Genetics

Genetics
Multiple Choice Questions (1.25 points each)

1. Which of the following constitutes the primary structure of a protein?
A) The folding of a polypeptide chain
B) The linear sequence of amino acids in a polypeptide chain
C) The polypeptide chains stacked on top of each other
D) A pleated sheet
E) Several polypeptide subunit
2. A pathway that makes flower pigment shows the following sequential color conversions (each
arrow represents a protein responsible for each conversion step).
Colorless ÆÆ yellow ÆÆ blue ÆÆ red
A B C
A plant is homozygous for a null mutation in protein “B”. The flowers of this plant will be:
A) orange
B) blue
C) yellow
D) red
E) white
3. You are planning a transformation experiment and deciding how to do the transformation and
how to select for transformed bacteria. The best plan is to:
A) extract DNA from an auxotroph and add it to prototrophic cells and grow on minimal
media
B) extract DNA from arg- cells and add it to arg+ cells and grow on minimal media
C) extract DNA from arg+ cells and add it to arg- cells and grow on minimal media
D) extract DNA from StrS cells and add it to StrR cells and grow on media with Streptomycin
E) All of the above are equally good strategies
4. A female rabbit of phenotype c’ is crossed to a male rabbit with c
ch. The F1 is comprised of
five rabbits with a c’ phenotype, two with c
ch phenotype, and three with c phenotype. Of the
phenotypically c rabbits, two are females and are backcrossed to their father. This cross
produces only rabbits with c
ch phenotype. These results suggest that:
A) c could be dominant or recessive to c’
B) c is dominant to c’, but recessive to c
ch
C) c is dominant to c
ch, but recessive to c’
D) c is dominant to both c’ and c
ch
E) c is recessive to both c’ and c

5. The A site, P site, and E sites each control _________________ (in order) during translation.
A) binding incoming tRNAs (A), retention of the peptide chain during elongation (P), exit of
deacylated tRNAs (E)
B) activation of tRNAs (A), protease-based release of the protein product (P), exit of empty
rRNAs (E)
C) activation of ribosome activity (A), processing of amino acids (P), elongation of the
peptide chain (E)
D) amino acid bond transfer (A), protein processing (P), and elongation of the peptide chain
(E)
E) both A and C are correct
6. What proportion of progeny from a monohybrid self are themselves monohybrid?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%
7. Which of the following reduce heterozygosity?
A) Migration
B) Directional selection
C) Large population size
D) Balancing selection
E) Mutation
8. Eye-color in wombats is controlled by gene Y. The dominant eye-color phenotype is yellow
(Y) and the recessive eye-color phenotype is white (y), however the Y allele is
haploinsufficient. A wombat that is heterozygous for a recessive null mutation the eye-color
gene is testcrossed. What proportion of progeny will have yellow eyes?
A) All
B) 3/4
C) 1/2
D) 1/4
E) None
9. Which of these principles of evolution, as described by Darwin’s theory, is correctly matched
with its role in evolution?
A) Principle of variation: Offspring must resemble their parents more than they
resemble unrelated individuals
B) Principle of selection: Variation in morphology, physiology, and behavior must be present
in a population for selection to occur
C) Principle of variation: Variation in morphology, physiology, and behavior must be present
in a population for selection to occur
D) Principle of heredity: Some genotypes of a population should be more successful at surviving and reproducing than other genotypes in a given environment

10. Drosophila eyes are normally red. Several purple-eyed strains have been isolated as spontaneous mutants, and the purple phenotype is autosomal recessive in each case. To
investigate allelism between these different purple mutations, two purple-eyed pure strains
were crossed. If the purple mutations are in different genes (i.e., they are not allelic), the
phenotypic ratios in the F1 are expected to be:
A) 100% red
B) 75% red : 25% purple
C) 50% red : 50% purple
D) 25% red : 75% purple
E) 100% purple
11. A high variance indicates that:
A) the mean value is very high
B) most values are very close to the mean
C) most values are higher than the mean
D) most values are lower than the mean
E) the variation among the values is high
12. The following pedigree shows the inheritance of attached earlobes (black symbols) and
unattached earlobes (white symbol). Both alternative phenotypes are quite common in human
populations.
If the phenotypes are determined by alleles of one gene, then attached earlobes are inherited as:
A) an autosomal dominant trait
B) an autosomal recessive trait
C) a dominant trait that could be either autosomal or X-linked
D) a recessive trait that could be either autosomal or X-linked
E) an X-linked dominant trait
13. Which diagram most accurately shows the arrangement of homologous chromosomes during
the first metaphase of meiosis?
A)
B)
C)
D)
E)
14. Given below are the genotypic frequencies for a single gene with two alleles for three
different populations:
Population 1
AA
0.25
Aa
0.50
aa
0.25
Population 2 0.35 0.56 0.09
Population 3 0.49 0.42 0.09
Which of the following is NOT a correct statement about these three populations?
A) Only two of the populations are in Hardy-Weinberg equilibrium
B) Population 1 is in Hardy-Weinberg equilibrium; the frequency of allele A is 0.5
C) Population 2 is not in Hardy-Weinberg equilibrium; the frequency of allele a is 0.37
D) Population 3 is in Hardy-Weinberg equilibrium; the frequency of allele A is 0.7
E) Population 3 is in Hardy-Weinberg equilibrium; the frequency of allele a is 0.3
15. Crosses between two ducks with spread-out tail phenotype always produce ducklings with
spread-out, normal, and pointy tails in a 2:1:1 ratio. If the tail shape phenotype is controlled
by a single locus, what is the expected phenotypic ratio in the progeny of the cross normal ×
pointy tail?
A) 0:0:1 of spread-out : normal : pointy
B) 1:0:0 of spread-out : normal : pointy
C) 2:1:1 of spread-out : normal : pointy
D) 1:2:1 of spread-out : normal : pointy
E) 1:1 of normal : pointy
16. A plant is heterozygous at three unlinked loci A, B, and C. How many different gamete
genotypes can it theoretically produce with respect to these three loci?
A) 2
B) 3
C) 4
D) 8
E) 16
17. A diploid plant is a dihybrid for flower color (gene “R”) and leaf size (gene “L”). Its
phenotype is red flowers and large leaves. This plant is crossed to a tester plant (which has
white flowers and small leaves). The progeny is as follows:
23 Red, large
25 White, small
230 Red, small
235 White, large
What can be concluded about the linkage relationship of the loci in the dihybrid parent?
A) Genes R and L are unlinked
B) The dihybrid parent’s genotype is RL/rl
C) The dihybrid parent’s genotype is Rl/rL
D) Genes R and L are linked and 9 map units apart
E) B, C, and D
18. Over the years, a persons skin cells are exposed to UV damage leading to pyrimidine dimers.
Which DNA repair mechanism repairs this damage?
A) Double-strand break repair
B) Non-homologous end joining
C) Mismatch repair
D) Base-excision repair
E) Nucleotide-excision repair
19. If the DNA template 5’-ATGCATGC-3’ were transcribed to RNA, the RNA would read:
A) 3’ TACGTACG 5’
B) 5’ AUGCAUGC 3’
C) 5’ UACGUACG 5’
D) 3’ UACGUACG 5’
E) 5’ATGCATGC 3’
20. A corn plant is homozygous for a mutant allele that results in no pigment in the seed (i.e.,
white). The mutant is caused by Ds insertion that often exits late in seed development, when
there is an active Ac element in the genome. If there is NO active Ac element, the seeds of
this plant will be:
A) no pigment (i.e., white)
B) pigmented all over
C) white with small spots of pigment
D) white with large spots of pigment
E) weakly pigmented
21. Oncogenes are genes that:
A) Are expressed and activated in cancer cells
B) Gain-of-function mutations that promote cancer
C) Dominant mutations only need to be present in one allele to promote cancer
D) Positively control the cell cycle
E) All of the above
22. The INTERACTOME is defined as:
A) the sequence and expression patterns of all transcripts
B) the sequence and expression of all proteins
C) the complete set of physical interactions (i.e., protein/DNA or protein/protein)
D) the complete set of all metabolites
E) none of the above
23. In the Sanger sequencing method, the use of dideoxy adenosine triphosphate stops nucleotide
polymerization
A) opposite A’s in the template strand
B) opposite T’s in the template strand
C) opposite G’s in the template strand
D) opposite C’s in the template strand
E) opposite any base selected randomly in the template strand
MCELLBI_X143_Form A 10
Final Examination
24. The neutral theory of evolution proposed that most genetic mutations are functionally
neutral. Neutral mutations are fixed in a population by
A) natural selection
B) random genetic drift
C) environmental selection
D) asexual reproduction
E) the invariance of DNA
25. In a diploid animal in which 2n = 24, what is the total number of bivalents present at
prophase of meiosis I?
A) 6
B) 12
C) 24
D) 48
E) 96
26. The most closely related genes between two different organisms are called:
A) homologs
B) orthologs
C) paralogs
D) heterologs
E) alleles
27. A small (one base pair) insertion in the middle of the coding region of a gene will cause a:
A) synonymous mutation
B) silent mutation
C) nonsense mutation
D) missense mutation
E) frameshift
28. The________ is a DNA element that binds regulatory proteins
A) promoter
B) operator
C) enhancer
D) upstream activating sequence
E) all of the above
29. What is chromatin remodeling
A) The first stage of mitosis
B) The changing of amino acid position
C) The changing of nucleosome position
D) The changing of nucleotide position
E) None of the above
30. HOX genes encode:
A) signal proteins
B) serine proteases
C) DNA-binding transcription factors
D) cell-to-cell junction proteins
E) transmembrane proteins
31. Of the three key building blocks of DNA, which type(s) of building block is/are structurally
different in RNA molecules?
A) Phosphate
B) Deoxyribose
C) Nitrogenouse bases
D) Both deoxyribose and nitrogenous bases
E) All of the above
32. GAL80 inhibits GAL gene expression by
A) specifically binding to Gal4 and preventing DNA-binding
B) specifically binding to Gal4 and preventing activation
C) specifically binding to Gal3 and preventing DNA-binding
D) specifically binding to Gal3 and preventing activation
E) inducing an allosteric change in Gal3
33. In one strand of DNA the nucleotide sequence is 5′-ATGC-3′. The complementary sequence
in the other strand must be
A) 3′-ATGC-5′
B) 3′-TACG-5′
C) 5′-ATCG-3′
D) 5′-CGTA-3′
E) 5′-TACG-3′
34. The lac repressor (LacI) binds to
A) lactose and DNA
B) RNA polymerase
C) RNA polymerase and DNA
D) £-galactosidase, permease, and transacetylase
E) RNA and DNA
35. A linear DNA molecule has n sites for restriction enzyme EcoRI. How many fragments will
be produces after complete digestion?
A) n – 1
B) n
C) n + 1
D) 2n – 1
E) 2n + 1
36. For many generations, the following genotypic frequencies were observed in a large
population of dinosaurs: 4 percent AA, 32 percent Aa, and 64 percent aa. The climate
changed abruptly and resulted in the death of all homozygous recessive dinosaurs. What is
the new genotype frequency of heterozygotes?
A) 4%
B) 32 %
C) 88.8%
D) 11.1%
E) 44.4%
37. Which of the following is an example of a CIS-ACTING ELEMENT?
A) £-galactosidase
B) Operator site
C) LacI repressor protein
D) Lactose
E) Permease
38. If Escherichia coli, grown for a period of time in 15N, is transferred to 14N for one
generation of DNA replication, the resulting DNA should have 14N added to all “new” DNA.
If conservative replication is occurring, the 14N-containing “new” DNA will compose
.
A) one strand of all bacterial chromosomes
B) both strands of DNA in half of all bacterial chromosomes
C) regions dispersed throughout all bacterial DNA
D) none of the new DNA, it will only be found in the old DNA
E) all of the above
39. RNA synthesis is always 5′ to 3′ because:
A) the unwinding of the double-stranded DNA can only move one direction.
B) nucleotides can only be added to an available 3′-OH group on the transcript terminus
C) nitrogenous bases cannot pair up in the 3′ to 5′ direction
D) the structure of ATP restricts 3′ to 5′ polymerization into RNA
E) RNA synthesis can move in the 3′ to 5′ direction
40. A fish of genotype a/a; B/b is crossed to a fish whose genotype is A/a; B/b. What
proportion of the progeny will be heterozygous for at least one of the genes? (Assume
independent assortment.)
A) 1/8
B) 2/8
C) 4/8
D) 5/8
E) 6/8
In-Depth Questions – SHOW ALL OF YOUR WORK for the following questions. Write in
complete and legible sentences. Answers that are unclear or illegible will not be graded. Note:
You can receive partial credit for answers but cannot receive partial credit if you give no answer
at all.
41. (6 pts) You are breeding two polyploid species of wheat. You cross the two different
species (parent 1 x parent 2) and obtain F1 hybrids. The following meiotic pairing was
observed in the F1 hybrid:
F1 hybrid = 12 bivalents + 12 univalents
(A) How many chromosomes are there in a non-dividing somatic cell of an F1 plant (i.e. what
is the organisms ploidy)?
(B) How many chromosomes are there in a non-dividing somatic cell of each parent (i.e. what
is the organisms ploidy)?
(C) For each parent, identify whether the plant is diploid, autopolyploid or allopolyploid.
42. (9 points) Red hair (autosomal recessive) is found in approximately 4 percent of the
people in Norway. If we assume that the Norwegian population is in Hardy-Weinberg
equilibrium with respect to hair color:
A) What are the frequencies of the red hair (r) and non-red hair (R) alleles?
B) What is the frequency of heterozygotes?
C) What is the proportion of matings that CAN NOT have a child with red hair?
43. (6 points) Darwin’s theory of evolution was developed around three central principles.
List and define each of these principles.
44. (4 points) In a particular species of plants, flower color is dimorphic: Some individuals
have red flowers, whereas others have yellow flowers. If flower color is controlled by a single
gene with two alleles (c
red and c
yellow):
A) What would be the simplest way to determine which allele is dominant?
B) What will be the genotypic ratio in the offspring of a cross between a monohybrid and a
pure-breeding individual?
45. (5 points) Below is the pedigree of a family where some individuals are affected with
skin disease.
I.
II.
III.
A) Based on the pedigree, what is the most likely mode of inheritance of this condition, and
why?
B) Indicate the respective genotypes of each individual represented.
C) What is the probability that individual III-2 is heterozygous for the condition?

46. (7 points) Vanessa has obtained two true-breeding strains of mice, each homozygous for
an independently discovered recessive mutation that prevents the formation of hair on the
body. One of the mutant strains is called naked, and the other mutation strain is called
hairless. To determine whether the two mutations are simply alleles for the same gene,
Vanessa crosses naked and hairless mice with each other (cross 1). All the offspring are
phenotypically wild-type.
A) What is the most likely explanation for the phenotypes of the F1 mice? Are the naked and
hairless mutations alleles for the same gene?
B) Vanessa next intercrosses the F1 mice (cross 2) and based in the F2 phenotype ratios,
concludes that there is no gene interaction. Fill out the following information for the F1
intercross and F2 progeny. Use these symbols: n = naked mutation, N = wild-type allele, h =
hairless mutation, H = wild-type allele.
Cross 1: parental genotypes:
F1 genotypes:
Cross 2: Write out a Punnett square for the F1 intercross and give the F2 phenotype ratio:
F2 phenotype ratio __________________

47. (6 points) A gene makes a very short polypeptide that is 7 amino acids long containing
an alternating sequence of phenyalanine and tyrosine. Assuming Phe = UUU, and Tyr = UAU in
the mRNA. What are the sequences of nucleotides corresponding to this sequence in the
following? Make sure to indicate the polarity of each sequence
A) The mRNA strand (sense)
B) The non-template DNA strand (sense or coding strand)
C) The template DNA strand (antisense or non-coding stand)
48. (7 points)
To determine the
linkage
relationship of
three genes (A, B,
and C) in
Drosophila
melanogaster you
perform a threepoint
testcross
with a fly that is
heterozygous for
all three genes.
The F1 progeny
genotypes are:
Number of F1
a/a . B/b . C/c 60
A/a . b/b . c/c 56
a/a . b/b . C/c 95
A/a . B/b . c/c 89
a/a . b/b . c/c 375
A/a . B/b . C/c 315
a/a . B/b . c/c 6
A/a . b/b . C/c 4
Total 1,000
A) What is the recombination frequency between loci A and B?
B) What is the recombination frequency between loci B and C?
C) What is the recombination frequency between loci A and C?
D) Are these loci linked?
E) Demonstrate the physical arrangement of the genes on the chromosome, i.e. the order of the
genes on a chromosomes as well as their map distances.
F) Write out the genotypes of the parents for this cross, indicating which alleles are linked with
one another for each parent?
___________________ X _________________________
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