SBI4U- Molecular Genetics
Name: _____Answer_________________________
Evaluation Allocation
Achievement Category
Knowledge/Understanding (K/U)
Thinking & Inquiry (I)
Communication (C)
Making Connections (MC)
TOTAL:
Marks Received
/72
Achievement Level
30
18
11
13
Part A Multiple Choice (10 marks) (K/U) Using the test page provided, answer each of the
following on the SCANTRON CARD provided. Make NO MARKS on this test page.
Part B Modified True/False (10 marks) (K/U) Indicate whether the sentence or statement is
true or false. If false, change the italicized term or phrase to make the statement true in the
space provided.
1. Telomeric DNA is added continuously to one’s chromosomes as cells go through progressive
divisions during the course of their lives. ______________reduced due to losses with each
replication (inactive telomerase)________
2. Selective replication gives a reason as to why one gene can end up coding for more than 1
polypeptide. This occurs during post-transcriptional processing, allowing multiple combinations
of exons to be possible.
____alternative splicing___________________________________________________
3. Chargaff's rules state that in DNA the totalled proportions of adenine and guanine always
equal the totalled proportions of thymine and cytosine.
__________________________________________________________________________
4. The tRNA anticodon, 3’GGC 5’ may be able to recognize the mRNA codon, 5’ CCA 3’.
__________________________________
5. The mRNA transcript is likely to be most similar to the coding stand found on the DNA.
___________________________________________________________________
6. In a healthy individual, a normal cellular response to a build up of DNA structural damage would
be to start replicating the DNA rapidly and continuously.
____suspend division and go through apoptosis_____________________________________
7. Diseases, like diabetes, illustrate how the inheritance of one gene is the sole reason
explaining one’s fate.
________environmental factors interacting with underlying genetics_____________________
8. A mass of cells is considered to be cancerous when it is deemed by medical professionals that
the tumour has the ability to continue growing.
_____________spreads through blood stream (metastasis)_______________________
9. Special enzymes used by bacteria to degrade foreign DNA (injected by viruses) and by
molecular biologists to make recombinant DNA are called exonucleases.
_________restriction endonucleases_______________________________________
10. Inducible operons are gene control systems that are usually off but can be turned on if an
environment trigger precipitates the change.
__________________________________________________________________________
Part D Communication (7 marks) (C)
11. Put the following events involved with protein synthesis into order by putting appropriate
numbers in the blanks. For some of you it may help to draw a picture before you tackle
this question. 1 will represent the first event; 10 will represent the last event. Part
marks will be allocated in case one answer throws others off.
____3__
A 5’ cap and 3’ poly A tail are added to the mRNA
____9__
The stop codon is reached on the mRNA; no tRNA binds, the ribosomes
and polypeptide are released.
____4__
The mRNA leaves the nucleus.
____8__
The ribosome shifts one codon down, opening up its A site. The next
tRNA binds.
____1__
The DNA double helix is unwound beginning at the promoter sequence.
____7__
The larger ribosomal unit clamps down on the tRNA carrying MET.
____2__
Ribonucleotides are added by RNA polymerase creating a single strand
of mRNA.
____5__
The 5’ cap binds to the small ribosomal unit, leading to the exposure of
the start codon; the first tRNA carrying MET binds.
____10__
The first amino acid is cleaved off; the polypeptide undergoes folding.
____6__
The anticodon of the second tRNA binds at the A site; peptidyl
transferase catalyzes a reaction to join the two amino acids together.
For the following classical genetic experiments, choose to complete ONLY ONE of the
following. Look at the picture demonstrated and communicate meaning to them as to 1) what
scientific procedures were carried out, 2) the outcome observed, and 3) the key conclusion
from the experiment. (4 marks)
12. Experiment One – Meselson and Stahl
13. Experiment Two – Hershey and Chase
Marks were allocated based upon 2 marks for explanation of procedure going above and beyond information already stated in the
diagrams. 1 mark was explaining the outcome observed (go beyond info already stated) and then 1 mark for the significance to
DNA research (going beyond info already stated). For example, if you said for experiment 1 that it proved semi-conservatism, then
you’d have to explain what that meant since semi-conservatism was already stated in the diagram.
Part D Short Answer (10 marks) (K/U) Answer the following in the space provided.
14.
The genetic code is often termed to be 'redundant’ or ‘degenerate’. What is the
meaning of this term, and elaborate on the importance of this characteristic as it applies
to protein synthesis? (3 marks)
This means that there is more than 1 codon for an amino acid. This is significant since if there is a mutation where a base pair is
changed, then there is the possibility that when the corresponding codon is read in protein synthesis, the same amino acid will be
placed in the polypeptide. Thus, the effect of the mistake will be nullified and a regularly functioning protein can be made.
15.
Explain why DNA replication is slightly slower in the lagging strand of DNA than in the
leading strand. (3 marks)
Many answered this question as to how DNA replication is slowed. I want to know why it ends up being slower. This then is a
questiion almost the same from the chapter 4 practice quiz.
Because DNA is antiparallel and replication needs to proceed in the 5’  3’ direction, one of the strands wll be synthesized
continuously, as the polymerase follows helicase in the opening of the double helix. On the lagging strand, the polymerase is
working away from the opening of the fork, and thus needs needs to be put together in fragments. Why? B/c as the fork opens up,
multiple polymerases need to come into fill in the gaps. Also, multiple enzymes (eg. Ligase, polymerase I) need to come in to join
together fragment as and seal the sugar-phosphate backbone.
16.
Cancer is known to arise when multiple mutations have accumulated in certain parts of
the DNA over time. What are names of two potential target genes where these
mutations can occur and what impact can these have on the cell based upon what we
know these genes normally do? (4 marks)
Name of Target Gene
Impact on Cell Operation if Gene is Mutated
ProtoOncogene
These genes often act as regulators of cell division; if mutated into an oncogene, this tends to
accelerate cell division
Tumor Repressor
These genes often act as brakes on cell division. If mutated, the brake is released allowing
cell division to increase
Other answers exist: Telomerase gene, DNA repair genes, etc
Part E Thinking and Inquiry (18 marks) (I) Answer the following in the space provided.
17.
Researchers learned about the lac operon due to observations made as a result of
mutations affecting lactose metabolism.
a) If a mutation occurred affecting the operator site such that a component could not
bind, what effect on the transcription of structural genes would one observe in the
presence and absence of lactose. Clearly state your response to both scenarios.
What would be the disadvantage to the organism? (2 marks)
The component that wouldn’t bind is the repressor protein to turn off the operon. Thus, the system will always be on resulting
in excessive transcription and translation of the genes to break down lactose. This would be a waste in energy.
b) A bacterium contains a mutation in the lacZ gene, such that beta-galactosidase
cannot be made. It also contains a mutation that prevents it from producing lacI, the
repressor protein. Comment on the impact of these mutations on the bacterium in
an environment with surrounding lactose. (2 marks)
If lacI (repressor) is not made, then the system is always on. However, since a mutation exists in the beta-gal gene, then the
enzyme will not be made. Lactose could not be used as an energy source and there would be an excessive waste of energy to
the organism.
18)
The following represents the DNA version (actually called cDNA) of the mRNA of the
proinsulin gene. The polypeptide produced from this is post-translationally processed to
make mature insulin, helping in the maintenance of blood sugar.
* the numbers to the left represent the numbering of bases beginning at each row
1
61
121
181
241
301
361
421
agccctccag
tggccctgtG
cagccgcagc
tagtgtgcgg
tgcaggtggg
ccctggaggg
ccctctacca
ccgcctcctg
gacaggctgc
gatgcgcctc
ctttgtgaac
ggaacgaggc
gcaggtggag
gtccctgcag
gctggagaac
caccgagaga
atcagaagag
ctgcccctgc
caacacctgt
ttcttctaca
ctgggcgggg
aagcgtggca
tactgcaact
gatggaataa
gccatcaagc
tggcgctgct
gcggctcaca
cacccaagac
gccctggtgc
ttgtggaaca
agacgcagcc
agcccttgaa
agatcactgt
ggccctctgg
cctggtggaa
ccgccgggag
aggcagcctg
atgctgtacc
cgcaggcagc
ccagcaaaa
ccttctgcca
ggacctgacc
gctctctacc
gcagaggacc
cagcccttgg
agcatctgct
cccacacccg
The amino acid sequence for the polypeptide produced is found below. It has 110
amino acids in all (numbers below correspond to the amino acid).
*
*
*
*
*
*
MALWMRLLPLLALLALWGPDPAAAFVNQHLCGSHLVEALYLVCGERGFFY
1
10
20
30
40
50
*
*
*
*
*
TPKTRREAEDLQVGQVELGGGPGAGSLQPLALEGSLQKRGIVEQCCTSIC
60
70
80
90
100
*
SLYQLENYCN
110
a) What is the direction of the cDNA sequence shown? (1 mark)
5’  3’ since this is coding DNA (most similar to mRNA).
b) Imagine you know this amino acid sequence and are looking for the start of the
proinsulin gene. Above, you will notice possible start codons underlined (locate 3
possible). Which of these start codons is correct? Support your choice with
reasoning (ie. Why you picked that one and ruled out another one?). (2 marks)
The first start codon. The next triplet after codes for alanine, and the triplet after that codes for leucine. This is the same as
the amino acid sequence below. When you look at the triplets after the other possible start codons, the amino acids don’t
match up with the known sequence.
c) When looking at the amino acid sequence, how many nitrogen bases would
correspond to this? (1 mark)
110 x 3 nitrogen bases/amino acid = 330 nitrogen bases
d) What can you look for to find the end of the translated portion of this gene? For a
bonus mark, circle it. (1 mark + 1 mark BONUS)
Look for a stop codon (3 choices)
e) Pertaining to the first 4 amino acids of this polypeptide:
i)
Show the double-stranded DNA corresponding to this (Label strand
directions) (2 marks)
5’ ATG GCC CTG etc. 3’
3’ TAC CGG GAC etc. 5’
ii)
Show the mRNA corresponding to this (Label strand direction) (2 marks)
5’ AUG GCC CUG etc. 3’
iii)
Show a possible tRNA anticodon for one of the codons found on the
mRNA by showing a picture of how they would match up on the ribosome.
(2 marks)
f) On the above DNA sequence, a guanine is written in a capital letter (second row,
first grouping). Imagine that this guanine was replaced by an adenine via mutation.
What type of mutation would you call this? Explain your rationale and the impact it
would likely have on the final polypeptide produced, and the person. (3 marks)
This creates a nonsense mutation – formation of a pre-mature stop codon
Results in a very short polypeptide formed – would have drastic effect as this occurs very early in the amino acid sequence so
one would expect the polypeptide to be malfunctional (since shape determines function); the person would not be able to
regulate their blood sugar levels as no functional insulin would be made
Part F Making Connections (13 marks) Answer the questions on the back of the last two
pages of the test OR where space is available.. NOTE: Everyone will complete question 23.
Please read carefully for questions with choice options.
For Question 21 and 22, choose ONLY ONE of the following.
21.
Recent research indicates that in some bacteria, when an anticodon attempts to
hydrogen bond to a codon, two parts of the ribosome (called A1492 and A1493) change
shape and check that the match is correct (ie. That the corresponding bases are
complementary). There is also evidence that an antibiotic, paromycin, causes the same
kind of shape change in A1492 and A1493 as complementary base pairs do, no matter
what bases pairs match up.
a) Hypothesize as to how the antibiotic might kill bacteria. (2 marks)
The antibiotic’s action would result in all kinds of amino acids being brought in in the wrong spots. This is because any tRNA
antiicodon will be accepted. The bacterial proteins made will be useless and the bacteria will die as a result.
b) Not every antibiotic that can affect ribosome operation can be given to humans
for treatment. Give a reason as to why. (2 marks)
You have to make sure the antibiotiic affecting ribosome operatiion does not target a key area involved with human ribosomes.
If the antibiotic is non-selective, then it could kill off human cells too
OR
22.
The recognition sites of two restriction endonucleases are shown below as well as a
DNA fragment with a gene of interest for cloning.
5'AAGGGCCCGAATTCGCGAATTCAGGGCCC---------gene------GAATTCT3'
EcoRI
SalI
Recognition
sites
5'-GAATTC-3'
3'-CTTAAG-5'
5'-GGGCCC-3'
3'-CCCGGG-5'
Discuss the steps that you would take to produce a bacterium that would express the
gene from a recombinant plasmid. (4 marks)
5'AAGGGCCCGAATTCGCGAATTCAGGGCCC---------gene------GAATTCT3'
EcoRI
SalI
Recognition
sites
5'-GAATTC-3'
3'-CTTAAG-5'
5'-GGGCCC-3'
3'-CCCGGG-5'
The bolded print on either side of the gene region indicate ECORI sites. By using this enzyme, you could
cut out the gene in one test tube. Then treat a plasmid with 1 ECORI site with the same enzyme. Make
sure the plasmid has a gene for antibiotic resistance. Combine the two allotments together using the
transformation steps. Plate bacteria on an antibiotic medium to select for only bacteria that have taken up
the plasmid (b/c they will grow given they will have the resistant gene.
***#23 is a compulsory question for everyone***
23.
Cystic fibrosis can come about due to various mutations that impact a chloride ion
transporter protein. A unique mutation occurs at the beginning of an intron region
that leads to a severe case. Provide a possible explanation as to the severity of this
mutation given that it occurs in an intron region. (3 marks)
This could affect the ability of a spliceosome to detect a place to cut out an intron during splicing. If this intron is missed, a
longer than normal mRNA could be translated into a non-functional protein.
For Question 24 and 25, choose ONLY ONE of the following.
24.
a) Only a small section of the mammalian genome is made of genes. The rest has
been called “Junk DNA” in the past. Describe a couple of examples of how noncoding DNA has been proven to be anything but JUNK. (4 marks)
b) How do terms like JUNK DNA help or hinder society’s understanding of the need
for scientific procedure and on-going genomic research. (2 marks)
It generally hinders science because it gives the connotation that junk DNA has nopurpose. This is likely to give some people less
incentive to study it further, orit may lead the public with less incentive to support research with public funding. We know that Junk
DNA is not junk as it has been shown that it is linked to gene regulation, alternative splicing, and telomeric DNA (protects the end of
chromosomes during replication)
OR
c) Huntington’s disease is a neurodegenerative disease that affects motor
coordination and is characterized by progressive dementia. The affected gene
coding for a protein called Huntingtin, is characterized by a repeat of 3 nitrogen
bases (CAG). In normal individuals, they can inherit 28-35 repeats of CAG
whereas affected Huntington’s patients will have a full-blown illness if they have
40 repeats of CAG. Below, is a gel electrophoresis pattern demonstrating the
alleles inherited by a son whose dad has Huntington’s disease and whose mom
does not. The gel provides a comparison of their alleles.
SON
DAD
MOM
Negative terminal
Positive terminal
a) Analyze the gel results and determine if the boy has inherited the allele for
Huntington’s disease. Explain your reasoning. (3 marks)
Gel electrophoresis separates based upon the lengths of DNA fragments. One would expect that the disease allele would be
the longest b/c it has the most number of repeats. Thus, this would be the one furthest up on the Father lane. The boy did not
get this allele but rather the other one from the dad. Thus, he would not have the disease.
b) Why does mom only have one band? (1 mark)
Since each person inherits 2 alleles, this would imply that she is homozygous for that allele.
c) What is the name of a mutation that could give rise to the situation described above?
Insertion or Duplication could work
d) What amino acid would be found in high abundance in the Huntingtin polypeptide?
glutamine