Latest one exam 3(BIO)Answered
In the Hershey and Chase experiment, the pellet was radioactive after bacteria had
been infected with 32P-labeled viruses and centrifuged. Why?
A. Viruses were centrifuged to form the pellet, and they had incorporated radioactive
proteins from the bacterial DNA.
B. Viruses were centrifuged to form the pellet, and they had incorporated radioactive
DNA.
C. Bacteria were centrifuged to form the pellet, and they had incorporated radioactive
DNA.
D. Bacteria were centrifuged to form the pellet, and they had incorporated radioactive
proteins into their cell membranes.
E. Bacteria were centrifuged to form the pellet, and they had incorporated radioactive
proteins into their DNA.
{{ANS}}c Bacteria were centrifuged to form the pellet, and they had incorporated
radioactive DNA.
Which of the following nucleotide sequences represents the complementary sequence
that would bind to the DNA strand
5' - TCATGG - 3'?
a. 5' - GGTACT - 3'
b. 5' - TTGCAG - 3'
c. 3' - CCATGA - 5'
d. 3' - TCATGG - 5'
e. 3' - AGTACC - 5'
{{ANS}}e. 3' - AGTACC - 5'
The central dogma describes the flow of information of gene expression as
a. protein → DNA → RNA.
b. DNA → RNA → protein.
c. RNA → DNA.
d. DNA → protein → RNA.
e. RNA → DNA → protein.
{{ANS}}b. DNA → RNA → protein.
During DNA replication, the _ strand is assembled in the _ direction that the DNA
double helix unwinds and is produced by _ replication.
a. lagging; opposite; discontinuous
b. leading; same; discontinuous
c. lagging; same; continuous
,d. leading; opposite; continuous
e. leading; opposite; discontinuous
{{ANS}}a. lagging; opposite; discontinuous
Substitution of one base pair for another in a coding region of a gene can result in a
____________ mutation where the changed codon still codes for the same amino acid
as the original codon did.
a. chromosomal
b. missense
c. frameshift
d. nonsense
e. silent
{{ANS}}e. silent
Small ribonucleoprotein particles (snRNPs, or "snurps") are involved in ________.
a. initiation of transcription
b. initiation of translation
c. mRNA splicing
d. termination of translation
e. aminoacylation of tRNA
{{ANS}}c. mRNA splicing
Stopping transcription in eukaryotes requires:
a. generation of a poly-A tail.
b. splicing introns out and exons together.
c. generation of a stop codon.
d. copying a terminator sequence in mRNA.
e. activation of gene repressors.
{{ANS}}a. generation of a poly-A tail.
As a ribosome translocates along an mRNA molecule by one codon, which of the
following occurs?
a. The tRNA that was in the A site moves into the P site.
b. The polypeptide enters the E site.
c. The tRNA that was in the P site moves into the A site.
d. The tRNA that was in the A site departs from the ribosome via a tunnel.
e. The tRNA that was in the A site moves to the E site and is released.
{{ANS}}a. The tRNA that was in the A site moves into the P site.
RNA polymerase moves along the template strand of DNA in the _ direction of the
template strand, and adds nucleotides to the _ end of the growing transcript.
a. 3' to 5'; 5'
b. 5' to 3'; 3'
c. 3' to 5'; 3'
d. 5' to 3'; 5'
{{ANS}}c. 3' to 5'; 3'
,The DNA of an organism is studied and found to contain 36% cytosine. This organism
should have _ % guanine and _ % adenine in its DNA.
A. 14; 36
B. 14; 14
C. 86; 14
D. 36; 14
E. 36; 36
{{ANS}}D. 36; 14
The process of adding the correct amino acid onto a tRNA molecule is catalyzed by
________.
a. the tRNA itself
b. an mRNA
c. the ribosome
d. an aminoacyl-tRNA synthetase
e. RNA polymerase
{{ANS}}d. an aminoacyl-tRNA synthetase
In roses assume that red or yellow flower color is controlled by a single gene with two
alleles. Crossing roses with yellow flowers with each other yields only offspring that
produce yellow flowers, but when you cross roses with red flowers with each other you
sometimes get offspring that produce yellow flowers. If you take a rose plant that is
heterozygous for the flower color gene and cross it with another rose plant with red
flowers that has had yellow-flowered offspring in the past, what are the predicted
fractions for the possible phenotypes of the offspring?
a. 4/4 red, 0/4 yellow
b. 1/2 red, 1/2 yellow
c. 1/4 red, 3/4 yellow
d. 3/4 red, 1/4 yellow
e. 0/4 red, 4/4 yellow
f. cannot be determined
{{ANS}}d. 3/4 red, 1/4 yellow
in humans, one form of polydactyly (extra fingers and toes) is controlled by a single
autosomal gene with two possible alleles; the dominant allele leads to polydactyly, the
recessive is for the normal number of fingers and toes. A woman with this form of
polydactyly whose extra digits were removed when she was a baby mates with a man
who also has this form of polydactyly and who also had his extra digits removed when
he was a baby. Their first child is born with the normal number of fingers and toes. What
the odds that their second child will also be born with the normal number of fingers and
toes? Give your answer as a decimal fraction (for example, you would enter 0.85 for an
85% chance).
a. .50
b. .25
c. 1.0
, d. .75
{{ANS}}b. .25
3
In peas, the allele for tall (T) is dominant over short (t), and the unlinked gene for seed
color has the allele for yellow seeds (G) dominant over green seeds (g). What is the
predicted phenotypic ratio for offspring from this cross: TtGg x TtGg
a - 1 tall, yellow: 1 tall, green
b - 3 tall, yellow: 1 tall, green
c - 1 short, yellow: 1 short, green
d - 3 short, yellow: 1 short, green
e - 1 tall, yellow: 1 short, yellow
f - 3 tall, yellow: 1 short, yellow
g - 1 tall, green: 1 short, green
h - 3 tall, green: 1 short, green
i - 1 tall, yellow: 1 tall, green: 1 short, yellow: 1 short, green
j - 3 tall, yellow: 3 tall, green: 1 short, yellow: 1 short, green
k - 3 tall, yellow: 1 tall, green: 3 short, yellow: 1 short, green
l - 9 tall, yellow: 3 tall, green: 3 short, yellow: 1 short, green
{{ANS}}l - 9 tall, yellow: 3 tall, green: 3 short, yellow: 1 short, green
In peas, the allele for tall (T) is dominant over short (t), and the unlinked gene for seed
color has the allele for yellow seeds (G) dominant over green seeds (g). What is the
predicted phenotypic ratio for offspring from this cross: Ttgg x TtGg
a. 1 tall, yellow: 1 tall, green
b. 3 tall, yellow: 1 tall, green
c. short, yellow: 1 short, green
d. 3 short, yellow: 1 short, green
e. 1 tall, yellow: 1 short, yellow
f. 3 tall, yellow: 1 short, yellow
g. 1 tall, green: 1 short, green
h. 3 tall, green: 1 short, green
i. 1 tall, yellow: 1 tall, green: 1 short, yellow: 1 short, green
j. 3 tall, yellow: 3 tall, green: 1 short, yellow: 1 short, green
k. 3 tall, yellow: 1 tall, green: 3 short, yellow: 1 short, green
l. 9 tall, yellow: 3 tall, green: 3 short, yellow: 1 short, green
{{ANS}}j. 3 tall, yellow: 3 tall, green: 1 short, yellow: 1 short, green
In peas, the allele for tall (T) is dominant over short (t), and the unlinked gene for seed
color has the allele for yellow seeds (G) dominant over green seeds (g). What is the
predicted phenotypic ratio for offspring from this cross: Ttgg x ttGG
a. 1 tall, yellow: 1 tall, green
b. 3 tall, yellow: 1 tall, green
c. 1 short, yellow: 1 short, green
d. 3 short, yellow: 1 short, green
e. 1 tall, yellow: 1 short, yellow
