Brown eyes in humans are dominant to blue eyes. A brown-eyed man, whose mother was blue-eyed, marries a brown-eyed woman whose father had blue eyes. What is the probability that this couple will have a blue-eyed child?
the man and woman are both hybrid brown with a blue recessive
do a Punnett Square with Bb and Bb
Probability is 1/4 or 25%
To determine the probability of this couple having a blue-eyed child, we need to consider the genetic inheritance of eye color.
Since brown eyes are dominant and blue eyes are recessive, we can represent the brown-eyed gene as "B" and the blue-eyed gene as "b."
Let's analyze the genetic makeup of the parents first:
- The brown-eyed man has two possible combinations: BB (homozygous dominant) or Bb (heterozygous).
- The blue-eyed mother has the genotype bb (homozygous recessive).
Now, we can determine the possible genotypes of the offspring by combining the parental alleles:
- For the brown-eyed man (BB or Bb):
- BB crossed with bb would result in all Bb (brown-eyed) offspring.
- Bb crossed with bb would result in 50% Bb (brown-eyed) and 50% bb (blue-eyed) offspring.
- For the blue-eyed mother (bb):
- bb crossed with BB would result in all Bb (brown-eyed) offspring.
- bb crossed with Bb would result in 50% Bb (brown-eyed) and 50% bb (blue-eyed) offspring.
Considering both possibilities, the chance of having a blue-eyed child is 50% for each combination. Therefore, the overall probability that this couple will have a blue-eyed child is 50%.
To determine the probability of the couple having a blue-eyed child, we need to understand the inheritance pattern of eye color. In this case, brown eyes are considered dominant, while blue eyes are recessive.
We can use Punnett squares to represent the possible combinations of genes from the parents and predict the probability of their offspring having certain eye colors.
Let's label the brown eye gene as "B" (dominant) and the blue eye gene as "b" (recessive).
The man's genotype is unknown, but since he has brown eyes, we can assume he is either homozygous BB or heterozygous Bb.
The mother's genotype is also unknown, but we know she has blue eyes, so she must be homozygous bb.
To determine the probabilities, let's consider both possibilities for the man's genotype:
1. If the man has a homozygous genotype (BB):
- The man's genotype: BB
- The woman's genotype: bb
In this case, all their offspring will be heterozygous (Bb) and have brown eyes. So the probability of having a blue-eyed child is zero.
2. If the man has a heterozygous genotype (Bb):
- The man's genotype: Bb
- The woman's genotype: bb
In this case, there is a 50% chance of passing on the "b" gene from the mother (bb) to the child, resulting in a blue-eyed child.
Therefore, the probability of the couple having a blue-eyed child is 50%.
It's important to note that these probabilities are based on the assumption that eye color is solely determined by these two genes and follows simple Mendelian inheritance. In reality, eye color is influenced by multiple genes and can be more complex.