108 ❯ STEP 4. Review the Knowledge You Need to Score High
to have received two recessive alleles and thus have contracted the disease. How can we
calculate the potential probability of the father being a carrier? We construct a Punnett
square for a monohybrid cross of the father’s parents (first generation) (Figure 10.8).
We know with certainty that he is not dd, otherwise, he would have the condition. This
leaves three equally likely possible genotypes for the father, two of which are “carrier” geno-
types (Dd). Thus, the probability of his being a carrier is 2 ⁄ 3.
What is the probability that the mother (second generation) is a carrier? We don’t even
need a Punnett square to determine this one. Her mother (first generation) died of the con-
dition, which means that she must have been dd, and thus must have passed along a d to
each of her children. The mother in question does not have the condition, so she must have
a D as well. Therefore her genotype mustbe Dd.
To determine the probability that bothparents are carriers, apply the law of multipli-
cation with probabilities (similar to tossing a coin) and use the following formula:
PF×PM=2⁄3 × 1 =2⁄3
(where PF, PM =probabilities of father, mother being carriers).
Now that we have determined the probability that they are both carriers, we need to
determine the probability that one of their offspring will have the condition. Their
Punnett square would be the same as that shown in Figure 10.8, and we can see that the
probability of having a child with the recessive condition is 1 ⁄ 4. Again, we use the law of
multiplication and see that the probability of this couple having a child with the condi-
tion is 2 ⁄ 3×1⁄4 =1⁄6.
If these two second-generation parents had a child with the recessive condition, what
would the probability of their next child having the condition be? It would no longer
be 1 ⁄ 6; once they have had a child with the condition, we would know with 100 percent
certainty that they are heterozygous carriers. Thus, the probability that their next child will
have the condition is 1 ⁄ 4, as shown in Figure 10.8.Common Disorders
There are many simple recessive disorders in which a person must be homozygous reces-
sive for the gene in question to have the disease. Some of the most common examples are
Tay-Sachs disease, cystic fibrosis, sickle cell anemia, phenylketonuria, and albinism. These
diseases are commonly used as examples on the AP Biology exam and could also aid you
in constructing a well-supported essay answer to a question about heredity and inherited
disorders.
Tay-Sachs diseaseis a fatal genetic disorder that renders the body unable to break
down a particular type of lipid that accumulates in the brain and eventually causes blind-
ness and brain damage. Individuals with this disease typically do not survive more than aD dD Ddd Dd ddDDFigure 10.8 A Punnett square.KEY IDEA