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AVIT 480 Final Questions and Answers (100% Correct Answers) Already Graded A+

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AVIT 480 Final Questions and Answers (100% Correct Answers) Already Graded A+ AVIT 480 Final Questions and Answers (100% Correct Answers) Already Graded A+

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BIO 325
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BIO 325











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Institution
BIO 325
Course
BIO 325

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Uploaded on
December 11, 2025
Number of pages
60
Written in
2025/2026
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BIO 325 Final Questions and Answers (100%
Correct Answers) Already Graded A+
Angelina Jolie's mother and maternal grandmother died of
ovarian cancer. Her maternal aunt died of breast cancer.
Angelina Jolie opted to have a blood test to see if she had
a BRCA1 mutation and found that one copy of the gene
was wild-type, and the other copy had a nonsense
mutation that would cause production of a nonfunctional
protein.

What is the genotype of the non-tumor cells of her
mother, maternal aunt, and maternal grandmother?

A. Heterozygous

B. Homozygous mutant

C. Homozygous Wild Type [ Ans: ] A. Heterozygous

What is the relationship between chromosomal instability
and cell proliferation characteristic of tumorigenesis?
(Select two that apply.)

A. The higher the frequency of cell division, the smaller
the chance for genomic instability because of activation of
DNA repair.

B. There is no relationship between chromosomal
instability and cell proliferation rate.

,C. The greater the chromosomal instability, the higher the
probability of increased cell proliferation due to driver
mutations.

D. The greater the chromosomal instability, the higher the
probability of increased cell death due to apoptosis.

E. The higher the frequency of cell division, the greater
the chance for genomic instability because of more
opportunities to create mutations. [ Ans: ] C. The greater
the chromosomal instability, the higher the probability of
increased cell proliferation due to driver mutations.

E. The higher the frequency of cell division, the greater
the chance for genomic instability because of more
opportunities to create mutations.

One problem associated with CRISPR/Cas9 technology is
off-target effects. Part of the reason is that single base-
pair mismatches between the target site and the sgRNA in
the 20 bp DNA/RNA hybrid do not prevent Cas9 cleavage
of the target site. What is the best strategy to use in order
to specifically target the mutant Htt allele that causes
Huntington disease without off-target effects?

A. Design sgRNA that can base pair with several regions in
the genome, including the Htt locus, to ensure efficient
cleavage of the mutant allele.

B. Use multiple sgRNAs, each of which has one mismatch
with the target sequence at different nucleotide positions.

,C. Make sure that no other sequences in the genome are
identical or too similar to the 20 bp sequence involved in
base pairing with the sgRNA.

D. Use mutant Cas 9 enzyme with decreased target
specificity.

E. Look for a unique target sequence in the Htt gene that
lacks a PAM si [ Ans: ] C. Make sure that no other
sequences in the genome are identical or too similar to the
20 bp sequence involved in base pairing with the sgRNA.

As discussed in lecture, CRISPR/Cas9 system evolved in
bacteria to provide anti-viral immunity. Bacterial genome
has the CRISPR locus that produces short CRISPR RNAs,
but it does not contain any PAM sites. However, the
bacteriophage genome sequence targeted by CRISPR does.
How is this fact crucial to keeping the bacterial genome
intact and selectively destroying the phage genome?
(Select two that apply.)

A. The presence of the PAM site in the target sequence of
the bacteriophage genome prevents the bacterial
chromosome from being cleaved by Cas9.

B. The absence of the PAM site in the bacterial CRISPR
locus causes the phage chromosome to be cleaved by
Cas9.

C. The absence of the PAM site in the bacterial CRISPR
locus prevents the bacterial chromosome from being
cleaved by Cas9.

, D. The presence of the PAM site in the target sequence of
the bacteriophage genome causes the phage chromosome
to be cleaved by Cas9. [ Ans: ] C. The absence of the PAM
site in the bacterial CRISPR locus prevents the bacterial
chromosome from being cleaved by Cas9.

D. The presence of the PAM site in the target sequence of
the bacteriophage genome causes the phage chromosome
to be cleaved by Cas9.

The MAP kinase pathway is activated by ligand binding to
a transmembrane receptor, which switches Ras on via GEF
(Guanine nucleotide Exchange Factor) that counteracts
GAP (GTPase Activating Protein) activity. Ras-GTP in turn
activates the downstream signaling intermediates
(kinases), which are activated by phosphorylation. Which
drug CANNOT be used to target cancer cells that have
elevated receptor activation?

A. Small molecule inhibitors that bind the ATP-binding
pocket of kinases

B. Antibodies specific for ligand to prevent receptor
binding

C. Small molecule inhibitors that interfere with GAP
activity

D. Small molecule inhibitors that interfere with GEF
activity

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