Question

open response: fur color in a specific animal is controlled by a sex - linked, codominant trait, where some animals have black fur (b), some animals have orange fur (r), and some animals have brown - patterned fur (br). a female with brown - patterned fur mates with a male that has black fur and has a female offspring. what is the probability that the female offspring will have brown - patterned fur?

Explanation:

Step1: Determine Genotypes

The female has brown - patterned fur (BR), so her genotype is \(X^{BR}X^{?}\)? Wait, no, let's re - examine. Wait, the male has black fur and a female offspring. Wait, the male has black fur (B) and a female offspring. Since it's sex - linked (X - linked, we assume), the male's genotype for X - linked trait is \(X^{B}Y\) (because males have X and Y, and black fur is B). The female has brown - patterned fur (BR), so her genotype is \(X^{BR}X^{?}\)? Wait, no, the problem says "a female with brown - patterned fur (BR) mates with a male that has black fur and has a female offspring". Wait, let's clarify the inheritance. Since it's X - linked codominant, females have two X chromosomes, males have X and Y.

The female's genotype: brown - patterned (BR) so \(X^{BR}X^{?}\)? Wait, no, the brown - patterned is BR, so her genotype is \(X^{BR}X^{R}\)? No, wait, black (B), orange (R), brown - patterned (BR). So the female with brown - patterned fur has genotype \(X^{BR}X^{?}\)? Wait, no, let's think about the male. The male has black fur, so his X chromosome has B, so his genotype is \(X^{B}Y\). The female has brown - patterned fur (BR), so her genotype is \(X^{BR}X^{R}\)? No, wait, brown - patterned is BR, so she has one X with B and one with R, so \(X^{B}X^{R}\) (since BR is codominant, so B and R on X chromosomes for female). Wait, maybe I misread. Let's re - state:

Fur color is X - linked codominant. Black (B), orange (R), brown - patterned (BR). Female with brown - patterned fur: genotype \(X^{B}X^{R}\) (because BR is codominant, so she has B and R on her two X chromosomes). Male with black fur: his X chromosome has B, so genotype \(X^{B}Y\).

We are looking at the probability that their female offspring has brown - patterned fur (BR).

Step2: Determine Gametes

Female (\(X^{B}X^{R}\)) produces gametes: \(X^{B}\) and \(X^{R}\).

Male (\(X^{B}Y\)) produces gametes: \(X^{B}\) and \(Y\). But for female offspring, we only consider the male's \(X^{B}\) (since female offspring gets X from male and X from female).

Step3: Punnett Square for Female Offspring

The cross for female offspring: female gametes (\(X^{B}, X^{R}\)) and male gamete (\(X^{B}\)).

- If female gamete \(X^{B}\) and male gamete \(X^{B}\): offspring genotype \(X^{B}X^{B}\) (black fur).
- If female gamete \(X^{R}\) and male gamete \(X^{B}\): offspring genotype \(X^{B}X^{R}\) (brown - patterned fur).

So out of 2 possible genotypes for female offspring (\(X^{B}X^{B}\) and \(X^{B}X^{R}\)), 1 is brown - patterned (BR).

Wait, but wait, the female's genotype: is she \(X^{B}X^{R}\)? Let's confirm. Brown - patterned is BR, which is codominant, so having B and R on X chromosomes (for female) gives BR. The male is \(X^{B}Y\) (black fur). So when they have a female offspring, the female gets one X from male (\(X^{B}\)) and one X from female. The female (mother) has \(X^{B}\) and \(X^{R}\) (since BR is B and R). So the possible X from mother: \(X^{B}\) or \(X^{R}\).

So the possible genotypes for female offspring:

- \(X^{B}\) (from father) and \(X^{B}\) (from mother): \(X^{B}X^{B}\) (black)
- \(X^{B}\) (from father) and \(X^{R}\) (from mother): \(X^{B}X^{R}\) (brown - patterned)

So there are 2 possible genotypes, each with probability \(\frac{1}{2}\). So the probability that the female offspring has brown - patterned fur is \(\frac{1}{2}\) or 50%.

Wait, but let's check again. The mother's genotype: brown - patterned (BR) means she has B and R on her X chromosomes, so \(X^{B}X^{R}\). Father's genotype: black fur, so \(X^{B}Y\). For female offspring, we take X from father (\(X^{B}\)) and X from mother (either \(X^{B}\) or \(X^{R}\)). So the two possible female genotypes: \(X^{B}X^{B}\) (black) and \(X^{B}X^{R}\) (brown - patterned). So the probability of brown - patterned (BR) is \(\frac{1}{2}\).

Answer:

The probability that the female offspring will have brown - patterned fur is \(\frac{1}{2}\) (or 50%).