E4U Level 2 Biology 2.7 (AS 91159) 2015 — page 1 of 5
Assessment Schedule – Biology 2.7: Demonstrate understanding of gene expression (AS 91159)
QUESTION ONE - Evidence
Achievement
Describes transcription as taking place in the nucleus and translation in the cytoplasm.
Explains the role of the promotor, coding and terminator regions in the expression of a
gene.
A gene has three regions. The coding region contains the base sequence for a protein. The
coding region is bounded by a promotor region and a terminator region. The promotor
region turns the gene ‘on’ or ‘off’. RNA polymerase binds to the promotor region to begin
transcription of the gene and the terminator region signals the end point of transcription.
The promotor and terminator regions therefore control transcription by telling RNA
polymerase where to start and stop so only one gene length of DNA is transcribed into
mRNA at a time to be translated into a protein.
Explains transcription explain how transcription and translation can make a protein from
the genetic information carried by a gene.
Transcription is the first stage of gene expression. Transcription is the rewriting of the
DNA code into mRNA, it occurs in the cell’s nucleus. The DNA consists of two strands,
the coding strand and the template strand. An enzyme (RNA polymerase) unwinds the
DNA molecule and adds free nucleotides to the single mRNA strand using the template
strand of DNA. The mRNA is complementary to the template strand except that uracil
replaces thymine in mRNA. Base pairs combinations between DNA-RNA during
transcription are: A-U, C-G, G-C and T-A. The codons on the mRNA carry the same
message as the triplets on the coding strand. The polymerase detaches and the mRNA is
transported from the nucleus into the cytoplasm for translation.
• Describes where transcription and translation
take place
• Describes one of: coding,
promotor or terminator
region.
• Describes the complimentary base pairing rule
• Describes mRNA as the
product of transcription.
• Describes one of: triplets,
codons or anti-codons
with their correct function.
• Describes that one polypeptide chain/protein as
the product of translation.
Merit
Excellence
• Explains the role of two of
the promotor or coding or
terminator regions in the
expression of a gene
• Explains transcription.
• Explains translation.

Full discussion of protein
synthesis, presented in a
logical sequence, includes
the role of the promotor,
coding and terminator
regions of the gene.
Translation is the second stage of gene expression. It is the process by which ribosomes
read the mRNA to make a protein. It occurs in the cytoplasm on the ribosomes.
Translation begins with a start codon. As each codon is read, a tRNA molecule with the
complementary sequence (anticodon) to the mRNA codon binds to the ribosome and
delivers its amino acid. Once added, the tRNA moves away allowing the next amino acid
to be added, forming a polypeptide chain. Translation continues until a ribosome reaches
the stop codon and the polypeptide is released, folding into the three-dimensional
structure of a functional protein.
NØ
N1
N2
A3
A4
M5
M6
E7
E8
No response or no
relevant evidence.
Provides any
ONE statement
from
Achievement.
Provides any TWO statements
from
Achievement.
Provides any
THREE
statements from
Achievement.
Provides any
FOUR statements from
Achievement.
Provides any
ONE
statement
from Merit.
Provides
TWO
statements
from Merit.
Partial
discussion for
Excellence.
Full
discussion
for
Excellence
E4U Level 2 Biology 2.7 (AS 91159) 2015 — page 2 of 5
Q
TWO
(a)
Expected Coverage
Coding
DNA
Template
DNA
mRNA
Amino
acid
Achievement
ATC
ATC
TTT
GGT
TAG
TAG
AAA
CCA
AUC
AUC
UUU
GGU
IIe
IIe
Phe
Gly
Coding
DNA
Template
DNA
mRNA
Amino
Acid
ATC
ATT
GGT
TAG
TAA
CCA
AUC
AUU
GGU
IIe
IIe
Gly
(c)
Explains how the deletion mutation affects the sequence of the bases
in the DNA and the final protein.
The deletion of three DNA nucleotides, CTT, results in a (reading)
frameshift. A frameshift typically alters all the triplets on the DNA
that follow and when transcribed into mRNA the codons produce a
different sequence of amino acids are translated. This may results in
early termination of protein synthesis when a STOP codon is transcribed on the mRNA. Therefore no protein is made, or a protein is
made that has no biological function. This occurs in the deletion
mutation that causes cystic fibrosis, the amino acid, Phe, is deleted
Excellence
• In (a) correctly completes the
normal template DNA
sequence in the table.
• In (a) correctly completes the
normal amino acid sequence.
• In (b) correctly completes
one row or column of the
table.
(b)
Merit
• In (c) describes that the deletion mutation causes a reading frame shift.
• In (c) describes a change to
the protein: e.g. a polypeptide will be shortened by a
premature STOP codon.
• In (c) describes that each
amino acid can have more
than one codon code or AUU
and AUC both code for IIe.
• In (b) correctly completes the
table.
• In (c) explains how the frame
shift causes a shift in the ‘reading frame’ of the triplet code.
• In (c) explains why the second
amino acid stayed the same.
• In (c) fully discusses links
between mutation in the
DNA (triplet) to corresponding codon, and amino acid in
the context of cystic fibrosis.
E4U Level 2 Biology 2.7 (AS 91159) 2015 — page 3 of 5
from the sequence and the protein cannot perform its transport function.
Explain why the second amino acid stayed the same and link this to
the degeneracy of the genetic code.
There are 64 different codons that code for 20 amino acid , so several
codons code for the same amino acids. In normal DNA the second
codon AUC codes for the amino acid, IIe. By chance, the deletion of
CCT changed the second codon by only a single letter to AUU. This
variation in the third letter of the codon did not change the amino
acid.
Coding
DNA
Template
DNA
mRNA
ATC
ATC
TTT
GGT
TAG
TAG
AAA
CCA
AUC
AUC
UUU
GGU
NØ
N1
N2
A3
A4
M5
M6
E7
E8
No response or no
relevant evidence.
Provides any
ONE statement
from Achievement.
Provides any
TWO statements
from Achievement.
Provides any
THREE statements from
Achievement.
Provides any
FOUR statements
from Achievement.
Provides any
ONE statement
from Merit.
Provides
TWO
statements
from Merit.
Discusses deletion
mutation.
Fully discuses the
CFTR mutation.
E4U Level 2 Biology 2.7 (AS 91159) 2015 — page 4 of 5
QUESTION THREE - Expected Coverage
Achievement
Discuss how the metabolic pathway for choline and the internal gut environment interact to influence fish odour syndrome.
• Describes the term metabolic
pathway, eg a series of enzyme
controlled reactions.
A metabolic pathway is a sequence of enzyme controlled reactions. The
product of one enzyme controlled reaction becomes the substrate for the
next enzyme controlled reaction in the metabolic pathway. An
interruption in the reaction series prevents the pathway from progressing
to the end and can result in a metabolic disorder. When this happens,
there might be too much of one substance or too little of another and the
phenotype is affected.
• Describes that the product(s) of
one reaction are the reactants
(substrates) for subsequent reaction(s).
In the gut foods rich in choline such as meat and eggs provide the substrate for enzymes made by the gut bacteria to produce trimethylamine
(TMA) and this compound has a strong fishy smell. People with a normal FMO3 gene can make the FMO3 enzyme so that the intermediate
substrate, TMA, is broken down to odourless trimethylamine N-oxide
(TMAO), the end-product of this metabolic pathway. If the FMO3
enzyme is missing or its activity is reduced because of a mutation in
the FMO3 gene, TMA builds up in the body to be excreted in sweat that
gives off a strong fishy odour.
Typically people with fishy odour syndrome avoid meat and eggs to
reduce the amount of choline substrate in their gut environment. Without
this substrate the smelly intermediate substrate TMA can not be made.
This is the recommended treatment because the genotype of a person
with fishy odour syndrome cannot be changed but the internal gut
environment and therefore the amount of choline substrate is able to be
modified by a change in diet.
• Describes that the genes contained in gut bacteria codes for
enzymes that catalyses the
reaction of choline to
trimethylamine.
Merit
Excellence
• Explains genes and enzymes control
a metabolic pathway.
Discusses how the metabolic
pathway for choline and the
internal gut environment
interact to influence fish odour
syndrome.
• Explains how the metabolic pathway
may be affected when any the FMO3
enzyme is non-functional.
• Explains why a diet low in meat and
eggs is the recommended treatment
• Describes that FMO3gene codes
for the FMO3 enzyme that catalyses the reaction of
trimethylamine to
trimethylamine N-oxide.
NØ
N1
N2
A3
A4
M5
M6
E7
E8
No response or no
relevant evidence.
Provides any ONE
statement from
Achievement.
Provides any TWO
statement from
Achievement.
Provides any
TWO statements from
Achievement.
Provides any
THREE
statements from
Achievement.
Provides ONE
statement from
Merit.
Provides TWO
statements from
Merit.
Partial
discussion
Full discussion
linked to diet.
E4U Level 2 Biology 2.7 (AS 91159) 2015 — page 5 of 5
Judgement Statement
Score range
Not Achieved
Achievement
Achievement
with Merit
Achievement
with Excellence
0–6
7 – 12
13 – 18
19 – 24