 0.08 (since 8% are color blind).
- Thus, p = 0.92, and q = sqrt(0.92), approximately 0.959.
- Using the Hardy-Weinberg equation, we can calculate the frequencies of different genotypes for females:
- Normal (X^mX^m): p^2 = (0.92)^2 = 0.8464 (84.64%)
- Carrier (X^mX): 2pq = 2 * 0.92 * 0.959 = 1.7752 (approx. 17.75%)
- Color blind (XX): q^2 = (0.959)^2 = 0.9196 (approx. 9.2%)
- For males, since the trait is X-linked, the proportion of color blindness is the same as the frequency of the mutant allele: 0.08 or 8%, as given.
To summarize:
- For males: Approximately 8% are color blind (X^mX^m), and the rest (92%) are not color blind (X^mY).
- For females: Approximately 84.64% are normal (X^mX^m), 17.75% are carriers (X^mX), and 9.2% are color blind (XX).
These calculations assume Hardy-Weinberg equilibrium and that there is no selection, mutation, migration, or genetic drift influencing the gene frequencies in the population.