In humans, the allele for brown eye color (B) is dominant to blue eye color (b). If two heterozygous parents produce seven children, what are the following probabilities?
The first three children have blue eyes.
The first four children have brown eyes, the last three have blue eyes.
The first child has blue eyes, and the rest have brown eyes
To solve these types of genetics problems, it's helpful to use a Punnett square to determine the possible combinations of alleles that the offspring can inherit from their parents.
First, let's write out the genotypes of the parents:
Parent 1: Bb (heterozygous for brown eyes)
Parent 2: Bb (heterozygous for brown eyes)
Now, we can use a Punnett square to determine the possible genotypes and phenotypes of the offspring:
| | B | b |
|---|---|---|
| B | BB | Bb |
| b | Bb | bb |
We can see that there are four possible genotypes for the offspring: BB, Bb, Bb, and bb.
Now, let's use these genotypes to answer the probabilities:
1. The first three children have blue eyes.
To have blue eyes, a child must inherit two recessive alleles (bb). We can see from the Punnett square that there is a 25% chance of a child being bb.
The probability of the first child being bb is therefore 0.25.
The probability of the second child being bb is also 0.25.
The probability of the third child being bb is also 0.25.
To find the probability of all three events occurring together (i.e. the first three children having blue eyes), we multiply the probabilities:
0.25 x 0.25 x 0.25 = 0.015625
Therefore, the probability of the first three children having blue eyes is 0.015625, or approximately 1.6%.
2. The first four children have brown eyes, the last three have blue eyes.
To have brown eyes, a child must inherit at least one dominant allele (B). We can see from the Punnett square that there is a 75% chance of a child having at least one B allele.
The probability of the first child having at least one B allele is therefore 0.75.
The probability of the second child having at least one B allele is also 0.75.
The probability of the third child having at least one B allele is also 0.75.
The probability of the fourth child having at least one B allele is also 0.75.
The probability of the fifth child having blue eyes (i.e. being bb) is 0.25.
The probability of the sixth child having blue eyes is also 0.25.
The probability of the seventh child having blue eyes is also 0.25.
To find the probability of all seven events occurring in the specified order, we multiply the probabilities:
0.75 x 0.75 x 0.75 x 0.75 x 0.25 x 0.25 x 0.25 = 0.01171875
Therefore, the probability of the first four children having brown eyes and the last three having blue eyes is 0.01171875, or approximately 1.2%.
3. The first child has blue eyes, and the rest have brown eyes.
To have blue eyes, a child must inherit two recessive alleles (bb). We can see from the Punnett square that there is a 25% chance of this occurring.
The probability of the first child having blue eyes is therefore 0.25.
The probability of the second child having at least one B allele is 0.75.
The probability of the third child having at least one B allele is also 0.75.
The probability of the fourth child having at least one B allele is also 0.75.
The probability of the fifth child having at least one B allele is also 0.75.
The probability of the sixth child having at least one B allele is also 0.75.
The probability of the seventh child having at least one B allele is also 0.75.
To find the probability of all seven events occurring in the specified order, we multiply the probabilities:
0.25 x 0.75 x 0.75 x 0.75 x 0.75 x 0.75 x 0.75 = 0.05493164
Therefore, the probability of the first child having blue eyes and the rest having brown eyes is 0.05493164, or approximately 5.5%.