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Bio102: Introduction to Cell Biology and Genetics
Key for BioSynthesis on “Mendelian Genetics and Extensions”
You are a primary-care physician in private practice. You have just met with a young couple whose
first child has just been diagnosed with LCA2.
1. Neither of the boy’s parents has the disease (obviously, since they’re adults). Assuming that it’s a
genetic disease, what does this fact tell you about the basis of the disease?
The disease must result from a recessive allele, so that the parents could be carriers but not show
the trait.
2. The parents would like to have another child. However, they are understandably very concerned
about the risk of having another child with LCA2. They would like to know how likely this is.
What would be the probability that their second child would be born with this disease?
Both parents must be carriers (because the child had to inherit one recessive allele from each). If
we use the symbol l for the LCA2 allele and L for the normal allele, their genotypes are both Ll.
Therefore, there’s a ½ chance that the next child will get the LCA2 allele from the father and a ½
chance that the next child will get the LCA2 allele from the mother. The total probability that the
child will get a l allele from each parent and have the disease is ¼ (½ × ½).
Each event is independent of what has happened before, so the fact that they have previously had
one LCA2 child is irrelevant. Each child will have a ¼ chance.
3. Brown eye color is dominant over blue in humans (well, it’s actually not quite that simple, but let’s
assume it is for the purpose of this problem!). The mother has blue eyes, and the father is the
brown-eyed son of a blue-eyed mother. What is the probability that their next child will have blue
eyes?
Let b = the blue-eye allele and B = the brown-eye allele. Then, the mother’s genotype must be bb,
because she has blue eyes. The father’s has to be Bb, because he has brown eyes and he’s the son
of a woman with blue eyes (who must have given him a b allele). In the cross bb × Bb, the child
will always get a b allele from the mother and will get a b allele from the father as well ½ the time.
Thus, the probability of having blue eyes is ½.
4. What is the probability that their next child will have normal vision and have blue eyes?
Mendel’s Law of Independent Assortment tells us that these two probabilities are independent, so
we can multiply them to get the total probability. The probability of LCA2 is ¼, so the probability
of a child with normal vision is ¾. The probability of blue eyes is ½. So, the total probability is ¾
× ½ = 3/8.