1.5- Nucleotides Flashcards
Describe the role of DNA polymerase in DNA replication (1)
Joins nucleotides (to form new strand).
Accept: joins sugar and phosphate / forms sugar-phosphate
backbone
Reject: (DNA polymerase) forms base pairs / hydrogen
bonds
Other than being smaller, give two ways in which prokaryotic DNA is different
from eukaryotic DNA. (2)
- Circular / non-linear (DNA);
Accept converse for eukaryotic DNA
Ignore: references to nucleus, binary fission, strands and
plasmids - Not (associated) with proteins / histones;
Accept does not form chromosomes / chromatin - No introns / no non-coding DNA.
Accept only exons
Q Neutral: no ‘junk’ DNA
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 (2)
- Have different genes;
Reject: different alleles - (Sobases / triplets)
are in a different sequence / order;
Accept: base sequence that matters, not percentage - (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
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. (2)
- 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 - (So) no base pairing;
- (So) DNA is not double stranded / is single stranded.
The events that take place during interphase and mitosis lead to the production of two
genetically identical cells. Explain how. (4)
- DNA replicated;
Reject: DNA replication in the wrong stage - (Involving) specific / accurate / complementary base-pairing;
Accept: semi conservative replication - (Ref to) two identical / sister chromatids;
- Each chromatid / moves / is separated to (opposite) poles / ends of cell.
Reject: meiosis / homologous chromosomes / crossing over
Note: sister chromatids move to opposite poles / ends = 2
marks for mp 3 and mp 4
Reject: events in wrong phase / stage
The student cut thin sections of tissue to view with an optical microscope.
Explain why it was important that the sections were thin. (2)
- To allow (more) light through;
Accept: transparent - A single / few layer(s) of cells to be viewed.
Accept: (thin) for better / easier stain penetration
What does the graph show about the growth of roots?
Use the data to explain your answer. (2)
- More / faster mitosis / division near tip / at 0.2 mm;
Neutral: references to largest mitotic index - (Almost) no mitosis / division at / after 1.6 mm from tip;
Accept: cell division for mitosis
Penalise once for references to meiosis - (So) roots grow by mitosis / adding new cells to the tip.
Accept: growth occurs at / near / just behind the tip (of the
root)
Accept: converse arguments
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.
(4)
- (Figure 1 shows) DNA has antiparallel strands / described;
- (Figure 1 shows)
shape of the nucleotides is different / nucleotides
aligned differently; - Enzymes have active sites with specific shape;
- Only substrates with complementary shape / only the 3’ end can bind
with active site of enzyme / active site of DNA polymerase.
During replication, the two DNA strands separate and each acts as a template for
the production of a new strand. As new DNA strands are produced, nucleotides can
only be added in the 5’ to 3’ direction.
Use the figure in part (a) and your knowledge of enzyme action and DNA replication
to explain why new nucleotides can only be added in a 5’ to 3’ direction. (4)
- Reference to DNA polymerase;
- (Which is) specific;
- Only complementary with / binds to 5’ end (of strand);
Reject hydrogen bonds / base pairing - Shapes of 5’ end and 3’ end are different / description of how different.
DNA is a polymer. What is the evidence from the diagram that DNA is a
polymer? (1)
Repeating units / nucleotides / monomer / molecules;
Allow more than one, but reject two
The gene for this polypeptide contains more than this number of bases.
Explain (1)
Some regions of the gene are non-coding / introns / start / stop code /
triplet / there are two DNA strands;
Allow addition mutation
Ignore unqualified reference to mutation
Accept reference to introns and exons if given together
Ignore ‘junk’ DNA / multiple repeats
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 Table 5.
- (Treatment D Antibody binds to cyclin A so) it
cannot bind to DNA/enzyme/initiate DNA
replication; - (Treatment E) RNA interferes with
mRNA/tRNA/ribosome/polypeptide formation
(so cyclin A not made); - 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;
Give two ways in which the arrangement of prokaryotic DNA is different from the
arrangement of the human DNA in Figure 1.
[2 marks]
(Prokaryotic DNA) is 1. Circular (as opposed to linear); 2. Not associated with proteins/histones ; 3. Only one molecule/piece of DNA OR present as plasmidsMax 1 if prokaryotic DNA only found as plasmids OR if prokaryotic DNA is single stranded. Ignore references to nucleus, exons, introns or length of DNA. Do not credit converse statements. Ignore descriptions of eukaryotic DNA alone.
v
Describe the role of two named enzymes in the process of semi-conservative
replication of DNA.
[3 marks]
- (DNA) helicase causes breaking of hydrogen/H
bonds (between DNA strands); - DNA polymerase joins the (DNA) nucleotides;
- Forming phosphodiester bonds;
In a eukaryotic cell, the base sequence of the mRNA might be different from the
sequence of the pre-mRNA.
Explain why.
[2 marks]
- Introns (in pre-mRNA);
2. Removal of sections of (pre-mRNA)/splicing;
