REVISION 1 (51 Marks)
IB Standard and Higher level Biology Dulwich College Shanghai
Topic 3 (SL): Chemistry of Life
Topic 7 (HL): Nucleic Acids and Proteins
3.5
3.5.1
3.5.2
3.5.3
3.5.4
7.3
7.3.1
7.3.2
7.3.3
7.3.4
7.4
7.4.1
7.4.2
7.4.3
7.4.4
7.4.5
7.4.6
7.4.7
Transcription and Translation (SL)
Compare the structure of RNA and DNA.
Outline DNA transcription in terms of the formation of an RNA strand complementary
to the DNA strand by RNA polymerase.
Describe the genetic code in terms of codons composed of triplets of bases.
Explain the process of translation, leading to polypeptide formation.
Transcription (HL)
State that transcription is carried out in a 5’  3’ direction.
Distinguish between the sense and antisense strands of DNA.
Explain the process of transcription in prokaryotes, including the role of the promoter
region, RNA polymerase, nucleoside triphosphate and the terminator.
State that eukaryotic RNA needs the removal of introns to form mature mRNA.
Translation (HL)
Explain that each tRNA molecule is recognized by a tRNA-activating enzyme that binds
a specific amino acid to the tRNA, using ATP for energy. The shape of tRNA and the
CCA at the 3’ end are important.
Outline the structure of ribosomes, including protein and RNA composition, large and
small subunits, three tRNA binding sites and mRNA binding sites.
State that translation consists of initiation elongation, translocation and termination.
State that translation occurs in a 5’  3’ direction. The start codon is at the 5’ end.
Draw and label a diagram showing the structure of a peptide bond between two amino
acids
Explain the process of translation, including ribosomes, polysomes, start codons and stop
codons.
State that free ribosomes synthesize proteins for use primarily within the cell, and that
bound ribosomes synthesize proteins primarily for secretion or for lysosomes.
Paper 1
Multiple Choice (8 Marks)
1.
What is removed to form mature eukaryotic mRNA?
A.
RNA primers
B.
Exons
C.
RNA polymerases
D.
Introns
2.
A certain gene codes for a polypeptide that is 120 amino acids long. Approximately
how many nucleotides long is the mRNA that codes for this polypeptide likely to be?
A.
30
B.
40
C.
360
D.
480
3.
Where do transcription and translation occur in eukaryotic cells?
Transcription
Translation
A.
Cytoplasm
Cytoplasm
B.
Cytoplasm
Mitochondria
C.
Nucleus
Cytoplasm
D.
Nucleus
Nucleus
4.
What is a difference between the sense and antisense strands of DNA?
A.
Nucleotides are linked to the sense strand by hydrogen bonding during
transcription, but not to the antisense strand.
B.
The sense strand has the same base sequence as tRNA, but the antisense
strand does not.
C.
Nucleotides are linked to the antisense strand by hydrogen bonding during
transcription, but not to the sense strand.
D.
The antisense strand has the same base sequence as mRNA but the sense
strand does not.
5.
What is an intron?
A.
The 3’  5’ strand of a DNA double helix
B.
The 5’  3’ strand of a DNA double helix
C.
A section of mRNA removed before translation
D.
A tRNA with a start anticodon
6.
How
A.
B.
C.
D.
would the following DNA sequence, ACGTTGCATGGCA, be transcribed?
UGCAACGUACCGU
TGCAACGTACCGT
ACGTTGCATGGCA
ACGUUGCAUGGCA
7.
Why are the messenger RNA molecules received by eukaryotic ribosomes shorter
than the messenger RNA molecules formed by transcription of DNA?
A.
Base deletion mutations make the mRNA shorter.
B.
Start codons are not at the end of the mRNA molecule.
C.
Introns are removed before the RNA is translated.
D.
Bases are removed from the ends of the mRNA each time it is translated.
8.
What enzyme is used in transcription but not in translation?
A.
DNA polymerase
B.
Helicase
C.
Protease
D.
RNA polymerase
Paper 2
Section A
Data Analysis (17 marks)
1.
It had always been assumed that eukaryotic genes were similar in organization to
prokaryotic genes. However, modern techniques of molecular analysis indicated that
there are additional DNA sequences that lie within the coding region of genes.
Exons are the DNA sequences that code for proteins while introns are the
intervening sequences that have to be removed. The graph shows the number of
exons found in genes for three different groups of eukaryotes.
Percentage of genes
100
80
Saccharomyces cerevisiae (a yeast)
60
40
20
0
40
30
Drosophila melanogaster (fruit fly)
20
10
0
20
15
Mammals
10
5
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 <30<40<60>60
Number of exons
(a)
Calculate the percentage of genes that have five or less exons in mammals.
47–49% (units are not needed)
(1)
(b)
Describe the distribution of the number of exons and the percentage of genes in D.
melanogaster.
(2)
D. melanogaster / Drosophila has few genes with one exon;
highest percentage has 2 exons;
most genes have 5 or fewer exons;
a few genes have 10 or more exons / more than 8;
maximum number of exons does not exceed 60;
(c)
(i)
(ii)
Compare the distributions of the number of exons found in genes of S.
cerevisiae and mammals.
S. cerevisiae / yeast has most genes with only 1 exon while mammals
5 exons is most frequent;
no yeast genes have more than 5 exons while some mammal genes
have greater than 60 exons;
mammal genes contain more exons on average;
with a wider distribution than yeast;
(2)
Suggest one reason for the differences in the numbers of exons found in
genes of S. cerevisiae and mammals.
(1)
S. cerevisiae / yeast is a unicellular organism / mammals are
multicellular / complex;
mammals have more transcriptional regulation;
S. cerevisiae smaller in size / more compact genome;
Human DNA has been analysed and details of certain genes are shown in the table
below.
Gene
Gene size / kb* mRNA size / kb
Number of
introns
Insulin
1.7
0.4
2
Collagen
38.0
5.0
50
Albumin
25.0
2.1
14
Phenylalanine hydroxylase
90.0
2.4
12
Dystrophin
2 000.0
17.0
50
*kilobase pairs
(d)
Calculate the average size of the introns for the albumin gene.
gene size – mRNA size = intron size / 25.0 – 2.1 = 22.9 kb;
average size of intron =
(2)
22.9
= 1.6 (± 0.1) kb; (unit required)
14 introns
(e)
Analyse the relationship between gene size and the number of introns.
smaller genes usually have less introns / larger genes have more introns /
relationship not clear;
dystrophin and collagen have same number of introns but the dystrophin
gene is larger;
albumin has more introns but is smaller than the gene for phenylalanine
hydroxylase;
(f)
Determine the maximum number of amino acids that could be produced by
translating the phenylalanine hydroxylase mRNA.
2.4 kb (1 amino acid / 3 bases) = 800 amino acids or 799 amino acids
Hemoglobin is a protein composed of two pairs of globin molecules. During the
process of development from conception to adulthood, human hemoglobin changes in
composition. Adult hemoglobin consists of two alpha- and two beta-globin molecules.
Two globin genes occur on chromosome 16: alpha- and zeta-globin. Four other globin
genes are found on chromosome 11: beta, delta, epsilon and gamma. The graph below
(2)
(1)
illustrates the changes in expression of the globin genes over time.
50
Key:
alpha-globin
gamma-globin
beta-globin
delta-globin
epsilon-globin
zeta-globin
40
Percentage of
hemoglobin 30
20
10
0
10
20
30
weeks of gestation
40
birth
2
6
4
month of age
(g)
State which globin genes are the first to be expressed after fertilization.
epsilon and zeta (globin)
(1)
(h)
Compare the expression of the gamma-globin gene with the beta-globin gene.
gamma genes (mostly) expressed before birth and beta genes expressed after
birth;
beta-globin levels rise at 28(±2) weeks of gestation while gamma
levels decrease / as one rises, the other falls;
gamma-globin expression starts at 0–2 weeks whereas
beta-globin starts at 26 (±2) weeks / gamma expression starts earlier;
one month after birth hemoglobin has equal
mixture of beta-globin and gamma-globin;
gamma levels go to zero while beta becomes a regular part of hemoglobin;
(3)
(i)
Deduce the composition of the hemoglobin molecules at 10 weeks of gestation and 2
months after birth.
(2)
10 weeks after gestation:
two alpha-globins with two gamma-globins / 49 (50)% alpha and 48 (49)%
gamma;
2 months after birth:
variety of molecules all containing alpha and two chains from the other
three types / 6% delta, 14% gamma, 35% beta, 50% alpha;
Paper 2
Section A
Short Structured (5 Marks)
1.
Define the term degenerate as it relates to the genetic code.
more than one codon / base triplet codes for an amino acid
2.
Compare DNA transcription with translation.
both in 5' to 3' direction;
both require ATP;
DNA is transcribed and mRNA is translated;
transcription produces RNA and translation produces polypeptides / protein;
RNA polymerase for transcription and ribosomes for translation / ribosomes
in translation only;
transcription in the nucleus (of eukaryotes) and translation in the cytoplasm / at ER;
tRNA needed for translation but not transcription;
Section B
Extended Response (21 Marks)
1.
The process of translation involves the use of transfer RNA (tRNA) and amino acids.
Outline the structure of tRNA.
tRNA is composed of one chain of (RNA) nucleotides;
tRNA has a position / end / site attaching an amino acid;
(Reject tRNA contains an amino acid.)
at the 3' terminal / consisting of CCA / ACC;
tRNA has an anticodon;
anticodon of three bases which are not base paired /
single stranded / forming part of a loop;
tRNA has double stranded sections formed by base pairing;
double stranded sections can be helical;
tRNA has (three) loops (sometimes with an extra small loop);
tRNA has a distinctive three dimensional / clover leaf shape;
Accept any of the points above if clearly explained using a suitably labelled diagram
2.
Describe the consequence of a base substitution mutation with regards to sickle cell
anaemia.
mutation is a change in DNA sequence;
changes the mRNA during transcription;
changes the amino acid sequence;
substitution mutation / changes to one codon;
glutamic acid is changed to valine / GAG to GTG;
changes the shape of hemoglobin /
hemoglobin becomes less soluble and crystallizes out;
cannot carry oxygen as well;
red blood cells sickle / impairs blood flow;
causes other health problems / anemia / tiredness;
sickle cell anemia caused by two mutated recessive alleles;
(1)
(4)
(5)
(7)
3.
Explain the process of translation.
consists of initiation, elongation and termination;
mRNA translated in a 5' to 3' direction;
binding of ribosome to mRNA;
small sub-unit then large;
first / initiator tRNA binds to start codon /
to small subunit of ribosome;
AUG is the start codon;
second tRNA binds to ribosome;
large subunit moves down mRNA after a second tRNA binds;
amino acid / polypeptide on first tRNA is transferred /
bonded to amino acid on second tRNA;
peptide bonds between amino acids / peptidyl transferase;
requires GTP;
movement of ribosome /
small subunit of ribosome down the mRNA;
loss of tRNA and new tRNA binds;
reach a stop codon / termination;
polypeptide released;
tRNA activating enzymes link correct amino acid to each tRNA;
(activated) tRNA has an anticodon and the
corresponding amino acid attached;
(9)