Questions:
1. Consider a hypothetical insect species that has red eyes. Imagine mutations in two different unlinked genes that can, in certain combinations, block the formation of red eye pigment yielding mutants with white eyes. In principle, there are two different possible arrangements for two biochemical steps responsible for the formation of red eye pigment. The two genes might act in series such that a mutation in either gene would block the formation of red pigment. Alternatively, the two genes could act in parallel such that mutations in both genes would be required to block the formation of red pigment. When mutant phenotype is not reverted back to wild type we can conclude that the
a. Mutant genes belong to same complementation group
b.
Mutant genes belong to different complementation group
c.
Mutant genes do complement each other
d. Mutant genes did not revert back to wild type by gene conversion
2. Most Border collie dogs are black and white. The allele for this is B
and the trait is autosomal dominant. Homozygous recessive dogs are chocolate
and white. When a test cross was done with black and white dog, we got all
black and white puppies, what does this result tell you about the genotype of
the black and white parent?
a. BB
b. Bb
c. bb
d. None of the above
3. Based on the following recombination frequencies between genes on the
same chromosome, determine the correct order of the genes on the chromosome.
O-I 25%; G-I 50%; A-I 30%; O-G 25% ; A-G 80%
Your
answer:
a. GOAI
b. GAOI
c. GIOA
4. In the human species, all somatic cells have 46 chromosomes. Which of
the following can also be true?
a. A
plant species (privet shrubs) has 46 chromosomes per cell.
b. Some
adult humans have 69 chromosomes per cell.
c. Some
adult humans have 23 chromosomes per cell.
d. A certain fungal species has only one chromosome per cell.
5. If a is linked to b, and b to c, and c
to d, does it follow that a recombination experiment would detect
linkage between a and d?
a. Yes
b. No
c. May
be
d. Can’t
be determined
6. The human X and Y chromosomes
a. Are
both present in every somatic cell of males and females alike.
b. Are of
approximately equal size and number of genes.
c. Are
almost entirely homologous, despite their different names.
d.
Include genes that determine an individual's sex.
7. If both children are of blood type M, which of the following is
possible?
a. Each
parent is either M or MN.
b. Each
parent must be type M.
c. Both
children are heterozygous for this gene.
d.
Neither parent can have the N allele.
8. In the cross AaBbCc × AaBbCc, what is the probability of producing
the genotype AABBCC?
a. 1/4
b. 1/8
c. 1/32
d. 1/64
9. Given the parents AABBCc × AabbCc, assume simple dominance for each
trait and independent assortment.What proportion of the progeny will be
expected to phenotypically resemble the first parent?
a. 1/4
b. 1/8
c. 3/4
d. 3/8
10. Which of the following describes the ability of a single gene to have
multiple phenotypic effects?
a.
incomplete dominance
b.
multiple alleles
c.
Pleiotropy
d.
Epistasi
11. Hydrangea plants of the same genotype are planted in a large flower
garden. Some of the plants produce blue flowers and others pink flowers. This
can be best explained by which of the following?
a. the
knowledge that multiple alleles are involved
b. the
allele for blue hydrangea being completely dominant
c. the
fact that a mutation has occurred
d.
environmental factors such as soil pH
12. Most genes have many more than two alleles. However, which of the
following is also true?
a. At
least one allele for a gene always produces a dominant phenotype.
b. Most
of the alleles will never be found in a live-born organism.
c. All
of the alleles but one will produce harmful effects if homozygous.
d. There
may still be only two phenotypes for the trait.
13. A scientist discovers a DNA-based test for one allele of a particular
gene. This and only this allele, if homozygous, produce an effect that results
in death at or about the time of birth. Of the following, which is the best use
of this discovery?
a.
Screen all newborns of an at-risk population.
b.
Design a test for identifying heterozygous carriers of the allele.
c.
Introduce a normal allele into deficient newborns.
d.
Follow the segregation of the allele during meiosis.
14. Phenylketonuria (PKU) is a recessive human disorder in which an
individual cannot appropriately metabolize a particular amino acid. The amino
acid is not otherwise produced by humans. Therefore, the most efficient and effective
treatment is which of the following?
a. Feed
them the substrate that can be metabolized into this amino acid.
b.
Transfuse the patients with blood from unaffected donors.
c.
Regulate the diet of the affected persons to severely limit the uptake of the
amino acid.
d. Feed
the patients the missing enzymes in a regular cycle, such as twice per week.
15. Recombination between linked genes comes about for what reason?
a.
Mutation on one homolog is different from that on the other homolog.
b.
Independent assortment sometimes fails because Mendel had not calculated
appropriately.
c. When
genes are linked they always "travel" together at anaphase.
d.
Crossovers between these genes result in chromosomal exchange.
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