SOME SAMPLE PROBLEMS
1. Tay-Sachs
is a recessive genetic disease in humans. If one parent is homozygous normal
and the other is a carrier (heterozygous), what are the chances that their
child will have this disease?
2. A
woman who can roll her tongue (presumably dominant) is married to a man who
cannot. Two of their four children can roll their tongues and two cannot. If A
= roll tongue and a = cannot roll tongue, then what is the genotype of the
parents?
3. Some
plants fail to produce chlorophyll, and this trait appears to be recessive. Many
plants also self-pollinate. If we locate a pea plant that is heterozygous for
this trait, self-pollinate it and harvest seeds, what are the likely phenotypes
of these seeds when they germinate?
4. In
1940, two researchers named Weiner and Landsteiner discovered that about 85
percent of the human population sampled possessed a blood cell protein labeled
Rh positive (Rh+) and was found to be dominant over the absence of the blood
factor (Rh-). Under normal Mendelian inheritance, which of the following
statements is/are false?
A. Two
Rh+ parents could have an Rh- child.
C. An
Rh- child would require that both parents be carriers of at least one Rh- gene.
D. It
is possible with just one pair of parents to have children where some siblings
are Rh- and some are Rh+.
E. All
of the above are false.
5. If
you had two guinea pigs of opposite sex, both homozygous, one black and one
brown, but you didn't know which was the dominant characteristic, how would you
find out the dominant color?
B. Mate
them and see what color the offspring are--the other will be the dominant
color.
C. Mate
them, then mate their offspring to see what color the next generation is--that
will be the dominant color.
D. Mate
them together, then mate their offspring to see what color the next generation
is--the other color will be the dominant color.
E. None
of the above are correct.
6. How
could you be certain that the guinea pigs used in the previous question are
truly homozygous?
A. The
guinea pigs would be homozygous for black (or brown) coat color if each strain
could be bred for many generations and only black (or brown) colored offspring
were produced.
B. If
the immediate parents of the black (or brown) guinea pigs were both of that
color, it proves they are homozygous.
D. Each
of the above results would prove the black guinea pig was homozygous.
E. Only
microscopic examination of the guinea pig's genes could absolutely confirm
homozygosity.
7. In
pea plants, the gene for round seed (R) is dominant, and wrinkled seeds (r) are
recessive. The endosperm of the pea is also either starchy, a dominant gene (S),
or waxy (s).What can be said of a fully heterozygous (or dihybrid) cross?
A. It
is impossible to secure offspring that are homozygous for both dominant genes.
B. It
is impossible to secure offspring that are homozygous for both recessive genes.
C. It
is impossible to secure offspring that are homozygous for one dominant gene
such as round seed and homozygous recessive for the other recessive waxy gene.
D. All
of the above are impossible combinations in a dihybrid cross.
8. Mr.
X has been accused by Miss Z of fathering her child. The child's blood type is
AB and Miss Z's blood type is A. Mr. X could NOT be the father if his blood type
is__
9. Lethal
genes are genes that result in the failure to develop a critical organ or
metabolic pathway. They are nearly always recessive. Animal breeders who
discover a unique trait and cross-breed to increase the occurrence of that
trait often encounter a noticeable increase in lethal genes. Why?
B. Spreading
the gene among offspring of both sexes will increase the likelihood it will be
sex-linked and expressed when alone.
C. The
cross-mating of closely related individuals, or inbreeding, increases chances
two recessive genes will "meet" in offspring.
D. Lethal
genes may switch from recessive to dominant.
E. Selecting
for the desired trait may result in "uncovering" the lethal gene.
10. A
man with blood type AB could NOT be the father of a child with
A. blood
type A.
B. blood
type B.
C. blood
type AB.
11. Nondisjunction
of the sex chromosome in a woman (assume normal meiosis in husband) could NOT
lead to a child with the genotype
A. XO.
B. XXX.
C. XXY.
12. When
Mendel established his mathematical model for pea plant inheritance, including
his laws of segregation and independent assortment, he used a small number of traits
that happened to be the same number as the numbers of chromosome pairs (yet to
be discovered). What is the relationship between his traits and the number of
chromosome numbers of peas?
B. Linkage
groups would not cause him any trouble except for instances of crossing over.
C. Linkage
groups would not cause him any trouble except for cases of inversion.
D. If
all traits were on the same chromosome, they would be inherited in identical
patterns without any variation.
E. There
would be no problem with linkage groups and he would have arrived at the same
mathematical models of heredity particles.
13. Animals
generally build an organism using the diploid number of chromosomes. However, insects in the order of ants, wasps
and bees use a haploid-diploid system in which adults of one sex are formed
with a haploid number of chromosomes.
This would mean that:
A. a
single set of chromosomes is sufficient to code for a functional individual.
B. a
female could determine the sex of the offspring by fertilizing or not
fertilizing an egg.
C. the
males and females are not equally related to their mothers, considering the proportion
of genes held in common.
E. This
system in no way differs from regular diploid organisms.
14. We
presume that meiosis evolved later than mitosis. What process(es) would have to evolve to make this transition?
A. Homologous
chromosomes would have to pair up.
B. An
additional division would have to occur but without a second DNA replication.
C. Sister
chromosomes do not separate at first anaphase; centromeres hold until second
division.
E. All
of the above occur in mitosis and none have changed in evolution of meiosis.
15. In
organisms that are parthenogenetic, females produce young that are female
without any fertilization of eggs occurring.
Which process(es) could theoretically accomplish this through alteration
of the normal meiotic cycle?
A. A
haploid egg develops into a female; females only complete meiosis I.
B. A
polar body fuses with the egg nucleus in place of the sperm and females
complete meiosis.
C. Meiosis
stops before anaphase, telophase or cytokinesis occurs, leaving a full diploid
individual.
17.Suppose
a color blind man with type O blood mates with a woman with normal vision
(homozygous) and
type AB blood. What are the possible genotypes and phenotypes of their
offspring? What are the probabilities of each?
18.A man has type A blood and his wife has type B. Their child has type O. What are the genotypes of the parents? AO, BO