BCHM 307 EXAM 2 QUESTIONS WITH ALL ANSWERS REVISED AND UPDATED
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Course
BCHM 307
Institution
BCHM 307
BCHM 307 EXAM 2 QUESTIONS WITH ALL ANSWERS REVISED AND UPDATED
A short nucleic acid sequence, such as is required by DNA polymerase, is called a(n) - Answer-primer
A:T base pairs contain ____ hydrogen bonds, and G:C base pairs contain ____ hydrogen bonds. - Answer-2:3
The first of three ste...
BCHM 307 EXAM 2 QUESTIONS
WITH ALL ANSWERS REVISED
AND UPDATED
A short nucleic acid sequence, such as is required by DNA polymerase, is called a(n) -
Answer-primer
A:T base pairs contain ____ hydrogen bonds, and G:C base pairs contain ____
hydrogen bonds. - Answer-2:3
The first of three steps in mRNA modification is addition of a - Answer-5' cap
Nucleotides are added to the ______ of the newly synthesized DNA strand - Answer-3'
end
The polymerase that is primarily responsible for replicating DNA is - Answer-DNA
Polymerase III
The small DNA sections formed during synthesis of the lagging strand are called -
Answer-Okazaki fragments
The amount of DNA in eukaryotic cells is significantly greater than in prokaryotes. With
this in mind, how is the eukaryotic DNA replicated in a timely, synchronous fashion? -
Answer-Eukaryotic DNA is replicated in a timely, synchronous fashion because it
consists of a very long linear double-stranded DNA molecule within the nucleus.
During DNA replication, how are the individual Okazaki fragments of the lagging strand
converted into a whole unit? - Answer-1. Removal of RNA primer by RNase H excising
nucleotides at primer end
2. Synthesis of replacement DNA by Polymerase I
3. Sealing of adjacent DNA fragments by DNA ligase.
Protein synthesis begins at _____ of the peptide strand. - Answer-the N-terminus
The third base position in the codon, which often pairs with inosine or another
nucleotide on the anticodon, is referred to as - Answer-the wobble base
When a stop codon is in place at the ribosomal A site, _____ binds to the site instead of
a new aminoacyl-tRNA. - Answer-a release factor
List two specific causes of DNA mutations. - Answer-1. Replication errors
2. Non-enzymatic cellular processes
, 3. from environmental factors.
Describe the steps involved in translation of mRNA to generate a protein, including the
all the important molecules involved and how they interact. You may answer using bullet
points if you find it easier, but make sure they are in the correct order! - Answer-1.
Initiation
a. Requires initiator tRNA
b. In prokaryotes, Shine-Dalgarno sequence signals start of translation.
i. mRNA and fMet-tRNA met f in complex with IF-2-GTP bind to the small ribosomal
subunit
ii. Large subunit joins small subunit and triggers IF-2 to hydrolyze GTP
iii. tRNA that has first fMet is positioned in P site of ribosome
2. Elongation
i. Aminoacyl-tRNA are delivered to ribosome in complex with elongation factor and EF-
Tu hydrolyses GTP and dissociates ribosome, leaving aminoacyl-tRNA in A site if tRNA
anticodon matches mRNA codon
1. If doesn't match-aminoacyl-tRNA dissociates before EF-Tu hydrolyzes GTP.
ii. Transpeptidation (peptide bond formed = aa in P site gets added to the aa the A site)
and tRNA moves to P site and new Aminoacyl-tRNA arrives in A site
iii. Translocation and Ef-G-GTP arrives in A site
iv. EF-G hydrolyzes GTP to GDP
3. Termination
i. A Release Factor (RF-1 or RF-2) binds a stop codon instead of Aminoacyl-tRNA in A
site
ii. RF-1/RF-2 causes the ribosome to transfer the peptidyl group to water to release the
polypeptide chain
iii. GTP Hydrolysis by RF-3 allows the Release Factor to dissociate.
Cells have a mechanism for tagging and destroying proteins containing a C-terminal
poly(Lys) sequence. What is the source of these proteins and why is destroying them
helpful for the cell? - Answer-The poly(Lys) peptide results from translation of the
poly(A) tail of a mRNA whose stop codon is missing as the result of a mutation or faulty
transcription. With the addition of the poly(Lys) segment, the protein is likely to be
nonfunctional, so it is best to destroy it and reuse its amino acids.
Cystic fibrosis is caused by a mutation in the 250,000 bp CFTR gene. The mature
CFTR mRNA is only 6129 nucleotides.
a. Why is the mature mRNA so much shorter than the gene from which it was
transcribed?
b. What is the minimum number of nucleotides required to encode the 1480 residue
CFTR protein? - Answer-a. it doesn't contain introns
b. 1480 x 3=4,440 nucleotides
The CFTR gene contains the sequence ATCATCTTTGGTGTT..., which codes for
residues 506-510 of the protein. (*Coding strand)
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