Two heterozygous individuals having the genotype AaBb are crossed with...
When AaBb and AaBb are crossed with each other, the probability of getting a genotype Aa will be ½. Same way the probability of getting genotype Bb will be ½.
Using the product rule, to get an individual with both Aa and Bb, we will multiply the individual probabilities. This gives us ¼ as the probability of getting a baby with genotype AaBb.
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Two heterozygous individuals having the genotype AaBb are crossed with...
The genotype of an individual refers to the combination of alleles (different forms of a gene) that they possess for a particular trait. In this case, the individuals have the genotype AaBb, which means they carry one copy of the A allele and one copy of the a allele, as well as one copy of the B allele and one copy of the b allele.
When two heterozygous individuals with the genotype AaBb are crossed, the offspring have the potential to inherit any combination of alleles from their parents. To determine the probability of a specific genotype, we need to consider the possible combinations of alleles that can be passed on.
- Possible allele combinations from the mother: Aa or a
- Possible allele combinations from the father: Bb or b
Now, let's consider all the possible combinations of alleles that can result in the genotype AaBb:
- AaBb: This can be achieved by inheriting the Aa allele combination from the mother and the Bb allele combination from the father.
The probability of inheriting a particular allele combination from each parent is 1/2, as there are two possible alleles for each gene and the parents are heterozygous. Therefore, the probability of inheriting the Aa allele combination from the mother is 1/2, and the probability of inheriting the Bb allele combination from the father is also 1/2.
To calculate the probability of obtaining the genotype AaBb, we multiply the individual probabilities together:
Probability of inheriting Aa allele combination from the mother = 1/2
Probability of inheriting Bb allele combination from the father = 1/2
Probability of obtaining genotype AaBb = Probability of inheriting Aa allele combination from the mother * Probability of inheriting Bb allele combination from the father
= (1/2) * (1/2)
= 1/4
= 0.25
Therefore, the probability of having a baby with the genotype AaBb is 0.25, or 25%.