1. You have identified a protein factor that when added to in vitro protein synt
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Question
1. You have identified a protein factor that when added to in vitro protein synthesis reaction increases the incorporation of 35S-methionine from 4 pmoles/minute to 8 pmoles/min. When you centrifuge these reactions and measure polyribosomes and free ribosomes, you observe that more of the ribosomes are present as polyribosomes for the reactions that contain your stimulatory factor. Which explanation is consistent with this result?
A. Your factor binds to the small subunit to promote translation initiation
B. Your factor binds to the large subunit to promote translation termination
C. Your factor binds to the ribosome to decrease the rate of elongation
D. Your factor binds to the ribosome to increase the rate of elongation
Questions 2 and 3 deal with the prokaryotic promoter sequence shown below
5’-ATCGGTACCGTTGACAATCGACTAGGCTAGCTTATATTGTACGTAGGTC-3’
3’-TAGCCATGGCAACTGTTAGCTGATCCGATCGAATATAACATGCATCCAG-5’
2. The bold sequences (TTGACA and TATATT) represent
A. TATA boxes recognized by RNAP II
B. -35 and -10 boxes recognized by the sigma factor of RNAP
C. Binding sites for general transcription factors
D. Binding sites for the small subunit of the ribosome
3. If transcription starts at the arrow, the sequence of the transcribed RNA will be:
a.5’-AGGTC…-3’
b.5’-AGGUC…-3’
c.5‘-TCCAG…-3’
d.5’-UCCAG…-3
4. The concentration of a particular protein X in a normal human cell rises gradually from a low point, immediately after cell division, to a high point, just before cell division, and then drops sharply. The level of its mRNA in the cell remains fairly constant throughout this time. Protein X is required for cell growth and survival, but the drop in its level just before cell division is essential for division to proceed. You have isolated a line of human cells that grow in size in culture but cannot divide, and on analyzing these mutants, you find that levels of X protein do not decrease. Which of the following mutations could explain these results?
A. A mutation that results in the loss of an enzyme that adds a ubiquitin tag to the protein.
B. a mutation in gene X that results in the loss of polyA addition to its mRNA
C. a mutation in gene X that changes the sequence that encodes sites at which ubiquitin can be attached to the protein
D. Both A and C
5. You want to amplify the DNA between the two stretches of sequence shown below. Of the listed primer pairs, choose the correct pair that will allow you to amplify the DNA by PCR.
5’-CTGGACGAAA----TGTAATGGTT-3’
3’-GACCTGCTTT----ACATTACCAA-5’
A. 5’-CTGGACGAAA-3’ and 5’-TGTAATGGTT-3’
B. 5’-CTGGACGAAA-3’ and 5’-AACCATTACA-3’
C. 5’-TTTCGTCCAG-3’ and 5’-TGTAATGGTT-3’
D. 5’-GACCTGCTTT-3’ and 5’-ACATTACCAA-3’
Explanation / Answer
question 2 Answer- B. -35 and -10 boxes recognized by the sigma factor of RNAP.
Explanation -The Sigma Subunit enables the rna polymerase to recognise promoter sites.Most ecoli promoters interact withthe major form of RNA polymerase ,which contains Sigma 70.now there are several form of sigma factors .The Alternative form of Sigma factors sigma 28 sigma 32 sigma 54 etc.they are recognize different consensus promoter sequence .-35 and -10 is consesnsus sequence.
Question -3 Answer b.5’-AGGUC…-3’
explanation-RNA synthesis are always the comlementry of Antisense strand.
Question No-5 Answer-A. 5’-CTGGACGAAA-3’ and 5’-TGTAATGGTT-3’
Primer design of any oligo we take the sequence of forword primer 5-3 and reverse primer is the complementry of 3-5.
QUESTION No-4 ANSWER- D. Both A and C
Because Mutation create loss of an enzyme that adds a ubiquitin tag to the protein
or Mutaion in sequence where ubiquitin going to be bind and proceed for proteosomal degradation.
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