iClicker Questions Flashcards
A
B
A – fragment 5 is the largest fragment; smaller fragments migrate faster through the gel and will therefore be further to the bottom
C – DNA polymerase synthesizes 5’ to 3’!
B – EcoRI cuts leaving a sticky end, whereas PCR leaves (mostly) blunt ends
C is also ok
D is defensible if you don’t know that E.coRI creates sticky ends
C – to transcribe insulin – vectors for protein expression need a promoter for RNA polymerase binding
B – this is a tricky question… in eukaryotes the AUG is embedded in the Kozak sequence and mRNA is recognized by the 5’cap but if we want to express a human protein in bacteria, we have to embed the AUG in the prokaryotic ribosome binding site (shine dalgarno sequence)
Both A and D are correct
All patients have about the same amount of GapDH – that is an internal control to make sure you put about the same amount of material into the reaction (each cell expresses GapDH)
Lower Ct means earlier begin of exponential phase in qPCR and therefore higher amounts of starting material / template – so higher virus load
E - 4 because the probe will also anneal to both of the normal chromosomes 9 and 22
B, C, and D
B has more cells with normal chromosomes, C only Philadelphia chr cells and D has more philad chr cells than normal cells
B – it doesn’t recognize the 5’cap and the EK mRNA doesn’t have the shine dalgarno sequence
Correct answer: A – it is most similar to the consensus sequence
B had the highest level of translation within the few mutants analyzed
C had high levels of translation but less than B (in the mutation analysis)
D would also work as G in pos -3 is also found more than by chance (36% vs 25% by chance) – however, A is twice as often in this position
Correct answer: C
IF3 prevents the 30S and 50S subunits to aggregate to the 70S
Correct answer: A
(c and d are nucleic acids, not polypeptides!!)
Correct answer: A
Keep in mind – in other processes, ATP can also be used to drive conformational change! This is not universally true
Correct answer: D
All steps of the pathway would occur. RF1/2 don’t need GTP hydrolysis of RF3 to transfer the polypeptide chain to an h2O – only needs RF3 to be released from ribosome
Correct answer C - Histone – nuclear proteins are synthesized in the nucleus before they are transported into the nucleus by importin
Antibody and hormones are secreted, surface receptors are in the plasma membrane. All of those proteins are synthesized in the ER
B – A is a purine, T is a pyrimidine – that is a transversion
D
B
B
A
D
D
C
C
B
B – lacZ was destroyed by integration of the insert into the MCS – bacterium does not have the ability to convert Xgal to the blue compound anymore
C – some viruses cause the host cell to lyse when the new virus particles are released, some won’t and can therefore stay in the genome and continue replicating
B – retrotransposons. They transcribe multiple molecules of RNA which can be reverse transcribed to multiple molecules of DNA
In a DNA transposon one molecule of DNA is cut out and placed in a different genomic location
Correct answer: B – 3
Possible anti-codon combinations: 5’-UAA-3’, 5’-IAG-3’, 5’-UAG-3’ or also 5’-UAA-3’, 5’-GAG-3’, 5’-UAG-3’
B
B
Answer is C – it needs to be T not U as it is DNA
Correct answer is B – 6 to the third power
(A is 4 to the third power, C is 6 to the 4th, D is 4 to the 6th)
Correct answers are A AND D
The researcher had different fragments and ordered them by radioactivity ratio. Only one fragment was affected by an enzyme that digests proteins from the C-terminus → he knew which fragment corresponded to the C terminus of Hb.
The fragments could be ordered in a gradient by their 3H:14C ratio.
If it was nonlinear synthesis, you would expect all fragments to have the same ratio
That way he found out that protein synthesis is linear AND proceeds from N to C terminus
C – there is only one band of mixed DNA molecules so conservative can be ruled out
B – all mixed at a lower Mw – the heavy DNA would become more and more dilute between DNA strands but evenly distributed
C - mixed and light – you would get two light-light and thow light-heavy strands
B
B – DNA polymerases can’t initiate synthesis of a new strand. They need a primer
D – only RNA polymerase can initiate synthesis of a new strand without a primer. RNA polymerase only synthesizes RNA
C
A – there are short and longer fragments that are isolated, in accordance with this model
The experiment was done by Reiji and Tsunebu Okazaki
Okazaki experiment – pulse chase (times are chase times)
D
B – DNA Ligase – so you only get short fragments, they never transition to longer fragments as in the original experiment
A – telomeres become shorter with age
Correct answer is C. B and D will not occur spontaneously bc they are endergonic. C has a lower activation energy than A so it will proceed faster.
Correct answer is D – SDS is a detergent that denatures proteins but has little effect on other substances like DNA or RNA
C
D
A – lots of positively charged residues!
C
D
B
C
Correct answer: B
The mature mRNA being translated into proteins is shorter than the recently transcribed hnRNA
D – the fact that there are discrete RNAs that are longer than the mature mRNA suggest that there are intervening sequences that are consistently removed from the hnRNA to arrive at mRNA.
If it was a gradual process without distinct intervening sequences, you would expect a ‘smear’ between pre-mRNA and mRNA
C
Answer: A
The 5′ splice site contains either a conserved GU or AU motif. If the 5′ splice site contains a GU, then the major spliceosome machinery will be recruited. In the initial reaction step, the phosphoryl group of the conserved G is attacked by the 2′ OH of a conserved A nucleotide at the branch site. If the 5′ splice site instead contains an AU, then the minor spliceosome machinery will be recruited. The 2′ OH of the branch site A nucleotide will recognize and attack the phosphoryl group of the conserved A. In both cases, the result is cleavage of the 5′ splice site.
The 3′ splice site contains either a conserved AG or AC motif. If the major spliceosome machinery was involved in cleaving the 5′ splice site, then the second step of the splicing process consists of the 3′ OH of the 5′ exon attacking the conserved G nucleotide in the 3′ splice site. the result is cleavage of the 3′ splice site, joining together of exons, and release of the intron in the form of a lariat.
Both A and C are correct
The proteins the antibodies are directed to are the snRNPs and are critical for splicing
B
Lariat is the intron that has the lariat shape
3’ splice site means the 3’ exon
Correct answer E – 12 x 48 x 33 x 2 = 38,016
B
B – no, because DNA does not have the 2’OH. This functional group of the branch site A is required for the first transesterification step
Exon shuffling can occur through homologous recombination or through transposons and is pretty rare
