A heterozygous brown mouse is crossed with a homozygous white mouse. What are the genotypes and phenotypes of the offspring? If these mice had 3 offspring, what is the probability that all 3 would be white?
To determine the genotypes and phenotypes of the offspring, we need to understand the inheritance pattern of the brown and white fur color.
In this case, the fur color is determined by a single gene with two possible alleles: B for brown and b for white. The mouse that is heterozygous (Bb) carries one brown allele and one white allele, while the homozygous mouse (bb) carries two copies of the white allele.
Now let's determine the possible genotypes and phenotypes of the offspring using a Punnett square:
Brown mouse: Bb
White mouse: bb
| B | b |
-----|------|------|
B | BB | Bb |
b | Bb | bb |
In the Punnett square, the top row represents the possible alleles from the brown mouse, and the left column represents the possible alleles from the white mouse.
From the Punnett square, we can determine that there will be two possible genotypes: BB and Bb. The corresponding phenotypes are brown and white, respectively.
Now, let's calculate the probability that all three offspring will be white, assuming random segregation and independent assortment of alleles:
Probability of white offspring = Probability of white allele (b) from the white parent raised to the power of the number of offspring
Since the white parent is homozygous (bb), the probability of passing on the white allele to each offspring is 1 (100%).
Therefore, the probability that all three offspring will be white is:
Probability = (1/2)^3 = 1/8 or 0.125 (12.5%)
So, there is a 12.5% chance that all three offspring will be white.
To determine the genotypes and phenotypes of the offspring, we need to understand the principles of Mendelian genetics and the inheritance of coat color in mice.
In this scenario, the brown coat color is most likely dominant, and the white coat color is recessive. Therefore, the genotype for the heterozygous brown mouse (Bb) would consist of one dominant allele (B) for brown color and one recessive allele (b) for white color. The homozygous white mouse would have two recessive alleles (bb) for white color.
Now, let's consider the offspring. Each parent will contribute one allele to each offspring.
When the heterozygous brown mouse (Bb) is crossed with the homozygous white mouse (bb), there are two possible outcomes for each offspring:
1. The offspring can receive the dominant brown allele (B) from the heterozygous brown mouse and the recessive white allele (b) from the homozygous white mouse, resulting in a genotype of Bb. This genotype would correspond to a brown phenotype, as the brown allele is dominant.
2. The offspring can receive the recessive white allele (b) from both parents, resulting in a genotype of bb. This genotype would correspond to a white phenotype, as both alleles are recessive.
To calculate the probability that all three offspring would be white, we need to determine the probability of getting a white offspring with each individual mating event and then multiply those probabilities together.
Since the white mouse is homozygous recessive (bb), the only possible genotype it can contribute is a recessive white allele (b). Therefore, for each individual mating event, the probability of getting a white offspring is 1 (100%).
To calculate the probability of all three offspring being white, we multiply the probabilities together: 1 * 1 * 1 = 1.
Therefore, the probability that all three offspring would be white is 100%.
bw x ww
bw, bw, ww, ww
Pr(3 ww)= 1/2*1/2*1/2=1/8