Author: Philip Colls
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Nucleic acids Booklet Revision questions         NAME:
Q1. (a)     Draw and label a single DNA nucleotide.

(2)
(b)     Give two features of DNA and explain how each one is important in the semi-conservative replication of DNA.
1. _________________________________________________________________
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2. _________________________________________________________________
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(c)     Replication of mitochondrial DNA (mtDNA) is different from that of nuclear DNA.
The replication of the second strand of mtDNA only starts after two-thirds of the first strand of mtDNA has been copied.
A piece of mtDNA is 16 500 base pairs long and is replicated at a rate of 50 nucleotides per second.
Tick (✓) the box that shows how long it would take to copy this mtDNA.
 
A | 330 seconds |
B | 440 seconds |
C | 550 seconds |
D | 660 seconds |
(1)
(Total 5 marks)
Q2.  (a)     The mRNA codon for the amino acid tyrosine is UAU.
(i)      Give the DNA triplet for tyrosine.
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(ii)     Give the tRNA anticodon for tyrosine.
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(b)     Give two ways in which the structure of a molecule of tRNA differs from the structure of a molecule of mRNA.
1. _________________________________________________________________
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2. _________________________________________________________________
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(Total 4 marks)
Q3. (a)     The diagram shows one pair of nucleotides of a DNA molecule.
 
Name
D _________________________________________
E  _________________________________________
F  _________________________________________
(3)
(b)     Complete the table to give two differences between the structure of DNA and the structure of RNA.
 
| DNA | RNA
1 | |
2 | |
(2)
(Total 5 marks)
Q4. The diagram shows a short section of a DNA molecule.
 
(a)     On the diagram draw a box round one nucleotide.
(1)
(b)     Use the letters in the diagram to indicate a part of the molecule which
(i)     is not a base and is different in an RNA molecule;
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(ii)     contains nitrogen.
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(c)     (i)      The sequence of bases on one strand of DNA is important for protein synthesis. What is its role?
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(1)
(ii)     How are the two strands of the DNA molecule held together?
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(1)
(iii)    Give one advantage of DNA molecules having two strands.
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(1)
(Total 6 marks)
Q5. (a)     Describe the role of two named enzymes in the process of semi-conservative replication of DNA.
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(b)     Scientists investigated the function of a eukaryotic cell protein called cyclin A. This protein is thought to be involved with the binding of one of the enzymes required at the start of DNA replication.
The scientists treated cultures of cells in the following ways.
C – Control cells, untreated
D – Added antibody that binds specifically to cyclin A
E – Added RNA that prevents translation of cyclin A
F – Added RNA that prevents translation of cyclin A and added cyclin A protein
They then determined the percentage of cells in each culture in which DNA was replicating.
Their results are shown in the table.
 
Cell treatment | Percentage of cells where DNA was replicating

C
Control

91

D
Antibody that binds specifically to cyclin A

11

E
RNA that prevents translation of cyclin A

10

F
RNA that prevents translation of cyclin A and added cyclin A protein

92

Suggest explanations for the results in the table.
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____________________________________________________(3)(Total 6 marks)
Q6.
Figure 1 shows one base pair of a DNA molecule.
Figure 1
 
(a)     Name part F of each nucleotide.
_________________________________________________________________(1)
(b)     Scientists determined that a sample of DNA contained 18% adenine.
What were the percentages of thymine and guanine in this sample of DNA?
 
Percentage of thymine |
Percentage of guanine |
(2)
During replication, the two strands of a DNA molecule separate and each acts as a template for the production of a new strand.
Figure 2 represents DNA replication.
Figure 2
 
(c)     Name the enzyme shown in Figure 2.
________________________________________________________________(1)
The arrows in Figure 2 show the directions in which each new DNA strand is being produced.
(d)     Use Figure 1, Figure 2 and your knowledge of enzyme action to explain why the arrows point in opposite directions.
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 (4)   (Total 8 marks)
Q7.  (a)     (i)      Describe the role of DNA polymerase in DNA replication.
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(ii)     Other than being smaller, give two ways in which prokaryotic DNA is different from eukaryotic DNA.
1. ____________________________________________________________
2. __________________________________________________________(2)
(b)     The table shows the percentage of each base in the DNA from three different organisms.
 
Organism | Percentage of each base in DNA
| Adenine | Guanine | Thymine | Cytosine
Human | 30.9 | 19.9 | 29.4 | 19.8
Grasshopper | 29.4 | 20.5 | 29.4 | 20.7
Virus | 24.0 | 23.3 | 21.5 | 31.2
(i)      Humans and grasshoppers have very similar percentages of each base in their DNA but they are very different organisms.
Use your knowledge of DNA structure and function to explain how this is possible.
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(ii)     The DNA of the virus is different from that of other organisms. Use the table above and your knowledge of DNA to suggest what this difference is. Explain your answer.
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 (2)       (Total 7 marks)
Q8. (a)     There are two forms of nitrogen. These different forms are called isotopes. ^15N is a heavier isotope than the normal isotope ^14N.
In an investigation, a culture of bacteria was obtained in which all the nitrogen in the DNA was of the ^15N form. The bacteria (generation 0) were transferred to a medium containing only the normal isotope, ^14N, and allowed to divide once. A sample of these bacteria (generation 1) was then removed. The DNA in the bacteria of generation 1 was extracted and spun in a high-speed centrifuge.
The bacteria in the ^14N medium were allowed to divide one more time. The DNA was also extracted from these bacteria (generation 2) and spun in a high speed centrifuge. The diagram shows the results of this investigation.
 
(i)      Which part of the DNA molecule contains nitrogen?
____________________________________________________________(1)
(ii)     Explain why the DNA from generation 1 is found in the position shown.
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(iii)     Complete the diagram to show the results for generation 2.
(2)
(b)     The table shows the percentage of different bases in the DNA of different organisms.
 
Organism | Adenine% | Guanine% | Thymine% | Cytosine%
Human | | 19 | |
Bacterium | 24 | 26 | 24 | 26
Virus | 25 | 24 | 33 | 18
(i)      Complete the table to show the percentages of different bases in human DNA.
(2)
(ii)     The structure of virus DNA is different from the DNA of the other two organisms. Giving evidence from the table, suggest what this difference might be.
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(2)
(Total 9 marks)

Mark schemes
Q1.
(a)     1.      Phosphate, deoxyribose and base correctly labelled;
Accept P in a circle / P_i / PO_4^3– for phosphate.
Do not accept phosphorus for phosphate.
Do not accept only pentose for deoxyribose.
Ignore references to sugar.
Accept a named base, (eg adenine, thymine, guanine, cytosine).
Do not accept uracil or only letters (eg A, T, G or C).
Ignore labelled bonds
2.      Correct shapes and bonds in the correct positions (as shown below);
 
Accept correct shapes with incorrect labels
Accept any orientation of diagram, eg inverted / mirror image
Accept any pentagon for deoxyribose
2
(b)     1.      Weak / easily broken hydrogen bonds between bases allow two strands to separate / unzip;
may appear in the same feature
2.      Two strands, so both can act as templates;
may appear in the same feature
3.      Complementary base pairing allows accurate replication;
Allow description of complementary base pairing and accurate replication.
2 max
(c)     C. 550 seconds;
1
[5]
Q2.
          (a)     (i)      ATA;
1
(ii)     AUA;
1
(b)     tRNA ‘clover leaf’ shape; (allow reference to loop / folded structure)
tRNA standard length;
tRNA has an amino acid binding site;
tRNA has anticodon available / three exposed bases;
tRNA has hydrogen bonds (between base pairs);
2 max
[4]
Q3.
(a)     D  phosphate;
E  pentose sugar/deoxyribose;
F  (nitrogenous) base/ organic base/ thymine/adenine/ cytosine/guanine;
In D reject phosphorous
In E
Accept 5-carbon sugar
Reject sugar alone
3
(b)     
| DNA | RNA
| double-stranded | single-stranded
| deoxyribose | ribose
| Thymine/T | Uracil/U
| very large/long | very small/short
Accept double helix for DNA 
Accept longer and shorter 
Need comparison but could be in one box
List rule applies.
2 max
[5]
Q4.
(a)     appropriately placed box;
1
(b)     (i)      B;
(ii)     A;
2
(c)     (i)      determines (sequence of) amino acids / specific protein
produced / mRNA formation;
1
(ii)     hydrogen bonds;
1
(iii)     stability / protects bases / replication;
1
[6]
Q5.
(a)     1.      (DNA) helicase causes breaking of hydrogen/H bonds (between DNA strands);
Reject ‘helicase hydrolyses hydrogen bonds’.
2.      DNA polymerase joins the (DNA) nucleotides;
Reject if suggestion that DNA polymerase joins the complementary nucleotides or forms H bonds.
Reject if joining RNA nucleotides or forming RNA.
3.      Forming phosphodiester bonds;
3
(b)     1.      (Treatment D Antibody binds to cyclin A so) it cannot bind to DNA/enzyme/initiate DNA replication;
For ‘bind to enzyme’ accept ‘activate’.
Idea of ‘initiate DNA replication’ must be linked to start not just less replication.
For ‘enzyme’ accept named enzyme.
2.      (Treatment E) RNA interferes with mRNA/tRNA/ribosome/polypeptide formation (so cyclin A not made);
3.      In Treatment F added cyclin A can bind to DNA/enzyme (to initiate DNA replication)
OR
Treatment F shows that it is the cyclin A that is being affected in the other treatments
OR
Treatment F shows that cyclin A allows the enzyme to bind (to DNA)
OR
(Some cells in D or E) can continue with DNA replication because they have a different cyclin A allele
OR
(Some cells in D or E) can continue with DNA replication because the antibody/RNA has not bound to all the cyclin A protein/mRNA
OR
(Some cells in E) can continue with DNA replication because they contain previously translated cyclin A;
Context needed for Treatment F but it does not need to be named.
For ‘enzyme’ accept named enzyme.
3
[6]
Q6.
(a)     Deoxyribose.
1
(b)     1.      Thymine 18 (%);
2.      Guanine 32 (%).
2
(c)     DNA polymerase.
1
(d)     1.      (Figure 1 shows) DNA has antiparallel strands / described;
2.      (Figure 1 shows) shape of the nucleotides is different / nucleotides aligned  differently;
3.      Enzymes have active sites with specific shape;
4.      Only substrates with complementary shape / only the 3’ end can bind with active site of enzyme / active site of DNA polymerase.
4
[8]
Q7.
(a)     (i)      Joins nucleotides (to form new strand).
Accept: joins sugar and phosphate / forms sugar-phosphate backbone
Reject: (DNA polymerase) forms base pairs / hydrogen bonds
1
(ii)     (Prokaryotic DNA)
1.      Circular / non-linear (DNA);
Accept converse for eukaryotic DNA
Ignore: references to nucleus, binary fission, strands and plasmids
2.      Not (associated) with proteins / histones;
Accept does not form chromosomes / chromatin
3.      No introns / no non-coding DNA.
Accept only exons
Q Neutral: no ‘junk’ DNA
2 max
(b)     (i)      1.      Have different genes;
Reject: different alleles
2.      (Sobases / triplets) are in a different sequence / order;
Accept: base sequence that matters, not percentage
3.      (So) different amino acid (sequence / coded for) / different protein / different polypeptide / different enzyme.
Unqualified ‘different amino acids’ does not gain a mark
Reject: references to different amino acids formed
Ignore: references to mutations / exons / non-coding / introns
2 max
(ii)     (Virus DNA)
1.      A does not equal T / G does not equal C;
Accept: similar for equal
Accept: virus has more C than G / has more A than T
2.      (So) no base pairing;
3.      (So) DNA is not double stranded / is single stranded.
2 max
[7]
Q8.
(a)     (i)      base / named bases;
reject nucleotide or uracil
1
(ii)     it has been produced by semi-conservative
replication / one old strand and one new; 
One strand has ^15N bases and the other ^14N;   
Accept light / heavy N (therefore) it is less dense / lighter;
2
(iii)     one band is in same position as generation 1;
one band higher;
accept a line. N.B. need a visible gap
2
(b)     (i)      A = 31 and JT = 31;
C = 19;
2
(ii)     viral DNA single-stranded / not double-stranded;
evidence from table e.g. not equal amount of A and T
/ C and G / all different;
2
ignore no base-pairing In this Question assume It’ means viral DNA
[9]