If a rare genetic disease is inherited on the basis of an X-linked gene, one would expect to find which of the following:
A. Affected fathers have 100% affected sons
B. Affected fathers have 100% affected daughters
C. Carrier mothers have 50% affected sons
D. Carrier mothers have 50% affected daughters
E. Carrier mothers have 100% affected sons

C

A drosophila with 4 X chromosomes, a Y chromosome and 3 sets of autosomes will be a
A. normal male
B. normal female
C. metafemale
D. intersex
E. metamale

C

What part of the Y chromosome allows it to pair with the X chromosome in meiosis?
A. PAR
B. SRY
C. TDF
D. PTA
E. MSY

A

B is a dominant giving black color. b is the recessive allele causing blue color. In a population of 100 B/b individuals, four blue
animals are seen. This is an example of
A. incomplete dominance
B. partial dominance
C. reduced penetrance
D. reduced expressivity
E. poor record keeping

C

Shell orientation in snails is due to a maternal effect gene. A true breeding sinistral (recessive) is crossed to a true breeding
Dextral (dominant). The offspring from that cross are self-crossed. What will be the expected ratio of shell types?
A. All sinistral
B. All Dextral
C. Half sinistral, half Dextral
D. ¾ Dextral, ¼ sinistral
E. ¾ sinistral, ¼ Dextral

B

A couple has five children. What is the probability that they would have five boys?
A) 1/2
B) 1/4
C) 1/8
D) 1/16
E) 1/32

E

In peas, axial (A) flower position is dominant to terminal (a), tall (L) is dominant to short (l), and yellow (Y) is dominant to
green (y). If a plant that is heterozygous for all three traits is allowed to self-fertilize, how many of the offspring would be dominant
for all three traits?
A) 3/64
B) 9/64
C) 27/64
D) 32/64
E) 64/64

C

Contact points between nonsister chromatids that mark the locations of DNA-strand exchange are called
A) synaptonemal complex.
B) metaphase plate.
C) chiasmata.
D) kinetochore.
E) centrosome.

C

A chromosome with a centromere at the end is called
A. metacentric
B. submetacentric
C. acrocentric
D. telocentric
E. centrocentric

D

Morgan was the first scientist to analyze the white-eyed male Drosophila mutation which helped explain
A) autosomal dominance.
B) random mutation.
C) crossing over.
D) independent assortment.
E) X-linked inheritance.

E

The product H substance is needed to express the blood antigen. A mating of IA IB Hh X IA IB hh should produce what ratio of
blood types?
a) 1/4 AB, 1/8 A, 1/8 B, 1/2 O
b) 3/8 AB, 3/16 A, 3/16 B, 1/4 O
c) 1/4 AB, 3/16 A, 3/16 B, 3/8 O
d) 3/16 AB, 3/8 A, 3/16 B, 1/4 O
e) 1/8 AB, 1/4 A, 1/8 B, 1/2 O

A

If a common slipper limpet Crepidula fornicata lands on top of a female limpet, it will
A. become a male and cannot change
B. become a female and cannot change
C. become a male and change to female if another limpet lands on top of it
D. become a female and change to a male if there are too many females
E. beg the limpets pardon and move to a different part of the ocean

C

True breeding white flowering plants are crossed to true breeding red flowering plants. The offspring are red flowered. When
these flowers are self crossed, the resultant offspring are 87 red, 32 pink and 39 white flowered plants. Which of the following
pathways is most likely the cause of this
A. white ---A--->pink---B--->red
B. pink ---A--->pink---B--->red
C. red<---A---white ---B--->red
D. white ---A--->pink white---B--->pink both genes producing pink together give red

A

If a trait is X-linked recessive, who would express the trait?
A) homozygous dominant females and hemizygous recessive males
B) heterozygous recessive females and hemizygous dominant males
C) homozygous recessive females and hemizygous recessive males
D) heterozygous dominant females and hemizygous dominant males
E) the same proportions of females and males

C

If purple (P) is dominant to white (p) and axial (A) is dominant to terminal (a), and in a cross of white, axial to purple, axial the
ratio of offspring is
6/16 white, axial;
2/16 white, terminal;
6/16 purple, axial;
2/16 purple, terminal,
then what is the genotype of the parents?
A. PPAA X ppaa
B. PpAa x PpAa
C. PpAa X Ppaa
D. PpAa X ppAa
E. Ppaa X Ppaa

D

Color blindness is X-linked recessive and blood type is autosomal. If two parents who are both Type A and have normal vision
produce a son who is color blind and type O, what is the probability that their next child will be a female who has normal vision
and is type O?
A. 0
B. 1/8
C. 1/4
D. 1/2
E. 1

B

26) Which of the following occurs only in Meiosis II:
A. Separation of homologues
B. Synaptonemal complex formation
C. Formation of spindle poles
D. Separation of centromeres of sister chromatids
E. None of the above

D

A non-color blind father is married to a woman who is also not color blind. They have four sons, two of whom are color blind.
They have a child with Klinefelter syndrome. This child suffers from color blindness. This occurred because of a normal gamete
fusing with a gamete which is the result of nondisjunction in:
A. Either Meiosis I or II in male
B. Meiosis II in male
C. Either Meiosis I or II in female
D. Meiosis I in female
E. Meiosis II in female

What is the sex of the color blind child in the previous question
A. male
B. female
C. intersex

E, A

In a cross of AaBbCcDdEeFf X AaBbccDdEeFf, what proportion will have the ABCDeF phenotype?
A. 27/64
B. 27/128
C. 27/512
D. 81/512
E. 81/2048

E

Mendel crossed purebred wrinkled, green-seeded plants with purebred round, yellow-seeded plants. The F1 progeny were selfcrossed.
Which of the following numbers are most likely to result from this cross.
A. 22 round, yellow; 26 round, green; 28 wrinkled, yellow, 24 wrinkled green
B. 91 round, yellow; 32 round, green; 29 wrinkled, yellow, 8 wrinkled green
C. 64 round, yellow; 26 round, green; 27 wrinkled, yellow, 27 wrinkled green
D. 11 round, yellow; 26 round, green; 27 wrinkled, yellow, 87 wrinkled green

What is the chi-square value of the product of the correct cross from problem 30 preceding
A. 2/5
B. 4/5
C. 7/25
D. 49/90
E. 26/45

32) What is the probability of this event occurring, chi-square table follows
A. >0.9
B. 0.9>0.5
C. 0.5>0.1
D. 0.1>0.05
Degrees of Freedom 0.9 0.5 0.1 0.05
===========================================================
1 0.02 0.46 2.71 3.84
2 0.21 1.39 4.60 5.99
3 0.58 2.37 6.25 7.82
4 1.06 3.86 7.78 9.49

B, E, A

If a male bird that expresses a recessive Z-linked mutation is crossed to a wild type female, what proportion of the total
progeny will be mutant females?
A. 0%
B. 25%
C. 50%
D. 75%
E. 100%

C

Pudgy Birds come in two forms, listless and active. Independent of who they mate with, listless mothers always give birth to
listless offspring and their daughters always give birth to more listless offspring of both sexes. This is most likely due to:
A. Sex-limited trait
B. Recessive lethal
C. Mitochondrial genes
D. Maternal effect genes

C

True breeding white flowering plants are crossed to true breeding red flowering plants. The offspring are red flowered. When
these flowers are self crossed, the resultant offspring are 92 red, 57 pink and 12 white flowered plants. Which of the following
pathways is most likely the cause of this
A. white ---A--->pink---B--->red
B. pink ---A--->pink---B--->red
C. red<---A---white ---B--->red
D. white ---A--->pink white---B--->pink both genes producing pink together give red

D

If a rare genetic disease is inherited on the basis of an Y-linked gene, one would expect to find which of the following:
A. Affected fathers have 100% affected sons
B. Affected fathers have 100% affected daughters
C. Carrier mothers have 50% affected sons
D. Carrier mothers have 50% affected daughters
E. Carrier mothers have 100% affected sons

A

The correct order in mitosis is:
A. Methaphase, Interphase, Telophase, Anaphase, Prophase
B. Interphase, Telophase, Metaphase, Prophase, Anaphase
C. Interphase, Prophase, Metaphase, Anaphase, Telophase
D. Telophase, Metaphase, Interphase, Prophase, Anaphase

C

A cross of two heterozygotes yields a 2:1 ratio of phenotypes. This is indicative of:
A. Incomplete dominance
B. codominance
C. Recessive lethal
D. dominant lethal
E. multiple alleles

C

A carrier mother with normal vision and a colorblind father have a child who is XXXXYY and colorblind. This is the result of:
A. Nondisjunction in Meiosis I and II in the father and Meiosis I in the mother
B. Nondisjunction in Meiosis I in the father and mother
C. Nondisjunction in Meiosis I in the father and Meiosis I and II in the mother
D. Nondisjunction in Meiosis II in the father and Meiosis I and II in the mother.
E. Nondisjunction in Meiosis I and II in the father and Meiosis II in the mother.

E

During meiosis I, when does homologous chromosome pairing and recombination occur?
A) prophase I
B) pro-metaphase I
C) metaphase I
D) anaphase I
E) telophase I

A

1) If you are given a recombination frequency of 34% between genes X and Y and 27% between X and Z, can you predict the order of
the three genes?
A) Yes; the order is X-Z-Y.
B) Yes; the order is X-Y-Z.
C) Yes; the order is Z-X-Y.
D) No; based on this data alone, the order could be Z-Y-X or X-Y-Z.
E) No; based on this data alone, the order could be X-Z-Y or Z-X-Y.

E

Which of the following has the largest overall affect on the allele frequency of a population?
A. mutation
B. migration
C. genetic drift
D. inbreeding

C

In a population of 20,000 individuals, 200 men are afflicted with a recessive, X-linked disease. How many woman would be
expected to be afflicted in this population?
A. 1
B. 2
C. 4
D. 8
E. 24

C

The clearing made by bacteriophages in a "lawn" of bacteria on an agar plate is called a ________.
A) clear zone
B) lysogenic zone
C) prophage
D) plaque
E) host range

D

Heritability of corn height is 0.5. Farmer Bill's crop averages 8 ft tall. He takes all of his 10 foot tall plants and crosses them. The
expected average height of the resultant offspring is:
A. 8 ft
B. 8.5 ft
C. 9 ft
D. 9.5 ft
E. 10 ft

C

Assume that a cross is made between AaBb and aabb plants and that the offspring occur in the following numbers: 106 AaBb, 48
Aabb, 52 aaBb, 94 aabb. These results are consistent with the following circumstance:
A) sex-linked inheritance with 30% crossing over.
B) linkage with 50% crossing over.
C) linkage with approximately 33 map units between the two gene loci.
D) independent assortment.
E) 100% recombination.

C

Outbreeding in a population tends to:
A. select for dominant alleles
B. decrease heterozygotes
C. eliminate recessive alleles
D. decrease homozygotes

D

If 4% of individuals in a population express a recessive gene, what percentage of the total population are heterozygous for the
gene?
A. 4%
B. 10%
C. 16%
D. 32%
E. 64%

D

14) The following progeny phenotypes result from a cross of an A/a B/b individual with an a/a b/b individual
A B 300
a b 300
A b 200
a B 200
What is the recombination rate between A and B?
A. 0.1
B. 0.2
C. 0.3
D. 0.4
E. 0.5

D

An allele affecting antler size has a frequency of .25 in one group of caribou, but is not found in another group living in a nearby
mountain valley. This is an example of:
A. inbreeding
B. mutation
C. genetic drift
D. gene flow
E. selection

C

Three years later, the same caribou groups are surveyed and both now have a frequency of .20 for the antler size allele. This is an
example of:
A. inbreeding
B. mutation
C. genetic drift
D. migration
E. selection

D

In a population of people, 8% of the males are red/green color blind (X-LINKED). Assuming the population is in equilibrium,
what proportion of the females would be expected to be afflicted?
A. 0.64%
B. 4%
C. 64%
D. 6.4%
E. 0.32%

A

In which form of bacterial recombination does the bacteria find and take up the DNA instead of having it injected into the cell?
A. Conjugation
B. Interference
C. Transformation
D. Transduction
E. Liposuction

C

Cheetahs are an extreme example of
A. mutation
B. migration
C. genetic drift
D. inbreeding

C

Villagers in isolated villages in the mountains tend to have higher rates of recessive genetic diseases because of:
A. lower ozone levels at high altitudes
B. less diverse diets
C. increased inbreeding
D. no cable TV

C

Natural selection:
A) eliminates alleles from a population
B) has no affect on allele frequency
C) increases heterozygosity
D) alters allele frequencies, but usually does not remove alleles from a population
E) increases homozygosity

B

In a population of 10,000 individuals, 100 express a recessive trait. How many heterozygotes are there expected to be?
A. 600
B. 1,000
C. 1,800
D. 2,400
E. 3,000

C

The power of bacterial genetics is
A. Large number of genes so mapping can be done more completely
B. Many varied phenotypes so markers are easy to determine
C. Haploid genome makes genetics easier to carry out
D. Very small organisms are easier to manipulate
E. Very fast generation time and large numbers of offspring

E

Why is mutation critical for change in the allele frequencies of a population
A. It creates the new mutations that are needed for variation
B. It drives the change of frequency of alleles
C. It can create the multiple alleles that are needed for true variation
D. It stops alleles from going into fixation by recreating lost alleles

A

Which of the following is not true of translocations?
A. They can cause speciation to occur
B. They usually alter expression of multiple genes
C. They can bring the control region from one gene next to a different gene
D. They reduce the number of viable gametes produced

B

Among the human autosomes, there are 3 viable trisomy conditions, all with developmental problems, but no viable monosomy
conditions. This is because:
A. Trisomies alter the balance of developmental signals and monosomies uncover recessive lethals
B. Both trisomies and monosomies uncover recessive lethals
C. Monosomies alter the balance of developmental signals and trisomies uncover recessive lethals
D. Both trisomies and monosomies alter the balance of developmental signals

A

In generalized transduction, a phage introduces a segment of donor DNA into the recipient cell. This is followed by
recombination of the donor fragment with the recipient chromosome. Which of the following must occur?
A) circularization of the donor fragment before recombination
B) a pair of crossovers between the donor segment and the recipient
C) degradation of one of the two strands of phage genome
D) replication of the donor segment before recombination
E) a single crossover between the donor segment and the recipient chromosome

B

If you wanted to release negative supercoiling in a DNA chromosome, which of the following would you mix with the DNA?
A. Shelterin
B. Histones
C. complementary DNA strands
D. topoisomerase

D

If a DNA molecule of 50 base pairs contains 15 cytosine bases (C), how many thymine bases will it have?
A. 10
B. 15
C. 30
D. 35
E. 50

D

Which of these sequences could form a hairpin?
Α. 5ʹ GGGGTTTTCCCC 3ʹ
Β. 5ʹ AAAAAAAAAAAA 3ʹ
Χ. 5ʹ ACACACACACAC 3ʹ
D. 5ʹ TTTTTTCCCCCC 3

A

Which activity is not associated with DNA polymerases?
A. Ability to “read” a template and incorporate appropriate nucleotides in the growing strand
B. 5’ to 3’ synthesis of the new strands
C. Ability to initiate DNA synthesis on a completely single stranded DNA molecule
D. 3’ to 5’ exonuclease
E. Proofreading

C

After sigma factor guides the RNA polymerase to the correct promoter site, it:
a) dissociates from the core enzyme
b) guides the polymerase through the transcription process
c) signals the polymerase when to terminate transcription and fall off the DNA
d) reassociates with the holoenzyme

A

The end of an RNA from a bacterial gene is likely to contain
a) Inverted repeats and a string of U’s
b) Direct repeats and a string of A’s
c) Inverted repeats and a string of A’s
d) Direct repeats and a string of U’s
e) Direct repeats and a string of G’s

A

Hershey and Chase differentiated between DNA and protein by:
A. labeling the DNA with 32Phosphorous, proteins with 35Sulfur
B. labeling the DNA with 35Sulfur, proteins with 32Phosphorous
C. labeling the DNA with cesium , proteins with chloride
D. labeling the DNA with 14Carbon, proteins with 3Hydrogen

A

Which is a mechanism that allows a single gene to encode more than one polypeptide?
A. Regulation of mRNA stability
B. Alternative RNA splicing
C. RNA interference
D. Reverse transcription
E. None of the above

B

Is the “replace RNA” enzyme A
A DNA Pol I
B DNA Pol III
C Helicase
D Primase
E Gyrase

A

12) Relieves torsional stress

A DNA Pol I
B DNA Pol III
C Helicase
D Primase
E Gyrase

E

13) Actually synthesizes RNA, not DNA

A DNA Pol I
B DNA Pol III
C Helicase
D Primase
E Gyrase

D

14) Has 5’-3’ exonuclease activity

A DNA Pol I
B DNA Pol III
C Helicase
D Primase
E Gyrase

A

15) Opens the double helix for replication machinery

A DNA Pol I
B DNA Pol III
C Helicase
D Primase
E Gyrase

C

16) Synthesizes the majority of the DNA during replication

A DNA Pol I
B DNA Pol III
C Helicase
D Primase
E Gyrase

B

After several generations of labelling in 15N media, Meselson and Stahl switched to 14N media for one generation. The DNA extracted
after one generation in 14N media was centrifuged in a cesium gradient. The DNA formed a single band with a density somewhere between
DNA grown in 15N media and DNA grown in 14N media. This data is consistent with:
A. conservative replication
B. semiconservative replication
C. dispersive replication
D. conservative and dispersive replication
E. semiconservative and dispersive replication

E

An in vitro transcription system that contains a bacterial gene initiates transcription, but from random points on the DNA. Which of the
following proteins most likely is missing from the reaction?
A. TATA-binding protein
B. RNA polymerase II
C. rho factor
D. sigma factor

D

The DNA replication enzyme whose function most closely resembles RNA polymerase is
A. DNA Polymerase I
B. DNA Polymerase III
C. Primase
D. Telomerase
E. Helicase

C

For double-stranded DNA, which of the following base ratios usually does not equal 1?
A. (A+T)/(G+C)
B. (A+G)/(C+T)
C. C/G
D. (G+T)/(A+C)
E. A/T

A

An siRNA is processed by which enzyme into the 21 base pair fragment?
A. Spicer
B. Cutter
C. Cleaver
D. Dicer
E. RISC

D

Given the DNA sequence 3’-ACGCTACGTC-5’ which is properly hybridized to its complementary strand, how many hydrogen bonds
hold the two strands together?
A. 10
B. 16
C. 24
D. 26
E. 30

D

A DNA molecule is 900 bp long and has 100 complete rotations. This molecule is
a. Positively supercoiled
b. Negatively supercoiled
c. Relaxed

A

Which of the following need(s) to match correctly an anticodon and an amino acid?
A. Codon
B. mRNA
C. Aminoacyl-tRNA synthetase
D. Ribosome
E. More than one of the above

C

The genetic code is universal except for
A. prokaryotes, which use a different genetic code than eukaryotes.
B. a few mitochondrial genes, which substitute one sense codon for another.
C. viruses, which use an entirely different genetic code.
D. archaebacteria, which have their own genetic code.

B

To translate a mRNA you require two other RNAs. These are
A. tRNA and mRNA
B. tRNA and miRNA
C. rRNA and tRNA
D. rRNA and snRNA

C

Which of the following ratios of molar quantities of histones in DNA would be the largest?
a) H3/H4
b) H2A/H2B
c) H1/H4
d) H2A/H1
e) H4/H3

D

What would Avery, Macleod, and McCarty have concluded if their results had been that only RNAse treatment of the heat-killed bacteria
prevented transformation of genetic virulence?
a. that DNA was the genetic material
b. that RNA was the genetic material
c. that protein was the genetic material
d. that DNAse or protease, but not RNase, stimulates transformation

B

Suppose that Hershey and Chase found that phage ghosts contained 32P label but the label was absent from infected E. coli.
Furthermore, suppose they found 35S lacking in the ghosts and present in the infected E. coli. They would have concluded:
a. that protein was the genetic material in phage.
b. that DNA was the genetic material in phage.
c. that somehow the radioactivity prevented DNA from getting into E. coli.
d. that protein and DNA together made up the genetic material.

A

What linkage is present at the 5’ cap of an mRNA?
a) 3’ to 3’
b) 5’ to 5’
c) 5’ to 3’
d) 3’ to 5’
e) 5’ to 2’

B

What is one difference between DNA replication in bacteria versus eukaryotes?
A) Eukaryotic chromosomes are replicated bidirectionally, while bacterial chromosomes are replicated in one direction.
B) Eukaryotic chromosomes have many origins of replication and replicate bidirectionally, while bacteria have only one origin of replication
and replicate unidirectionally.
C) Bacterial chromosomes are replicated bidirectionally, while eukaryotic chromosomes are replicated in one direction.
D) Eukaryotic chromosomes have many origins of replication, while bacteria have only one origin of replication.
E) The process is identical in bacterial and eukaryotic DNA replication.

D

What enzyme(s) is/are responsible for removal of RNA primers and joining of Okazaki fragments?
A) DNA polymerase I
B) DNA polymerase III
C) DNA ligase
D) DNA polymerase I and DNA ligase
E) DNA polymerase III and DNA ligase

D

During initiation of DNA replication in E. coli, what is the role of helicase?
a. It binds to the origin and causes a short section of the double helix to unwind.
b. It binds to and stabilizes the single-stranded DNA.
c. It reduces torsional strain by making double-stranded DNA breaks ahead of the replication fork.
d. It breaks hydrogen bonds between the two DNA strands.

D

In human beings, which sequence will have the largest Cot 1/2?
A. telomeres
B. parasitic DNA
C. ribosomal RNA gene
D. a gene involved in eye color

D

RNA interference blocks further transcription of the target gene by
A. Methylating the histones
B. Methylating the cytosines in the gene
C. Degrading the nucleotides of the gene
D. Binding on to the DNA to block transcription

B

A chromosome with a centromere in the middle is called
A. metacentric
B. submetacentric
C. acrocentric
D. telocentric
E. centrocentric

A

Which of the following is the indivisible unit of genetic information
a) gene
b) allele
c) chromosome
d) ribosome

A

A man expressing a Y-linked trait marries a woman who does not express the trait. What do the rules of probability indicate
for the ratio of their offspring?
a) All sons and all daughters affected
b) Half of sons and half of daughters affected
c) Half of sons and all daughters affected
d) All sons and half of daughters affected
e) All sons and no daughters affected

E

In a cross of AaBbCcDdEe X AaBbccDDEe, what proportion will have the ABCde phenotype?
a) 0
b) 1/4
c) 27/64
d) 27/128
e) 37/128

A

B is a dominant giving black color. b is the recessive allele causing blue color. In a population of 100 B/b
individuals, ten purple (blue mixed with black) animals are seen. This is an example of
A. incomplete dominance
B. partial dominance
C. reduced penetrance
D. reduced expressivity
E. poor record keeping

D

In a cross of two double heterozygotes A/a B/b, what proportion will be dominant for a, recessive for b?
A. 1/16
B. 3/16
C. 4/16
D. 5/16
E. 9/16

B

The chromatids in a pair of chromosomes are held together at a specific region of the chromosome called the
A. Centromere
B. Mitotic spindle
C. Centrosome
D. Chromosome binding site

A

A cross of two heterozygous individuals produces 78 dominants and 22 recessives. What is the chi-square value
for these results?
A. 8/25
B. 16/75
C. 4/75
D. 8/75
E. 12/25

E

Which of the following results from a physical exchange between chromatids of homologous chromosomes?
A. Bivalent
B. Meiotic division
C. Chiasma
D. Tetrad
E. Synapsis

C

In a cross of two flies +/vg Cy/+ +/se +/ri X +/vg Cy/+ se/se +/ri what proportion of the offspring will be
mutant in phenotype for all four markers?
A. 0
B. 1/4
C. 1/32
D. 1/64
E. 3/128

E

Which of the following occurs only in Meiosis II:
A. Separation of homologues
B. Synaptonemal complex formation
C. Formation of spindle poles
D. Division of centromeres
E. None of the above

D

Women who are heterozygous for an X-linked condition causing the absence of sweat glands exhibit patches of
skin with sweat glands and patches lacking sweat glands. This patchwork effect is the result of
A. Mosaicism
B. Mitotic segregation
C. All heterozygotes would be expected to display this phenomenon regardless of sex
D. Dosage compensation
E. Both A and D

E

Color blindness is X-linked recessive and blood type is autosomal. If two parents who are both Type A and have
normal vision produce a son who is color blind and type O, what is the probability that their next child will be a son
who has normal vision and is blood type A?
A. 0
B. 1/8
C. 1/4
D. 3/16
E. 1

D

A drosophila with 3 X chromosomes, a Y chromosome and 2 sets of autosomes will be a
A. normal male
B. normal female
C. metafemale
D. intersex
E. metamale

C

On average, what proportion of the X-linked genes in a daughter are the same as her father?
a) 0
b) 1/4
c) 1/2
d) 3/4
e) 1

C

XYY can only occur because of a normal gamete fusing with a gamete which is the result of nondisjunction in:
A. Meiosis II in male
B. Meiosis I and II in male
C. Meiosis I in male
D. Meiosis II in female
E. Meiosis I and II in female

A

Which of the following is not a reason that the garden pea was a good genetic organism
A. rapid breeding rate
B. true-breeding lines
C. easily identified traits
D. multiple chromosomes
E. ability to control mating

D

white (w) is an X-linked recessive and tinman (tn) is an autosomal recessive mutation. What proportion of white,
tinman females (relative to whole population) is expected in the F2 starting with a true breeding white female which is
wild type for tinman mating with a true breeding male mutant only for tinman.
a) 0
b) 1/32
c) 1/16d) 1/8
e) 3/16

C

A drosophila with 2 X chromosomes, a Y chromosome and 3 sets of autosomes will be a
A. normal male
B. normal female
C. metafemale
D. intersex
E. metamale

D

The mitotic stage in which the chromosomes condense, nuclei disappear, and the mitotic spindle forms is known
as
A. Metaphase
B. Prophase
C. Anaphase
D. Cytokinesis

B

A colorblind mother and normal vision father have a child who is XXXXYY and colorblind. This is the result of:
A. Nondisjunction in Meiosis I in the father
B. Nondisjunction in Meiosis I in the father and mother
C. Nondisjunction in Meiosis I in the father and Meiosis I and II in the mother
D. Nondisjunction in Meiosis II in the father and Meiosis I and II in the mother.
E. Nondisjunction in Meiosis I and II in the mother

D

A husband and wife have normal vision, although both of their fathers are colorblind. What is the
probability that their first child will be a daughter who is colorblind?
A. 0
B. 1/8
C. 1/4
D. 1/2
E. 1

A

A cross of two heterozygous individuals produces 77 dominants and 23 recessives. What is the chisquare
value for these results?
A. 8/25
B. 16/75
C. 4/75
D. 8/75
E. 16/25

B

The situation in which an expected phenotype is not expressed in the numbers predicted is referred to
as
A. Pleiotropy
B. Variable expressivity
C. Incomplete penetrance
D. Recessivity

C

A colorblind mother and normal vision father have a child who is XXXXYY and colorblind. This is the
result of:
A. Nondisjunction in Meiosis I in the father
B. Nondisjunction in Meiosis I in the father and mother
C. Nondisjunction in Meiosis I in the father and Meiosis I and II in the mother
D. Nondisjunction in Meiosis II in the father and Meiosis I and II in the mother.
E. Nondisjunction in Meiosis I and II in the mother

D

Mating of two organisms produces a 1:1 ratio of phenotypes in the progeny. The parental genotypes are
A. Aa x Aa
B. Aa x aa
C. AA x aa
D. AA x AA

B

Color blindness is X-linked recessive and blood type is autosomal. If two parents who are both Type A
and have normal vision produce a son who is color blind and type O, what is the probability that their next
child will be a female who has normal vision and is type O?
A. 0
B. 1/8
C. 1/4
D. 1/2
E. 1

B

Two parents with normal vision have a child with only one sex chromosome. The child is colorblind.
This is the result of nondisjuntion in:
A. Either Meiosis I or II in male
B. Meiosis II in male
C. Either Meiosis I or II in female
D. Meiosis I in female
E. Meiosis I in male

A

36) Metaphase
A. Chromosome replication occurs
B. Sister chromatids begin to separate
C. Chromosomes become aligned
D. Replicated chromosomes begin to condense
E. Cytoplasm divides

C

37) Anaphase
A. Chromosome replication occurs
B. Sister chromatids begin to separate
C. Chromosomes become aligned
D. Replicated chromosomes begin to condense
E. Cytoplasm divides

B

38) Telophase E
A. Chromosome replication occurs
B. Sister chromatids begin to separate
C. Chromosomes become aligned
D. Replicated chromosomes begin to condense
E. Cytoplasm divides

E

39) Interphase
A. Chromosome replication occurs
B. Sister chromatids begin to separate
C. Chromosomes become aligned
D. Replicated chromosomes begin to condense
E. Cytoplasm divides

A

40) Prophase

A. Chromosome replication occurs
B. Sister chromatids begin to separate
C. Chromosomes become aligned
D. Replicated chromosomes begin to condense
E. Cytoplasm divides

D

A chromosome with a centromere close to the end is called
A. metacentric
B. submetacentric
C. acrocentric
D. telocentric
E. centrocentric

C

Which of the following can result in dicentric chromosomes during meiosis?
A. duplication
B. paracentric inversion
C. pericentric inversion
D. deletion
E. translocation

B

Mendel's second law or fourth postulate states that alleles of different genes segregate independently of each
other. At the chromosome level this is because:
a) Synthesis of chromosomes in Interphase
b) Chiasma forming during Prophase I
c) Chromosomes aligning in metaphase I
d) Chromosomes aligning in metaphase II
e) chromosomes separating in anaphase II

C

Agouti yellow and Siamese coat color are examples of
A. interdominance
B. codominance
C. pleitropy
D. incomplete dominance
E. lethal genes

C

The following progeny phenotypes result from a cross of an A/a B/b individual with an a/a b/b individual
A B 400
a b 400
A b 100
a B 100
What is the recombination rate between A and B?
A. 0.1
B. 0.2
C. 0.3
D. 0.4

B

In mammalian cells, the only autosomal chromosome abnormality found at any frequency is
A) monosomy
B) trisomy
C) autopolyploidy
D) allopolyploid

B

True breeding white flowering plants are crossed to true breeding red flowering plants. The offspring are red flowered.
When these flowers are self crossed, the resultant offspring are 147 red and 12 white flowered plants. Which of the following
pathways is most likely the cause of this
A. white ---A--->pink---B--->red
B. pink ---A--->pink---B--->red
C. red<---A---white ---B--->red
D. white ---A--->pink white---B--->pink both genes producing pink together give red

C

Gene A converts compound A (green) to compound B (blue). Compound A is lethal to the organism if not broken down
through this process. Gene B converts compound B to compound C (red). These compounds determine the color of the
individual. In a self cross of A/a B/b individuals, what proportion of green to blue offspring should be seen?
A. 1:16
B. 3:1
C. 9:3:3:1
D. 0:4
E. 4:3

D

Which of the following is the most likely description of a trait which shows up rarely in a family tree equally in both sexes
and only after a consanguineous mating?
A. Recessive autosomal
B. Dominant Autosomal
C. X-Linked
D. Y-linked
E. Either X or Y-linked

A

An individual has four X chromosomes and no Y but appears to be a male. This is likely due to:
A. Love of remote controls
B. Transferred testosterone receptor
C. Transferred PAR
D. Transferred SRY

D

In a population of 10,000 individuals, 200 men are afflicted with a recessive, X-linked disease. How many woman would be
expected to be afflicted in this population?
A. 1
B. 2
C. 4
D. 8
E. 24

D

Disease A affects .0004 of people, Disease B affects .0009 of people, X-linked Disease C affects .0025 of males. The allele
frequencies for the recessive, disease causing alleles are:
A. .0004 for a, .0009 for b, .0025 for c
B. .02 for a, .03 for b, .05 for c
C. .04 for a, .02 for b, .0025 for c
D. .02 for a, .03 for b, .0025 for c
E. .03 for a, .02 for b, .05 for c

D

Joe the Farmer decides to increase the weight of the eggs produced by his chickens. He knows that the heritability of egg
weight is 0.6. If he selects for chickens producing eggs which are 5 grams heavier than the average, how much heavier will be
the eggs produced by their offspring?
A. 1
B. 2
C. 3

D. 4
E. 5

C

True breeding white flowering plants are crossed to true breeding red flowering plants. The offspring are red
flowered. When these flowers are self crossed, the resultant offspring are 87 red and 72 pink flowered plants. Which
of the following pathways is most likely the cause of this
A. white ---A--->pink---B--->red
B. pink ---A--->pink---B--->red
C. red<---A---pink ---B--->red
D. white ---A--->pink white---B--->pink both genes producing pink together give red

B

In a population at Hardy-Weinberg equilibrium, 25% of the individuals have the recessive phenotype. What is the percentage
of heterozygotes in the population?
a. 24%
b. 36%
c. 48%
d. 50%
e. 72%

D

n a population of 10,000 individuals, 100 express a recessive trait. How many homozygotes for the dominant allele are there
expected to be?
A. 600

B. 1,000
C. 1,800
D. 6,400
E. 8,100

E

11) Maintains the single-strand form of DNA
A SSB
B DNA Pol III
C Helicase
D Primase
E Telomerase

A

12) Is often overactive in cancer cells
A SSB
B DNA Pol III
C Helicase
D Primase
E Telomerase

E

4) Coordinate replication requires two
A SSB
B DNA Pol III
C Helicase
D Primase
E Telomerase

B

15) Opens the double helix for replication machinery A SSB
B DNA Pol III
C Helicase
D Primase
E Telomerase

C

XXXXY can only occur because of a normal gamete fusing with a gamete which is the result of nondisjunction in:
A. Meiosis II in male
B. Meiosis I and II in male
C. Meiosis I in female
D. Meiosis II in female
E. Meiosis I and II in female

E

In the Meselson Stahl experiment differentiating the modes of replication by labeling with different forms of nitrogen and spinning in a
cesium gradient, how many rounds of replication were needed to determine if the following modes of replication were not used in E. coli?
A. one round of replication for dispersive; one round of replication for conservative
B. one round of replication for dispersive; two rounds of replication for conservative
C. two rounds of replication for dispersive; one round of replication for conservative
D. two rounds of replication for dispersive; two rounds of replication for conservative

C

In what form of bacterial recombination is a double recombination event not required to integrate the DNA into the chromosome?
A. Conjugation
B. Transformation
C. Transduction
D. Required in all forms
E. Not required in any forms

D

The form of DNA normally found in our cells is:
A. A
B. B
C. C
D. D
E. E

B

The function of DNA ligase is to
A. Catalyze the formation of hydrogen bonds between adjacent 5'-P and 3'-OH termini
B. Relax supercoiling of DNA
C. Add methyl groups to DNA
D. Facilitate base pairing between single-stranded molecules of DNA
E. Catalyze the formation of covalent bonds between adjacent 5'-P and 3'-OH termini

E

The pyrimidine bases are
A. Thymine and cytosine
B. Thymine and guanine
C. Adenine and guanine
D. Cytosine and adenine
E. Cytosine and guanine

A

Acts in a distance and orientation independent fashion
a)CAAT box or GC box
b) enhancer
c) TF IID
d) RNA Polymerase I
e) RNA Polymerase III

B

32) Produces large quantities of a few large polymers
CAAT box or GC box
b) enhancer
c) TF IID
d) RNA Polymerase I
e) RNA Polymerase III

D

33) upstream elements
CAAT box or GC box
b) enhancer
c) TF IID
d) RNA Polymerase I
e) RNA Polymerase III

A

34) Makes tRNA
CAAT box or GC box
b) enhancer
c) TF IID
d) RNA Polymerase I
e) RNA Polymerase III

E

35) a protein which binds to TATA boxes

a)CAAT box or GC box
b) enhancer
c) TF IID
d) RNA Polymerase I
e) RNA Polymerase III

C

produces mRNA
A) middle repetitive DNA
B) unique DNA
C) A site
D) EF-Tu
E) RF I

B

38) a protein shaped like an RNA
A) middle repetitive DNA
B) unique DNA
C) A site
D) EF-Tu
E) RF I

E

39) Required for translocation of ribosome
A) middle repetitive DNA
B) unique DNA
C) A site
D) EF-Tu
E) RF I

D

40) tRNA entry site
A) middle repetitive DNA
B) unique DNA
C) A site
D) EF-Tu
E) RF II

C

41) ribosomal RNA genes

A) middle repetitive DNA
B) unique DNA
C) A site
D) EF-Tu
E) RF Imiddle repetitive DNA

A

4) needed for splicing
a) CAAT box or GC box
b) snRNP
c) TF IID
d) 5’ cap
e) telomerase

B

5) contains a triphosphate phosphodiester linkage
a) CAAT box or GC box
b) snRNP
c) TF IID
d) 5’ cap
e) telomerase

D

6) upstream elements
a) CAAT box or GC box
b) snRNP
c) TF IID
d) 5’ cap
e) telomerase

A

7) needed for translation
a) CAAT box or GC box
b) snRNP
c) TF IID
d) 5’ cap
e) telomerase

D

8) a protein which binds to TATA boxes
a) CAAT box or GC box
b) snRNP
c) TF IID
d) 5’ cap
e) telomerase

C

9) contains an RNA which is used as a template

a) CAAT box or GC box
b) snRNP
c) TF IID
d) 5’ cap
e) telomerase

E

The F factor is a:
a) genome
b) plasmid
c) plaque
d) prophage
e) lawn

B

Which of the following is not an RNA involved in transcription or translation?
a) tRNA
b) mRNA
c) gRNA
d) hRNA
e) snRNA

D

The genetic code is said to be degenerate because
A. mRNA is rapidly degraded
B. The code is not universal among organisms
C. Several codons direct the insertion of the same amino acid into a polypeptide chain
D. Frameshift mutations are tolerated
E. Stop codons may have corresponding tRNA molecules

C

Which level of protein structure deals with just the sequence of amino acids?
a) primary
b) secondary
c) tertiary
d) quaternary

A

The P site in the ribosome is named for
A. P-Diddy B. Protein C. Peptidyl D.Polysome

C

Which of the following describes the efficiency of an F’ strain in causing:
F- => F+ transfer of genetic
information
A) High High
B) High Low
C) Low High
D) Low Low

A

The signal for polyadenylation of the mRNA is
a) AAAAAA
b) AAUUUU
c) AAUAAA
d) AAAUAA
e) AUAUAU

C

27) Participates in addition of an amino acid to a growing polypeptide chain

a) aminoacyl synthetase
b) EF-Tu
c) tRNA
d) RFII
e) IF2

B

28) Interacts with A site when ribosome is on a stop codon
a) aminoacyl synthetase
b) EF-Tu
c) tRNA
d) RFII
e) IF2

D

29) helps assemble large and small ribosomal subunits
a) aminoacyl synthetase
b) EF-Tu
c) tRNA
d) RFII
e) IF2

E

30) carries the anticodon
a) aminoacyl synthetase
b) EF-Tu
c) tRNA
d) RFII
e) IF2

C

31) charges the tRNA a) aminoacyl synthetase
b) EF-Tu
c) tRNA
d) RFII
e) IF2

A

The A site in the ribosome is named for
A. Amino end B. Attila the Hun C. Aminoacyl D. Amino Acid

C

The snRNP which binds the 5’ splice junction is
a) U1
b) U2
c) U3
d) U4
e) U5

A

Discontinuous replication
A. Occurs on the DNA strand synthesized overall in the 3' to 5' direction
B. Occurs on the DNA strand synthesized overall in the 5' to 3' direction
C. Results in the formation of Okazaki fragments
D. Both A and C
E. Both B and C

D

Which level of protein structure includes alpha helix and beta pleated sheet?
a) primary
b) secondary
c) tertiary
d) quaternary

B

Which of the following is not different between eukaryotic and prokaryotic transcription?
a) Splicing
b) 5' capping
c) poly-adenylation
d) termination
e) all are different

E

The correct order of activity in replication is
a) helicase, SSB, Pol III, primase, Pol I, ligase
b) primase, helicase, SSB, Pol III, Pol I, ligase
c) SSB, helicase, primase, Pol III, Pol I, ligase
d) helicase, SSB, primase, Pol I, Pol III, ligase
e) helicase, SSB, primase, Pol III, Pol I, ligase

E

Which one of the following is not a stop codon?
A. UAA B. UGG C. UAG D. UGA

B

A DNA molecule of 500 bp will have approximately how many turns?
A)50
B)500
C)5,000
D)250
E)25

A

Which of the following statements is true?
a) Pol II promoters and Pol I promoters are internal
b) Pol II promoters and Pol III promoters are internal
c) Pol I promoters are at the 5’ end and Pol III promoters are internal
d) Pol III promoters are at the 5’ end and Pol I promoters are internal

D

Synthetic mRNAs were used to predict the genetic code. Which of the following is not true?
a) Repeats of the same nucleotide produce a single amino acid polymer
b) repeats of two nucleotides produce a polymer with a two amino acid repeat unit
c) repeats of three nucleotides produce a polymer with a three amino acid repeat unit
d) repeats of four nucleotides produce a polymer with a four amino acid repeat unit

C

Which level of protein structure deals with interactions of subunits?
a) primary
b) secondary
c) tertiary
d) quaternary

D

Mutant bacteria strains called auxotrophs can only grow in _______ medium.
a. complete
b. minimal
c. partially supplemented
d. no

A

The following figure shows the results of interrupted-mating experiments with 3 different Hfr strains. What is the
order of the genes, starting with C?
Hfr strain Order of transfer
1 A, B, E, D, F
2 D, F, C, G, A
3 D, E, B, A, G
a. C, G, A, D, F, B, E
b. C, F, D, B, A, E, G
c. C, B, E, D, F, G, A
d. C, G, A, B, E, D, F
e. C, D, F, G, A, B, E

D

Most cases of Down syndrome arise from:
a. inversions.
b. deletions.
c. X-rays.
d. maternal nondisjuction.
e. unequal crossing over.

D

The most common aneuploidy seen in living humans has to do with ___________.
a. autosomes
b. sex chromosomes
c. chromosome 21
d. chromosome 13
e. None of the above

B

Which of these RNA sequences could form a hairpin also called a stem loop?
a. 5′ GGGGTTTTCCCC 3′
b. 5′ AAAAAAAAAAAA 3′
c. 5′ ACACACACACAC 3′
d. 5′ TTTTTTCCCCCC 3′

A

Which of the following describes the efficiency of an Hfr strain in causing:
F- => F+ transfer of genetic
information
A) High High
B) High Low
C) Low High
D) Low Low

C

An in vitro transcription system transcribes a bacterial gene but terminates inefficiently. What is one possible
problem?
a. There is a mutation in the –10 consensus sequence, which is required for efficient termination.
b. Rho factor has not been added.
c. Sigma factor has not been added.
d. Spliceosomes have not been added

B

If the sequence of an RNA molecule is 5’-GGCAUCGACG-3’, what is the sequence of the template strand of
DNA?
a. 5’-GGCATCGACG-3’
b. 3’-GGCATCGACG-5’
c. 5’-CCGTAGCTGC-3’
d. 5’-CGTCGATGCC-3’
e. None of the above

D

Which is a mechanism that allows a single gene to encode more than one polypeptide?
a. Regulation of mRNA stability
b. Alternative RNA splicing
c. RNA interference
d. Reverse transcription
e. None of the above

B

25. Okazaki fragments
a. supercoil removal
b. RNA primer synthesis
c. 3′ -->5′ exonuclease activity
d. lagging strand
e. leading strand

D

26. DNA primase

a. supercoil removal
b. RNA primer synthesis
c. 3′ -->5′ exonuclease activity
d. lagging strand
e. leading strand

B

27. DNA gyrase
a. supercoil removal
b. RNA primer synthesis
c. 3′ -->5′ exonuclease activity
d. lagging strand
e. leading strand

A

28. continuous synthesis

a. supercoil removal
b. RNA primer synthesis
c. 3′ -->5′ exonuclease activity
d. lagging strand
e. leading strand

E

29. DNA proofreading

a. supercoil removal
b. RNA primer synthesis
c. 3′ -->5′ exonuclease activity
d. lagging strand
e. leading strand

C

30. heteroduplex DNA
a 5’ -> 3’ polymerase activity
b. strand invasion
c. phosphodiester bonds at DNA nicks
d. 5’->3’ exonuclease activity
e. bidirectional circular replication

B

31. DNA ligase
a 5’ -> 3’ polymerase activity
b. strand invasion
c. phosphodiester bonds at DNA nicks
d. 5’->3’ exonuclease activity
e. bidirectional circular replication

C

32. theta replication
a 5’ -> 3’ polymerase activity
b. strand invasion
c. phosphodiester bonds at DNA nicks
d. 5’->3’ exonuclease activity
e. bidirectional circular replication

E

33. DNA synthesis
a 5’ -> 3’ polymerase activity
b. strand invasion
c. phosphodiester bonds at DNA nicks
d. 5’->3’ exonuclease activity
e. bidirectional circular replication

A

34. Primer removal

a 5’ -> 3’ polymerase activity
b. strand invasion
c. phosphodiester bonds at DNA nicks
d. 5’->3’ exonuclease activity
e. bidirectional circular replication

D

5) Actually synthesizes RNA, not DNA
A DNA Pol II
B Telomerase
C Helicase
D Primase
E Gyrase

D

6) Maintains ends of chromosomes
A DNA Pol II
B Telomerase
C Helicase
D Primase
E Gyrase

B

7) Opens the double helix for replication machinery
A DNA Pol II
B Telomerase
C Helicase
D Primase
E Gyrase

C

8) Has no required function in DNA chromosome replication
A DNA Pol II
B Telomerase
C Helicase
D Primase
E Gyrase

A

9) contains an RNA molecule which is required for function

A DNA Pol II
B Telomerase
C Helicase
D Primase
E Gyrase

B

10) Relieves torsional stress

A DNA Pol II
B Telomerase
C Helicase
D Primase
E Gyrase

E

Termination of transcription in prokaryotes is carried out in part by a
A) trinucleotide repeat
B. LINE repeat
C. stem loop or hairpin structure
D. Bruce Bowen

C

23) Which of the following statements regarding gene expression is true?
A. Messenger RNA is translated in the 5' to 3' direction
B. Transcription involves the association of mRNA with ribosomes
C. mRNA undergoes proof-reading
D. Prokaryotic RNA usually undergoes nuclear processing
E. Polypeptides are synthesized by addition of amino acids to the amino terminus

A

The genetic code is said to be redundant because
A. mRNA is rapidly degraded
B. The code is not universal among organisms
C. Several codons direct the insertion of the same amino acid into a polypeptide chain
D. Frameshift mutations are tolerated
E. Stop codons may have corresponding tRNA molecules

C

Which of the following is not present as a dimer in a histone core?
A. H1
B. H2A
C. H2B
D. H3
E. H4

A

A DNA molecule of 500 bp will have approximately how many turns?
A)50
B)500
C)5,000
D)250
E)25

A

A drosophila with 4 X chromosomes, a Y chromosome and 4 sets of autosomes will be a
A. normal male
B. normal female
C. metafemale
D. intersex
E. metamale

B

Bacterial cells containing an F plasmid that has acquired bacterial chromosomal genes are called:
a. F+.
b. F′.
c. F–.
d. Hfr.

B

Translation is ended because of the action of
A. Initiation factors
B. Elongation factors
C. Termination factors
D. Release factors
E. Dissociation factors

D

In watchamakallits black fur is dominant to white. In a certain population of watchamakallits, white fur occurs in a frequency of
1/100. What is the frequency of the recessive allele in the population, assuming Hardy-Weinberg equilibrium?
a. 0.01
b. 0.1
c. 0.5
d. 0.9
e. 0.99
47. What is the frequency of heterozygous watchamakallits in the population described in question 46, assuming Hardy-Weinberg
equilibrium?
a. 0.01
b. 0.02
c. 0.18
d. 0.82
e. 0.99

B,C

bacterial cell transfers chromosomal genes to F– cells, but it rarely causes them to become F+. The bacterial cell is:
a. Hfr.
b. lysogenic.
c. auxtrophic.
d. lytic.

A

Which agent of evolution would tend to increase the frequency of all recessive phenotypes in a population relative to what is
expected from Hardy-Weinberg equilibrium?
a. Genetic drift
b. Inbreeding
c. Natural selection
d. Positive assortative mating
e. Negative assortative mating

B

A chromosome with a centromere near but not at the end is called
A. metacentric
B. submetacentric
C. acrocentric
D. telocentric
E. centrocentric

C

Pick the statement that is false.
a. A gene is the fundamental unit of heredity.
b. A genome is the complete set of DNA carried in an organism.
c. Alleles are alternative forms of a gene.
d. Genotype refers to the genes of an organism that determine a particular trait.
e. Gene pool is the total of all genes in an individual.

E

If adenine makes up 15 percent of the bases in a specific DNA molecule, what percentage of the bases are cytosine.
a. 15%
b. 30%
c. 35%
d. 60%
e. 70%

C

Assume that a cross is made between AaBb and aabb plants and all of the offspring are
eitherAaBb or aabb. These results are consistent with the following circumstance:
A. Complete linkage
B. Alternation of generations
C. Codominance
D. Incomplete dominance
E. Hemizygosity

A

DNA polymerase requires a 3’ OH group provided by a primer before it can begin DNA synthesis. How is it that the primer can be synthesized on the DNA without a 3’ OH group?

The primer is RNA-synthesized by RNA polymerase, which does not require a 3’ OH group.

Mismatch repair requires the ability to distinguish between template and newly synthesized DNA strands. How can E. colidistinguish between these two strands?

Template DNA is methylated

Eukaryotic DNA replication is similar to prokaryotic DNA replication. However, eukaryotes require unique replication strategies because of unique features of eukaryotic DNA. Which is NOT a unique feature linked to replication strategies?

Eukaryotic DNA contains introns.

Why is it that low-fidelity eukaryotic DNA polymerases are able to replicate DNA that contains abnormal bases, distorted structures, or bulky lesions, whereas high-fidelity DNA polymerases stall at these areas?

Low-fidelity DNA polymerases have large active sites that can bind to these regions.

You identify a mutant whose chromosomes shorten after each round of replication. A mutation in which gene would explain this observation?

telomerase

Understanding DNA recombination is important in understanding ___________.

a. DNA repair

b. gene linkage traits

c. genetic variation

d. all of the above

D

How would DNA replication be affected in a cell that is lacking topoisomerase?

Torsional strain ahead of the replication fork would eventually cause replication to stop

Which of the following is true of RNA compared to DNA?

RNA has a hydroxyl group on the 2’-carbon atom of its sugar component....

When RNA is transcribed from a gene, which strand of DNA is used?

the template strand

The transcription unit includes three essential regions. What is the proper order of these regions?

promoter, RNA coding sequence, terminator

What would the result be if a specific sigma subunit were mutated?

RNA polymerase would fail to initiate transcription at the promoter specific to the sigma subunit

The bacterial holoenzyme binds to which part of the promoter?

–10 and –35 consensus sequence

In rho-dependent transcription termination, the rho factor binds to ___________.

mRNA

Which is NOT one of the DNA sequences known to regulate gene transcription?

basal transcription apparatus

In eukaryotes, what initially binds to the TATA box on the DNA template?

TFIID

RNA polymerase I, II, and III terminate transcription differently. Which of the following statements is most correct?

a. RNA polymerase I requires a termination factor (like rho factor).

b. RNA polymerase II synthesizes hundreds of bases beyond what is needed for the mRNA, which is degraded by the protein Rat1.

c. RNA polymerase III terminates transcription after a termination sequence is transcribed.

d. All of the above statements are correct.......

D.

What is the most likely result from mutating a prokaryotic Shine–Dalgarno sequence?

The ribosome would not be able to bind to the mRNA.

Which mRNA processing event adds stability to the mRNA?

-addition of a 5’ cap -addition of a poly(A) tail to the 3’ end

What kind of RNA functions in splicing and is associated with the spliceosome?

small nuclear RNA (snRNA)

Which intron component is the first to be cleaved during the splicing process?

5’ splice site

What is it called when mRNA splicing occurs that results in variously sized mRNAs?

alternative 3’ cleavage sites

What process causes a protein amino acid sequence NOT to correspond with its DNA sequence?

RNA editing

Which arm of tRNA binds to mRNA?

anticodon arm

Which is NOT true of ribosomes?

They are made up of only ribosomal RNA molecules.

Small interfering RNA (siRNA) molecules block gene expression by degrading mRNA or inhibiting transcription. Which genes do they target?

genes from which they were transcribed

a. 5' untranslated region

ribosome binding site or Shine-Dalgarno sequence is found within the 5'untranslatedregion. However, eukaryotic mRNA does not have the equivalent sequence, and aeukaryotic ribosome binds at the 5' cap of the mRNA molecule.

b. Promoter

b. The promoter is the DNA sequence that the transcription apparatus recognizes and binds to initiate transcription.

c. AAUAAA consensus sequence

c. The AAUAAA consensus sequence lies downstream of the coding region of the gene. It determines the location of the 3' cleavage site in the pre-mRNA molecule.

d. Transcription start site

d. The transcription start site begins the coding region of the gene and is located 25 to 30 nucleotides downstream of the TATA box.

e. 3' untranslated region

e. The 3' untranslated region is a sequence of nucleotides at the 3' end of the mRNA that is not translated into proteins. However, it does affect the translation of the mRNA molecule as well as the stability of the mRNA.

f. Introns

f. Introns are noncoding sequences of DNA that intervene within coding regions of a gene.

h. Poly(A) tail

h. A poly(A) tail is added to the 3' end of the pre-mRNA. It affects mRNA stability.

i. 5' cap

i. The 5' cap functions in the initiation of translation and mRNA stability.

What kind of chemical bond holds adjacent amino acids together?

a peptide bond

The wobble rules account for non–Watson and Crick nucleotide pairing at which position of the codon and anticodon, respectively?

third and first

Any given DNA sequence has ______ possible reading frames, and the correct one is set by a(n) ___________________.

3; initiation codon

What is the final component of the initiation complex to be added?

a large subunit of the ribosome

The formation of peptide bonds during elongation is catalyzed by an enzyme called the ribozyme. The ribozyme is made of what macromolecule?

rRNA in the large subunit of the ribosome

Simultaneous transcription and translation does NOT occur in which of the following locations?

cytoplasm of eukaryotes

What are isoaccepting tRNAs?

Isoaccepting tRNAs are tRNA molecules that have different anticodon sequences but accept the same amino acids.

1. Which of the following chromatin changes will repress gene transcription?

DNA not methylated

which of the following is true of transcriptional activator proteins?

Some have acetyltransferase activity

What is the role of insulators?

They block the role of enhancers.

Which two mRNA structures interact with each other to provide mRNA stability?

5’ cap and 3’ poly(A) tail

siRNA and miRNA inhibit gene expression by all of the following EXCEPT

stabilizing translation machinery

When present in small amounts in sequencing reactions, dideoxyribonucleoside triphosphates (ddNTPs) terminate the sequencing reaction at different positions in the growing DNA strands. ddNTPs stop a sequencing reaction because they:

c. lack a hydroxyl (–OH) group at their 3′ end

All of the following are true of single nucleotide polymorphisms (SNPs) EXCEPT

they are rare among persons of the same ethnic group.

Shotgun sequencing does NOT:

depend on physical or genetic maps

Genetic and physical maps differ

a. in the order of the genes.

b. in that physical maps are based on recombination frequencies.

c. in that genetic maps are based on base pairs.

d None of the above.

D

Epigenetics is the study of:

inheritance of traits not coded by DNA sequence

How would you expect DNA methylation to alter gene expression

Measurably decrease expression

Cancer is a heterogeneous group of disorders resulting from

a. rapid uncontrolled cell division.

b. a multistep process that requires several different mutations.

c. Both a and b. :

C

Considering the clonal evolution of tumors, mutations in which kinds of genes would speed up the rate of accumulation of additional mutations?

DNA-repair genes

Which of the following is true of tumor-suppressor genes?

Their normal function is to promote cell proliferation.

b. Their mutant forms are typically dominant

c. They were the first cancer-causing genes to be identified.

d. None of the above is true of tumor-suppressor genes

D

In an experiment you mutate the retinoblastoma gene (RB) such that its gene product behaves as if hyperphosphorylated. The result would be a ______ association between RB and E2F, with _______ transcription of genes necessary for DNA replication.

weaker; continuous stimulation of

Which of the following is the major event associated with the retinoblastoma cancer?

Both copies of a tumor-suppressor gene are inactivated

Which of the following result(s) directly from metastasis?

Secondary tumors

Which of the following statements is false?

Most tumors arise from germ-line mutations that accumulate during our life span

Normal cellular genes whose products are involved in facilitating cell division to occur under appropriate conditions are called?

Proto-oncogenes

Most body (nonreproductive) cells of humans and other multicellular eukaryotes have two sets of each chromosome. Such cells are ______ and the matching pairs of chromosomes are called _________.

a. diploid; homologous chromosomes

3. Meiosis provides genetic variability by:

a. crossing over during prophase I.

b. random assortment of homologs during anaphase I.

C. Both a and b

C

Are mutations that change chromosome numbers and structure a cause of cancer or the result of cancer

a. cause

b. result

c. both

C

According to Mendel’s second law, when the different alleles for one trait separate into gametes, their separation:

is independent of how different alleles for other traits separate

When Morgan crossed a red-eyed female with a white-eyed male, which results made Morgan think that the locus affecting eye color was on the X chromosome?

In the F2, all females had red eyes, and half of the males had red eyes and the other half had white eyes.

he modification of the amount of protein produced by a sex chromosome is called:

dosage compensation.

The difference between dominance and epistasis is that:

epistasis masks genes at different loci.

In a certain species of plant, flowers occur in three colors: blue, pink, and white. A pure-breeding pink plant is mated with a pure-breeding white plant. All of the F1 are blue. When the blue F1 plants are selfed, the F2 occur in the ratio 9 blue:3 pink:4 white. What is the name for this type of interaction?

Recessive epistasis

A mother with blood type AB has a child with blood type B. Give all possible blood types for the father of this child

A, B, AB, O

You are studying cystic fibrosis (CF). While looking at a pedigree you notice that the CF phenotype is not present in a set of parents, but one out of their five children has CF. What can you conclude about CF?

The trait is autosomal recessive.

If genes A and B are linked, what is the maximum percentage of recombinant gametes that can be produced if a single crossover occurs during gametogenesis?

50%

What is the most likely order of the linked genes R, S, and T if the distance between R and S is 22 m.u., the distance between S and T is 8 m.u., and the distance between R and T is 14 m.u.?

S T R

A testcross include

one parent who shows the dominant phenotype for one or more genes and a second parent who is homozygous recessive for these genes.

Which is NOT a type of chromosomal mutation?

point mutation

After the first round of replication, Meselson and Stahl saw only one DNA band of density intermediate to DNA containing only 15N or 14N. After this observation, which hypothesis for DNA replication could be eliminated?

conservative

When RNA is transcribed from a gene, which strand of DNA is used?

the template strand

When a eukaryotic mRNA is hybridized to the complementary DNA, loops of unhybridized DNA are seen. These loops:

a. correspond to noncoding regions of the gene.

c. demonstrate that genes and proteins are not colinear.

d. Both a and c.

D

ou are studying a biochemical pathway and isolate mutants I, II, and III. Mutant I can grow if you supplement the medium with Z. Mutant II can grow if you supplement the medium with X, Y, or Z. Mutant III can grow if you supplement the medium with X and Z, but not if you supplement the medium with Y. What is the order of X, Y, and Z in this biochemical pathway?

Y, X, Z

A mutation in gene X overrides the effect of a previous mutation also in gene X and restores wild-type phenotype. The second mutation in gene X is called

intragenic suppressor mutation.

You want to design a repressor protein mutant. Which protein domain is the best target for preventing binding of the
corepressor?
A. A) DNA-binding domain
B. B) allosteric domain
C. C) promoter domain
D. D) helix-turn-helix domain
E. E) activator binding site

B

Given the DNA sequence 5′-TAC AAA ATA CAG CGG-3′, which of these sequences represents a frameshift mutation?
A) 5′-TAG AAA ATA CAG CGG-3′
B) 5′-TAC AAA TAC AGC GGG-3′
C) 5′-TAC AAG ATA CAG CGG-3′
D) 5′-TAC AAA ATA CAC CGG-3′
E) 5′-TAC AAA ATA CAG AGG

B

Which position of a codon evolves at the highest rate?
a. All positions of a codon evolve at the same rate.
b. first position
c. second position
d. third position
e. The first and third positions evolve at the same rate.

D

A mutant E. coli strain, grown under conditions that normally induce the lac operon, does not produce ß-galactosidase.
What is a possible genotype of the cells?
a. lacI+ lacP+ lacO+ lacZ+ lacY– lacA+
b. lacI+ lacP+ lacOc lacZ+ lacY+ lacA+
c. lacl+ lacP+ lacO+ lacZ+ lacY+ lacA+
d. lacI- lacP+ lacO+ lacZ+ lacY+ lacA+

A

Which type of mutation converts a nucleotide to an alternative structure with the same composition but slightly different placement
of hydrogen bonds with a rare, less stable form that causes base-pair mismatch?
A) depurination
B) deamination
C) tautomeric shift #####
D) transition
E) transversion

C

You have conducted an Ames test on a given compound. Which of the following would be classified as a positive result on the Ames
test?
A) His- strain grows on an his- plate.
B) His- strain grows on an his+ plate.
C) His+ strain grows on an his- plate.
D) His+ strain grows on an his+ plate.
E) His+ strain grows on either an his- or an his+ plate.

A

Which of these haploid strains produce β-galactosidase constitutively but do not produce permease?
A) I- P+ O+ Z+ Y+
B) I+ P+ O+ Z- YC)
I- P+ O+ Z-Y+
D) I+ P+ O- Z+ Y+
E) I- P+ O+ Z+ Y-

E

During the attenuation of the trp operon, which stem loop leads to polycistronic mRNA synthesis during tryptophan starvation?
A) 1-3 (antitermination) stem loop
B) 3-4 (termination) stem loop
C) 1-2 (pause) stem loop
D) 2-3 (antitermination) stem loop
E) 2-4 (termination) stem loop

D

In sporadic cases of retinoblastoma, how many gene mutations are thought to be necessary in the Rb genes of the same
cell for a tumor to develop?
A) one
B) two
C) four
D) six
E) There is insufficient information to answer this question

B

Nutritional mutations can be defined as ________.
A) those mutations that do not allow a bacterium or fungus to grow on minimal medium but do allow growth on complete
medium
B) those mutations that change the composition of the medium
C) those mutations belonging to the group called prototrophs
D) those mutations caused by site-specific mutagenesis
E) all strains that are not auxotrophic

A