Second year Uni Molecular biology exam 1

100 mark

Exam. [71 marks]

Multiple Choice Questions. [43 marks]

Question 1.

Which of the following does RNA not flow into according to the central dogma

• a) Proteins
• b) DNA
• c) Polypeptides
• d) amino acids

Question 2.

DNA polymerases supports the formation of DNA by catalyzing the linking of triphosphates into a DNA strand. Which of these is a false reason why this can be exploited for anticancer and antiviral drugs

• a) DNA replication only occurs in rapidly dividing cells
• b) Only DNA replication requires TTP
• c) The enzyme commits suicide upon irreversible binding with the incorrect base
• d) The cell runs out of thyamine

Question 3.

If we have equal number of ssDNA fragments as dsDNA fragments, of the same nucleotide length and sequence in two different solutions of the same volume. It would be impossible to differentiate the two using UV-Vis.

• a) True
• b) False

Question 4.

Gene expression can be increased through:

• a) HATs making histones less positive, by adding an amine group to arginine in the histone, therefore, reducing how tightly DNA is bound to the histone.
• b) HATs making histones less positive, by adding an acetyl group to lysine in the histone, therefore, reducing how tightly DNA is bound to the histone.
• a) HDACs making histones less positive, by adding an amine group to arginine in the histone, therefore, reducing how tightly DNA is bound to the histone.
• b) HDACs making histones less positive, by adding an acetyl group to lysine in the histone, therefore, reducing how tightly DNA is bound to the histone.

Question 5.

Which of the following statements is true regarding euchromatin

• a) Only compact during interphase
• b) Many regions containing satellite sequences
• c) Contains more genes then the heterochromatin regions
• d) Mostly an inactive part of the DNA

Question 6.

There is a mutation in the lac operon repressor gene of a bacterium as a result it can’t bind to the operator site. Which of the following will not change the rate of transcription of the lactose transporter transacetylase gene.

• a) A mutation in the lac promoter, which makes it bind stronger
• b) An increase in the amount of sucrose in the cell
• c) An increase in the amount of cAMP in the cell
• d) A decrease in the available ribonucleoside triphosphates

Question 7.

Which of the following will not change the rate of transcription of tryptophan by a cell

• a) An inability of tryptophan to be a co-factor to the trp-repressor
• b) The number of trp-repressors present
• c) The number of tryptophan amino acids present in the cell
• d) A mutation in RNA Polymerase

Question 8.

Which of the following is a reason why DNA contains thiamine as opposed to uracil

• a) It’s more energy efficient to form thiamine as opposed to uracil
• b) Uracil spontaneously deaminates into cytosine
• c) Thiamine is less chemical reactive then uracil
• d) Cytosine undergoes spontaneous deamination into uracil

Question 9.

DNA contains deoxyribose instead of ribose because:

• a) It is formed from NTP which are converted into dNTPs when they join together
• b) DNA is in translation and a hydroxyl group on the 2′ carbon would prevent the DNA from entering the A site of the ribosome
• c) It makes the DNA less susceptible to base attacks on the 2′ carbon
• d) DNA is short lived hence doesn’t need to be as stable as RNA

Question 10.

An experiment modelled off Hersey and Chase’s experiment want to demonstrate whether bacteria phages convey information using proteins or genetic information. Which of the following is/are true – you can select more then one

• a) The DNA is labelled with radioactive phosphorus
• b) The protein is labelled with radioactive Sulphur
• c) The RNA is labelled with radioactive phosphorus
• d) The proteome is labelled with radioactive Sulphur

Question 11.

Which of the following one of these statements is false.

• a) Centromeres are specialized DNA code which links two chromatids together
• b) Chromatids are half a duplicated chromosome
• c) Telomerases are repeating sequences on the 3′ tail of chromatids to protect it during DNA replication
• d) The p arm of the chromatid is the smaller arm and the q arm of the chromatid is the larger arm

Question 12.

The universal genetic code means:

• a) We can insert a gene coding region from one animal to another of a different species e.g human gene into a bacterium
• b) Only one type of RNA polymerase is required for transcription of DNA for both eukaryotes and prokaryotes
• c) Each codon of DNA codes for the same amino acid no matter what organism it is for
• d) DNA is redundant

Question 13.

In eukaryotic cells, the 5′ region of the pre-mRNA transcript is capped and the 3′ end is polyadenylated, adding a poly(A) tail. Which of the following regarding post-transcriptional processing are incorrect.

• a) All eukaryotic mRNA is poly adenylated
• b) The poly(A) tail increases stability and the 5′ capping protects from phosphatases
• c) Both increase the rate of translation
• d) Methylated guanine is added via 5′-5′ linkage and poly(A) tail is added with ATP as substrate

Question 14.

Positive super coiling makes the DNA more tightly coiled, as such it is harder to melt. Therefore, DNA replication requires topoisomerase to ‘cut’, ‘pass’ and ‘rejoin’ the DNA over and over to prevent this.

• a) True
• b) False

Question 15.

The accelerators of the cellular replication are ……….. genes and the breaks are ……….. genes. These mutating can result in cancerous cell division.

• a) Oncogenes and tumor suppressor genes
• b) Protooncogenes and tumor suppressor genes
• c) tumor suppressor genes and Oncogenes
• d) break genes and accelerator

Question 16.

RNA polymerase II is found in eukaryotic DNA transcription, which of the following is not an initiator element.

• a) 5′ TATA 3′
• b) 5′ CAAT 3′ and 5′ GC 3′
• c) Initiator element –> Defines the start site
• d) Down stream core promoter elements
• e) 5′ CACA 3′

Question 17.

A base insertion on the 3′ side of an initiator element will have more of an effect then a base insertion changing TATA to TATC.

• a) True
• b) False

Question 18.

What form is a cell’s genetic information required to be in, in order for a oligo dTs to be used to purify RNA.

• a) ssDNA
• b) dsDNA
• c) pre-mRNA
• d) mRNA

Question 19.

Which of the following is not a reason a DNA double Helix is the shape it is.

• a) Hydrophobic interactions ‘burying’ the negative charges deep inside the helix
• b) Pi electron charges interacting allowing for base stacking
• c) Hydrogen bonding allowing for complimentary bases to stick together
• d) Ionic interactions between positive bases and negative sugar-phosphate backbone
• e) Van der walls forces between neighboring bases

Question 20.

Aminoacyl-tRNA synthetase has:

• a) An ability to catalyze the formation of a polypeptide bond between amino acids
• b) An activation site which doubles as a proof reading site
• c) An ability to recognize amino acids, expend 1 ATP to bind aminoacyl to tRNA
• d) An editing site which uses steric hindrance to check correct tRNA and amino acid pairing

Question 21.

Protein synthesis is inhibited through, more then one may be correct:

• a) Diphtheria toxin irreversibly binding to the host receptor for the diphtheria toxin
• b) Ricin toxin adding ADP-ribose to dipthamide, thereby, blocking EF2’s ability to carry out translocation
• c) Diphtheria adds a methyl group to dipthamino, blocking protein synthesis
• d) Ricin on the 28S rRNA, removes adenine on an adenosine. Preenvtning the bonding of transcription factors

Question 22.

The bacterial promoter consensus sequence is TATAAT. Which of the following will least reduce the rate of transcription for bacteria cell.

• a) A drug which inhibits the polymerase sigma subunit from binding to TATAAT
• b) A mutation in the TATAAT sequence, changing it to TATAAA
• c) Moving the TATAAT closer to the start site
• d) A mutation which prevents the release of the polymerase sigma sub unit from TATAAT

Question 23.

RNA polymerase acts in the replication bubble for RNA synthesis in bacteria and has:

• a) Proof reading functionality
• b) The ability to only start with a primer
• c) The ability to unwind DNA
• d) An ability to add bases 3′ to 5′

Question 24.

If the start codon is analogous to a capital letter, then a stop codon is analogous to a full stop.

• a) True
• b) False

Question 25.

Which of the following are incorrect in creating and testing recombinant DNA molecules.

• a) This is the process by which the intended gene of interest is inserted into a screenable marker
• b) Selecting and amplifying chromatids – colony PCR
• c) Sequencing the DNA sequence of an extracted recombinant plasmid
• d) Restriction digesting the plasmid and using gel electrophoresis to determine whether successful insertion had occurred.
• e) When plasmid is prepared, placing it into competent E. Coli cells and heat shocking the solution

Question 26.

No-sense mediated RNA decay:

• a) mRNA stalls and recruits proteins to cleave it off and destroy mRNA
• b) Uses exon junction complex’s to detect and destroy codons with premature stop codons leading to cleavage of 5′ cap and RNases degrading mRNA
• c) mRNA runs through to the poly A tail and recruits proteins to cleave it off
• d) Uses intron junction complex’s to detect and destroy codons with premature stop codons leading to cleavage of 5′ cap and RNases degrading mRNA

Question 27.

Eukaryotic promoters are:

• a) Cis acting elements that attract RNA polymerases
• b) Proteins which act as recruitment factors
• c) Can act from the complimentary strand
• d) Increase the accuracy of transcription by attracting RNA polymerases

Question 28.

A mutation has occurred in splicing machinery components which are recruited by RNA polymerase II. This means there are more alternative splicing re-arrangements then what would normally occur for the protein. The effect on the cell will be.

• a) The cell will have to many different proteins to organize, therefore, it will stop producing the mRNA
• b) The cell will have lower numbers of functional proteins then normally expected
• c) There will only be dysfunctional proteins.
• d) The cell will be overloaded with dysfunctional proteins and as such will not be able to function

Question 29.

Which of the following statements is incorrect regarding enhancer sequences:

• a) Replace promoter sequences, if the promoter sequence is not present
• b) Act over thousands of base pairs
• c) Act from the non-coding strand, enhancing transcription on the coding strand
• d) Can be down stream, up stream or on the promoter sequence and still exhert an effect

Question 30.

A purine rich sequence which doesn’t code for any amino acids, which is required for transcription in prokaryotic cells is known as a:

• a) Sanger sequence
• b) AUG start codon
• c) Shine-Dalgarno sequence
• d) Fmet sequence

Question 31.

Select the false statement regarding histones:

• a) DNA is packaged around histones through electrostatic interactions between negatively charged sugar phosphate backbone and positively charged histone molecule
• b) A nucleosome is a histone octamer and DNA
• c) Histones are made up of two sets of H1, H2, H3 and H4
• d) Treating DNA wrapped around histones with a highly ionic solution can unwrap DNA from histones

Question 32.

Select the incorrect statement below:

• a) Low Iron levels results in low ferritin since the stem loop structures form in the 5′ region, preventing protein synthesis.
• b) High iron levels result in low transferrin since iron response elements are not bound in the non-coding 3′ region so the mRNA is rapidly degraded.
• c) High iron levels increases the rate of transcription of transferrin mRNA since the body doesn’t have to transport iron around
• d) Low iron levels will not affect the rate of transcription of mRNA for transferrin and ferritin

Question 33.

Internal ribosomal entry sites allows for multiple promoters to promote the transcription of a protein.

• a) True
• b) False

Question 34.

IRES allows for initiation factor (elF4F) or 40S ribosome to translate DNA which is lacking a 5′ cap

• a) True
• b) False

Question 35.

A false reason as to why DNA is a good store of genetic information is:

• a) DNA contains two strands so there are two copies always saved
• b) DNA contains thymine as opposed to uracil so spontaneous deamination of cytosine can be easily identified
• c) DNA is easy and low energy to replicate so it can be kept in every cell of the body
• d) DNA is deoxyribose so it is not susceptible to base attack.

Question 36.

Exogenous dsRNA can be introduced into a cell from a virus or experimental technique. This is cleaved by …… forming small interfering (siRNA) which form a ……….. with argonaute proteins, after perfect complimentary binding, translation is inhibited.

• a) dicer & RNA induced silencing complex
• b) ssRNA & stop codon
• c) RISC & translation complex
• d) mitochondrial DNA & rRNA

Question 37.

micro RNA (miRNA) from larger transcripts from pol II & III. This turns dsRNA fragments into ………….. which form a ……….. with argonaute proteins, after perfect complimentary binding, translation is inhibited.

• a) siRNA & RNA inhibiting ligate
• b) ssRNA & RISC complex
• c) RISC & translation complex
• d) mitochondrial DNA & rRNA

Question 38.

If the transformylase enzyme becomes defective in a prokaryotic organism, then the organsim will.

• a) Be unable for the initiator tRNA to bind to the methionine amino acid
• b) Be unable to complete protein synthesis and the mRNA chain will reach the poly A tail
• c) Be unable to form fMET and therefore be unable to synthesis mRNA
• d) Will be unable to synthesize proteins since translation can not be initiated

Question 39.

Which of the following is incorrect with regards to the discovery of nucleic acid

• a) Griffith’s transforming principle relating to the transformation of rough pathogens to smooth pathogens through the exposure of genetic information.
• b) Avery, McCarty and MacLeod discovered that genetic information was common to all organisms
• c) Hershey Chase and the Waring determining whether DNA o proteins were apart of the genetic information
• d) Chargaff’s rules relating to the numbering of amino acids

Question 40.

Which of the following key properties of DNA arguable does the least for storing genetic information

• a) Phosphodiester bonds
• b) N-glycosidic bonds
• c) Purine = pyramidine
• d) A = T, G = C
• e) They are all integral for information storage

Question 41.

The correct order for Cell cycle phases:

• a) G1, G0, S, M with an opt out at after M to G2
• b) G1, S, G2, M with an opt out at after M to G0
• c) G0, G1, S, M with an opt out at after M to G2

Question 42.

Quiescence cells can reenter the cell cycle, while senescence cells can not easily reenter the cell cycle

• a) True
• b) False

Question 43.

The nuclear hormone receptor has an antagonist bind to it, whilst DNA is still in the zinc fingers – this will result in the expression of the DNA.

• a) True
• b) False

Short answer questions. [28 marks]

Question 1. [8 marks]

Compare and contrast eukaryotic transcription and prokaryotic transcription

Question 2. [8 marks]

Compare and contrast eukaryotic translation and prokaryotic translation

Question 3. [4 marks]

Explain how transcription and translation can be exploited in order to generate recombinant plamsids

Question 4. [4 marks]

Compare and contrast DNA replication in prokaryotic and eukaryotic organisms.

Question 5. [4 marks]

Explain how a recombinant plasmid is made, and how this is inserted into and amplified in bacteria.

Answers.

1. d) [flows into DNA by reverse transcriptase and proteins and polypeptides by translation]
2. a) [DNA replication occurs in most cells, however, the reason senescence and quiescence cells are not effected is because they are not using TTP. The cell runs out of thymidylate synthetase (forced to commit suicide) hence the cell can’t produce thymine from uracil any more.]
3. b) [The hyperchromic effect means single stranded DNA molecules absorb more UV/Vis light then dsDNA molecules]
4. b) [HATs – (histone acetyl transferase), HDACs – (histone deacetylases)]
5. c) Euchromatin vs heterochromatin
6. c)
7. b)
8. d)
9. c)
10. a & b)
11. c) [Telomeres are these, telomerases catalyze the formation of a telomere]
12. c)
13. a) [Exception being histone proteins.]
14. a)
15. b)
16. e)
17. a) Initiator elements denote where the sequence begins.
18. d) Oligo dTs are fixed onto beads so they catch the mRNA which has been proceesed and a poly A tail has been added to it.
19. a) The DNA double helix buries the bases – not the negative charge
20. d)
21. d) Diphtheria toxin adding ADP-ribose to dipthamide, thereby, blocking EF2’s ability to carry out translocation
22. d) All the other options decrease rate of transcription. The polymerase sigma sub unit is released following the start of transcription.
23. c) RNA polymerase does not require a primer (de novo), only has proof reading functionality with another primer present and adds bases 5′ to 3′
24. a)
25. e)
26. b) – a is for no-go and c is for no-sense
27. b)
28. b)
29. a)
30. c) AUG codes for a modified methionine, the shine-Dalgarno sequence codes for no amino acid, is found upstream of start site.
31. c) A histone has two sets of H2A, H2B,H3,H4 and one H1 which ‘staples’ the DNA to around the histone at the linker DNA sites.
32. c) The rate of transcription is not affected by Fe levels, the rate of translation is the one that is affected.
33. False – The IRES provide an alternative site for transcription and allows one promoter to initiate the transcription of multiple mRNAs.
34. b) The 40S ribosomal subunit and the elF4F can bind to the IRES, however, this is for transcription – not translation.
35. c)
36. a)
37. b)
38. d)
39. b & d)
40. e) Purines include A and G bases while Pyrimidines include A, T, G
41. b)
42. a)
43. b)

Short answer Questions.

Question 1.

Transcription is the process of transcribing DNA into RNA, this can form mRNA, rRNA and tRNA for both eukaryotes and prokaryotes. The template, or antisense strand, is a template strand for RNA polymerase which adds the corresponding ribonucleosides to create the RNA. There are three stages initiation, elongation and termination.

In prokaryotic mRNA transcription:

• Specific promoter sequences resemble a consensus sequence – TATAAT, which is found 10 nucleotides upstream of start site – the closer the resemblance the stronger the promoter sequence is. The sigma sub unit of the RNA polymerase recognizes the start site and is released after initiation has occurred.
• The RNA polymerase then unwinds the DNA and moves in the 5′ to 3′ position, this occurs de novo. However, RNA polymerase will only be able to proofread, hence detect errors, if there are accessory proteins present.
• In order for this to be terminated, hairpin loop structures or RHO proteins must occur in the transcribed RNA sequence which allows for the RNA to be released.

This is controlled through a regulatory gene expressing a protein which binds to the control site, blocking it unless there is the correct signal e.g lactose. This will result in the lac operon being exposed in the example of the production of enzyme production to break down lactose.

This occurs in the cytoplasm of the cell. This means transcription and translation occure in the same area.

In eukaryotic mRNA transcription:

There are three types of RNA polymerases – RNA polymerase I which functions in the nucleosome synthesizing rRNA, RNA polymerase II which functions in the nucleoplasm synthesizing mRNA and snRNA and RNA polymerase III which also functions in the nucleoplasm synthesizing tRNA and rRNA.

• Likewise, initiation in eukaryotes requires cis acting promoters, however, these differ from the TATAAT box. There is the initiator element, CAAT and GC consensus sequence, TATA box and a downstream promoter element.
  • Additionally, prokaryotes also have enhancer sequences
• Analogous to the sigma sub unit of RNA polymerase for prokaryotic transcription, Transcription Factors bind to the promoter and enhancer sequences, unwinds DNA and is used to promote RNA polymerase II and phosphorylates it.
• This forms pre-mRNA which is then capped at the 5′ end, a poly A tail is added and the introns are spliced out of it.

This is then transported out of the nucleus through nuclear pores to then be translated at a later time and in a different area.

Question 2.

Translation:

• Similarities

Initiation, elongation and termination

• Diferrences
  • Prokayotes

Uses FMet to initiate translation. Uses a shane-dalgarno sequence. mRNA is linear. Uses EF-G

• Differences
  • Eukaryotes

Start site is the AUG codon. mRNA is circular. Uses EF-G

Question 3.

Question 4.

DNA replication is a semi-conservative method used by cells to replicate DNA. This occurs in the S phase of the cells cycle.

Both lay down the RNA primer, and polymerase forms the corresponding chain 5′ to 3′. With primers placed on the opposite chain in small fragments.

Prokaryotes

oriC sequence is recognized by multiple DnaA proteins, the initiator proteins. These recruit helicases, DnaB proteins.

Primase synthesizes an RNA primer, DNA pol II synthesizes DNA strand using the primer.

DNA pol I replaces RNA and DNA ligase finishes by filling in the small gaps.

DNA polymerase I has : – 5’ to 3’ exonuclease– 3’ to 5’ exonuclease(proof reading)– 5’ to 3’ polymerase (new strand)•DNA polymerase III has;– 3’ to 5’ exonuclease(proof reading)– 5’ to 3’ polymerase (new strand)

Eukaryotes

Slower process, multiple origins of replication – forms pre-RC in the G1 phase, waits unitl the s phase to use these.

Histone repackaging

–DNA pol δ

–DNA pol α/primase – primer –> initiation

These are capped with ribnucleoproteins – the telomerase.

Question 5.

Recombinant plasmids are made by lysing a cell. The RNA is then removed and oligo(T)s connected to beads are then used to catch the mRNA which contains a poly A tail. DNA polymerase will then form a double stranded DNA structure. The gene of interest is then inserted into the multiple coding sites using a restriction digest enzyme which will makes the lac Z gene defective – so plasmids which were successfully made can be identified later.

The plasmid contains an origin of replication and an antibiotics gene so it can be replicated and provides the cell a selective advantage. The competent E. Colli are heat shocked to promote the uptake of the plasmid.

After this point the bacteria are incubated on nutrient rich agar which contains antibiotics. The plasmids which have not changed the colour of X-Gal is then selected to be amplified and Sager sequencing is used to confirm whether the gene of interest has successfully been taken up.