Albinism is inherited as a simple recessive trait. If a woman and her husband are both heterozygous for albinism then what is the probability that their first five children will have these phenotypes? Child 1 - albino; child 2 - normal; child 3 - normal; child 4 - albino; child 5 - normal?
To determine the probability of each child having a particular phenotype, you need to understand the principles of Mendelian inheritance. In this case, albinism is inherited as a simple recessive trait, which means that an individual must inherit two copies of the albino allele to express the albino phenotype.
Given that both the woman and her husband are heterozygous for albinism (Aa), we can represent their genotypes as follows:
Woman (Aa) → A a
Husband (Aa) → A a
To predict the phenotypes of their offspring, we can use a Punnett square. This tool allows us to combine the possible alleles from each parent to determine the probability of different genetic outcomes.
The Punnett square for this situation would look like this:
A a
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A| AA | Aa |
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a| Aa | aa |
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The top row represents the possible alleles from the mother (A and a), while the left column represents the possible alleles from the father (A and a).
Now, let's analyze each child individually:
Child 1 - Albino
For the child to be albino, they must inherit two copies of the albino allele (aa). The probability of this happening is 1/4 or 25%, as represented by the box in the bottom right corner of the Punnett square.
Child 2 - Normal
To be normal, the child can either inherit two copies of the normal allele (AA) or be heterozygous (Aa). The probability of this happening is 3/4 or 75%, represented by the three boxes in the top row and left column of the Punnett square.
Child 3 - Normal
Following the same logic as Child 2, the probability of being normal is also 75%.
Child 4 - Albino
Using the same reasoning as Child 1, the probability of being albino is 25%.
Child 5 - Normal
Again, the probability of being normal is 75%.
Therefore, the probability that their first five children will have the given phenotypes is as follows:
Child 1 - Albino: 25%
Child 2 - Normal: 75%
Child 3 - Normal: 75%
Child 4 - Albino: 25%
Child 5 - Normal: 75%
To calculate the probability of the phenotypes of their first five children, we can use the Punnett square method.
First, let's represent the genotypes of the parents. The woman and her husband are both heterozygous for albinism, so their genotypes can be represented as Aa.
When two individuals with the genotype Aa have children, there are four possible combinations of alleles that the children can receive: AA, Aa, aA, and aa.
Now, let's fill in the Punnett square:
A a
---------------------
A | AA Aa
a | aA aa
From the above Punnett square, we can see that there is a 25% chance for each child to inherit the homozygous recessive genotype (aa), which corresponds to being albino.
Therefore, the probability that their first child is albino is 25% or 0.25.
Similarly, the probability that their second child is normal is also 25% or 0.25.
Applying the same calculation for the remaining children, the probability for each child to have the specified phenotypes is as follows:
Child 1 - albino: 25% or 0.25
Child 2 - normal: 25% or 0.25
Child 3 - normal: 25% or 0.25
Child 4 - albino: 25% or 0.25
Child 5 - normal: 25% or 0.25