Practice Problems in Genetics
NOTE: When a genotype is listed as, for example, A_, it means that the second allele is unknown. The actual genotype may
be AA or Aa.
Problems Involving One Gene
First assign symbols for the alleles, following the rules and conventions. For example, the letter “L” could be used
for this gene. The symbols would thus be L for the dominant short allele and l for the recessive long allele. The
problem tells you that the parents are both homozygous (the short-haired one because you are told it is true-
breeding, the long haired one because it has the recessive characteristic, which automatically makes it
homozygous). So here’s your mating:
All Ll; All short-haired
From problem 1 you already know the dominance here. Go ahead and use the same symbols.
Short haired: Ll
Since the known parent is ll, it contributes a recessive allele to all offspring. So the final phenotype of the kittens is
actually determined by the unknown parent. Since some of the kittens are short-haired, they must have received the
dominant allele from this unknown parent. The kittens who are long-haired must have received the recessive allele
from the unknown parent. So that parent must be Ll, and must be short-haired.
Again, assign symbols. A good choice here would be W for widow’s peak and w for no widow’s peak. As both parents
here have the dominant trait, they must each have at least one W. Their first child obviously inherited this allele
from at least one of them, as he also possesses a widow’s peak. The second child does not, and thus must be ww. But
note that there is no evidence here that prevents us from assuming that both of these parents could be carrying
hidden recessive w alleles, so Mr. Smith is not justified. Here’s the Punnett’s Square, assuming these two are truly
the parents of both children:
Genotypic ratio predicted: 1 WW : 2 Ww : 1 ww
Phenotypic ratio predicted: 3 Widow’s Peak : 1 No Widow’s Peak
Assign symbols. A good choice: E for free earlobes and e for attached earlobes. This problem is a simple exercise in
pedigree analysis. All of the individuals with attached earlobes must be ee. All of the individuals with free earlobes
must be E_. In this particular case, the E_ parent has children who have attached earlobes, so that parent must
have a little e, and is thus Ee. And all of the children have one parent with attached earlobes, so they must also all
have at least one little e, and are thus Ee. So the answer is, Mr. Jones is ee, Mrs. Jones is Ee. The three children
with attached earlobes are ee, and the three with free earlobes are Ee.