wgu biochemistry 785 final exam 100q questions wit
wgu biochemistry 785 final exam
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WGU BIOCHEMISTRY 785 FINAL EXAM 100Q
QUESTIONS WITH VERIFIED CORRECT ANSWERS/A+
GRADE 2023 VERSION
DNA replication is ___________, which allows each of the two strands to serve as a
_______ for the new strands.
b. semiconservative, template - ANSWER: DNA replication is semiconservative,
meaning that each new duplex has one original (parent) strand and one new strand.
Because the two parent strands are separated during replication and the base
pairing is predictable, each parent strand can serve as a template for the new strand
synthesis.
Which of the following enzymes does NOT assist the DNA polymerase on the lagging
strand to overcome its two problems? (Recall that the DNA polymerase can only
make DNA in the 5'->3' direction, and it must bind a double-stranded nucleotide
polymer before it can start making its own DNA polymer.)
c. Helicase - ANSWER: Helicase unwinds the double stranded DNA to allow for
replication, but this is not a problem for the DNA polymerase.
Several components of cigarette smoke, including benzopyrene, insert themselves
(intercalate) into the DNA and lead to several types of mutations such as frameshift
mutations, including both insertions and deletion. Which of the following repair
pathways would be used to repair this type of damage?
c. Nucleotide Excision Repair - ANSWER: Nucleotide excision repair is used to repair
deletions, insertions, and helix-distorting lesions, such as thymine dimers.
Maternal smoking during pregnancy is hazardous yet common in many places. Many
studies have associated prenatal smoking to unhealthy physical and psychological
outcomes for the baby. Researchers know that maternal smoking affects are
epigenetic in nature. Which of the following events can be considered epigenetic in
nature?
a. Changes in chromatin structure - ANSWER: Frame shift mutations are a kind of
mutations which result from addition of deletion of a nucleotide base resulting in an
altered reading frame and ultimately a different protein, than the one the gene
originally encoded. Frameshift mutations are genetic changes because they alter the
DNA sequence, whereas epigenetic changes do not alter the DNA sequence.
Epigenetic changes are modifications to genomic structure (not sequence) that are
caused by the external environment. These environmental factors affect the overall
chromatin structure to allow more or less "access" to the DNA by gene expression
machinery to turn the genes "on" or "off". In other words, epigenetics can alter gene
,expression without changing the underlying DNA sequences. The changes may or
may not be heritable, depending on the location and circumstances.
Blood type is an example of what type of inheritance?
a. Codominance - ANSWER: The genes that produce the A and B antigen proteins can
both be expressed independently, and a heterozygote (someone with both genes)
will be produce both A and B proteins - neither will dominate the other. The is an
example of codominance.
What is the expected probability that a child will have an autosomal dominant
disease if their father is heterozygous for the allele and their mother is homozygous
for the normal allele?
a. 50% - ANSWER: If D is the disease-conferring dominant allele and d is the normal
allele, the father has the genotype Dd and the mother's genotype is dd. Each child
can only inherit a d allele from their mother, and they have a 50% chance of
inheriting the D allele from their father. As a result, the expected probability that
their child will inherit the disease is 50%.
The physical trait of lip protrusion exhibits a characteristic type of inheritance, as
shown by the pedigree above. What type of inheritance best describes this
inheritance pattern?
a. Incomplete dominance - ANSWER: The correct answer is incomplete dominance.
The blending of the large and small lip protrusion into an intermediate, medium lip
protrusion, as well as the presence of all three variations in the offspring,
demonstrate a clear example of incomplete dominance.
The normal sequence of a section of the HLA-B27 gene, a genetic marker of the
inflammatory disease Ankylosing spondylitis, is given below. Match each mutation of
the sequence to the type of mutation it exhibits. A genetic code table is provided for
your use in answering the question.
5'- CGG CAG AAU UUA -3'
5'- CAG CAG AAU UUA -3' - Missense mutation
5'- CGG CAG AAA UUU A-3' - Insertion
5'- CGG CAG AAC UUA -3' - Silent mutation
5'- CGG CAG AAU UA -3' - Deletion
5'- CGG UAG AAU UUA-3' - Nonsense mutation - ANSWER: Silent mutations are those
in which the amino acid encoded doesn't change as a result of the mutation.
Missense mutations are those in which the amino acid encoded changes to a
different amino acid as a result of the mutation.
, Nonsense mutations are those in which the amino acid encoded changes to a stop
codon as a result of the mutation, yielding a smaller protein.
Insertions are the inclusion of extra nucleotides compared to the original sequence.
They can result in other mutations, such as nonsense mutations.
Deletions are the removal of nucleotides compared to the original sequence. They
can result in other mutations, such as nonsense mutations.
PCR is a powerful tool that can do all of the following....
b. detect mutations that lead to disease
c. copy small segments of DNA, less than 6kb
d. amplify DNA from samples that have just a few cells - ANSWER: PCR's ability to
amplify is powerful, and products can even be generated from samples with just a
few cells. PCR is less reliable for amplifying large segments of DNA greater than 6 kb,
though some careful changes to the techniques can allow it.
Which of the following is a required "ingredient" in a PCR
reaction?
a. DNA nucleotides
b. DNA primers
c. DNA polymerase - ANSWER: The primers used in PCR are made from DNA, rather
than RNA. RNA primers are used in DNA replication inside the cell, but the quick
degradation of RNA makes it less useful for PCR reactions. Instead, PCR reactions
contain primers made of DNA to anneal to the region of DNA that will be amplified
and serve as a starting point for DNA polymerase.
Which of the following changes can be detected using PCR?
a. Differences in DNA sequence
b. Insertions
c. Deletions - ANSWER: Differences in DNA sequence can be detected since these
changes can alter the ability of primers to anneal to the DNA. They can also be
detected by DNA sequencing of the PCR product. Epigenetic changes are not
detectable via PCR because they don't affect the DNA sequence.
A small segment of Kevin's green opsin gene is shown below. What would be the
resulting mRNA sequence?
Kevin's opsin gene at nucleotide positions 936 to 941.
5'-G.C.C.T.A.G-3' (coding strand)
3'-C.G.G.A.T.C-5' (template strand)
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