SBI4U- Molecular Genetics
Name: __ANSWER______________________________
Evaluation Allocation
Achievement Category
Knowledge/Understanding (K/U)
Thinking & Inquiry (I)
Communication (C)
Making Connections (MC)
Marks Received
TOTAL:
/85
Achievement Level
31
25
18
11
Part A Multiple Choice (15 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. HIV virus is known to replicate without exonuclease activity. You would expect individual HIV
viruses to be different.
_________________________________________________________________________
2. DNA in the nucleus of a human eukaryotic cell is found on one circular chromosome.
_________46 linear chromosomes_______________________________________________
3. You isolate DNA from a bacterium on the ocean floor of a deep sea vent. You measure the
adenine content to be 31%. According to Chargaff’s rules, a correct estimate of the other bases
would be: 31% Guanine, 38% Thymine, and 38% Cytosine. ___31% T, 19% G, 19% C___
4. There would not be a tRNA anticodon 3’ AUU 5’ to bond with the codon
5’ UAA 3’ on the mRNA _________________________________________________________
5. The mRNA transcript is likely to be most similar to the coding strand/anti-template 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.
___________apoptosis or suspend growth or stop replication________________
7. DNA rich in guanine/cytosine content would denature at a higher temperature than sample that
is high in adenine/thymine content.
__________________________________________________________________________
8. Ribosomes move along the mRNA, accepting incoming tRNAs in at the P site. __________A
site (except first tRNA but ribosome not moving then)_______________________________
9. The linkage connecting a phosphate and a deoxyribose sugar along the rails of the DNA ladder
is called a glycosyl bond.
__________phosphodiester linkage________________________________________
10. The control of gene expression in prokaryotes is more clearly understood since structural and
regulatory genes are found in clusters called promoters.
________operons_____________________________________________________
Part C Communication (C) (5 marks) Draw a small sketch of a DNA replication fork (with
accompanying explanation) showing why DNA synthesis is different on the leading and lagging
strands. Please name key components (eg. enzymes, DNA terminology) in your response.
Marks allocated for accuracy of replication fork with appropriate strand directions
Key enzymes demonstrating idea bolded above would be helicase and polymerase
Need to show that b/c replication is 5’-3’ and the DNA strands are antiparallel, as helicase
opens up the helix, replication on one strand is continuous, while on the other strand the
polymerase works away from the replication fork; therefore as helicase opens up more DNA,
need more polymerases to make further Okazaki fragments
Part C Completion (C) (13 marks) Fill in the appropriate term/phrase to complete each of the
following statements. All blanks carry a weight of 1 mark, except where indicated in brackets.
13.
DNA is a double-stranded linear polymer made up of ___nucleotide_______________,
each of which consists of a ____pentose sugar (deoxyribose)___ sugar, a
_____phosphate_________________ and one of four __nitrogen_____ bases.
14.
The sequence of __nitrogen bases______ in DNA determines the sequence of
___amino acids_____ in the polypeptide for which the gene codes.
15.
Post-transcriptional processing occurs in the __nucleus_____ (pick location) of the cell.
___Intons______ are removed by _spliceosomes_____ allowing coding regions (called
_exons___) to be fused back together. A ___modified guanine cap_____ is added to
the 5’ end and a ___poly A tail__ made of many ___adenine____ nucleotides is added
to the 3’ end.
Part D Short Answer (6 marks) (K/U) Answer the following in the space provided.
16.
a) Explain what is meant by the “wobble hypothesis” and how this concept is useful in
understanding the nature of the genetic code (ie. What codes for what). (2 marks)
1 tRNA can recognize more than 1 codon b/c of irregular base pairing at third position
on mRNA; thus this can help explain why there might be more than 1 codon per aa.
b) How can what you mentioned in a) be a benefit for possible faults (mistakes) in other
processes. (1 mark)
If a mutation occurs or mistake in transcription leading to an improper triplet, there is the
possibility that the same aa might be placed in the polypeptide
16.
Recent research has suggested about 80% of human genes are alternatively spliced.
Explain what this itatlicized term means, and indicate the significance of this for genetic
research. (3 marks)
Alternative splicing refers to the fact that when introns are cut out, the exons (coding
regions) can be rearranged or even omitted in different ways producing different mRNA
transcripts, thus affecting the resulting aa sequence produced in the polypeptides
This is significance b/c this helps to explain how one gene could correspond to more
than 1 polypeptide refuting the previous thought of a 1:1 relationship
Part E Thinking and Inquiry (25 marks) (I) Answer the following in the space provided.
For Question 17 and 18, choose ONLY ONE of the following.
17.
a) Imagine that you are a geneticist and are investigating a species of
bacteriophage about which little is known.
How might a technique from Hershey and Chase's experiment be used to
discover whether the nucleic acid in the bacteriophage is DNA or RNA? (2 marks)
Use radioactive thymine and uracil to distinguish between DNA or RNA. Label
one test tube where you use only radioactive thymine; the other test tube is radioactive
uracil
By seeing which one contains radioactivity inside the bacterium, can tell if it is DNA or
RNA
b) In Griffith and Avery’s experiment with R and S pneumococcus was used. As an
extension of their experiment, heat-treated S cells were combined with a polysaccharide
digesting enzyme and with a protease (enzyme that breaks down protein). This solution
was then combined with R cells and injected into a mouse. Predict whether or not the
mouse will survive or not. As always, whether you choose living or dead, the mice were
sacrificed and an autopsy was performed. Give a supportive reason of what one would
find. (2 marks)
Mouse would die b/c the DNA is still intact; from these experiments, one learns that
DNA is the transforming principle so as long as it is still intact it has the ability to
transform R cells into S cells
OR
18.
Suppose mammalian cells are cultured in a medium containing radioactive thymine.
They grow and divide many times, until eventually every chromosome contains
radioactive thymine. The cells are then REMOVED and allowed to replicate several
more times in a culture medium containing normal thymine. Daughter chromosomes
were tested with each successive generation to determine whether they carry the
radioactive thymine.
a) Predict the radioactive status of the daughter chromosomes after one, two, and
three rounds of division in the normal medium. (2 marks)
After one round – expect 2 hybrid DNA molecules each with a normal strand and a
radioactive one
Two rounds – expect 2 hybrids, 2 normal DNAs
Three rounds – expect 2 hybrids, 6 normal DNAs
b) Explain how your predictions are consistent with what we know about the way DNA
replicates itself. (2 marks)
DNA replicates semi-conservatively – means that there will always be an old strand
and new DNA strand in the daughter DNAs produced
19.
a) A mutation occurs such that no functional enzyme can be made from the lacZ gene
(B-galactosidase). Describe the possible ramifications of this mutation with respect
to: a) the ability to turn on or off transcription, and b) the ability to use lactose from
the surrounding environment for energy. (2 marks)
Does not affect switch b/c B galactosidase is not connected to the
repressor/operator interaction
Bacterium will not be able to use lactose for energy since a functional B-gal enzyme
cannot be made to break down the lactose (if present)
b) Two mutations of the lacI (repressor) gene have different effects on the repressor
protein and its function in the operon. Mutant A and B represent these different
examples; the last bacterium is normal.
Lactose Availability Lactose Absent
Lactose Present
Mutant A
Transcription occurs
Transcription occurs
Mutant B
No transcription
No transcription
Normal
No transcription
Transcription
* Transcription infers that beta-galactosidase and permease are made
Provide a possible explanation for the different effects of these two mutations on the
repressor protein molecule (lacI). (2 marks)
Mutant A – makes a repressor that changes shape and cannot bind to operator
Mutant B- makes a repressor that changes shape such that lactose can’t remove it
from the operator
20.
The following represents the DNA sequence close to where the beta-globulin gene
(polypeptide from hemoglobin) begins on chromosome #11 of a normal patient and
Maria’s cousin. The actual gene region covers over 1700 base pairs.
Normal_Patient_1
Maria’s Cousin
GAGCCAGGGCTGGGCATAAAAGTCAGGGCAGAGCCATCTATTGCTTACATTTGCTTCTGA
GAGCCAGGGCTGGGCATAAAAGTCAGGGCAGAGCCATCTATTGCTTACATTTGCTTCTGA
************************************************************
Normal_Patient_1
Maria’s Cousin
CACAACTGTGTTCACTAGCAACCTCAAACAGACACCATGGTGCACCTGACTCCTGAGGAG
CACAACTGTGTTCACTAGCAACCTCAAACAGACACCATGGTGCACCTGACTCCTGTGGAG
******************************************************* ****
a) What is the name of the strand of DNA shown? What direction is it read (left to
right)? (1 mark)
Coding strand/antitemplate
5’-3’ is universal way that geneticists store sequences
b) Use this information to find the start of the coding region of this gene. Explain your
selection. (2 marks)
ATG is the start codon, representing the beginning of the first exon
c) The beginning of the coding region is not where transcription would begin. Highlight
a sequence that you would hypothesize would be close to where transcription would
start. Provide a rationale for your answer. (2 marks)
Any sequence resembling a TATA sequence- these sequences make it easier for
opening the DNA double helix b/c less H bonds between A and T
d) Beginning from what you defined as the start of the coding region in b), write this
sequence as mRNA. Show appropriate strand directions. (2 marks)
5’ AUG GUG CAC CUG ACU CCU GAG GAG 3’
e) Translate the sequence from the normal patient into amino acids. (2 marks)
Met- val – his – leu – thr – pro – glu - glu
f) How does this differ from the Maria’s cousin? (1 mark)
Maria’s cousin has valine in the place of glutamic acid
g) What type of mutation would you use to describe this situation? Circle an
appropriate word from each column. (2 marks)
Column#1 – Nature of Mutation
Substitution
Insertion
Deletion
Column#2- Impact of Mutation
Silent
Frameshift
Nonsense
Missense
h) Locate the base pair underneath the star. Imagine if this base was deleted via a
mutation in the family line from Normal Patient #1. What would the severity of this
mutation be like? Provide a rationale for your answer, indicating the impact on the
polypeptide and person. (3 marks)
The insertion would not only change the order of the triplets downstream of the
insertion but this would create a premature stop codon in the next triplet. This would
terminate translation earlier resulting in a severely malfunctional polypeptide made.
This would dramatically influence the health of the person – in this case affect the
ability to transport oxygen
i) Select a possible codon from the mRNA above. Show a possible tRNA anticodon
(with attached amino acid) that would match up with a codon on the ribosome. (2
marks)
As shown in class on board
Part F Making Connections (11 marks) Answer the questions on another sheet of paper.
NOTE: For question 23 and 24, you have a choice to do ONE of two questions.
21.
In Europeans, it has been found that Type I diabetes is more likely to occur in people
who have only low number repeats (<50 repeats) at a particular gene locus (near the
insulin gene) compared to those people who have at least one high number repeat (200
repeats). Below is a gel electrophoresis pattern of the alleles of a family, where the dad
is a Type I diabetic.
a) By examining the gel, determine if the daughter is homozygous or heterozygous.
Explain. (1 mark)
Heterozygous – 2 separate bands
b) Based upon the analysis of the gel, will the daughter be likely to develop diabetes by
this possible method of inheritance? Explain why or why not. (2 marks)
This was marked based upon how it was perceived. Answer that follows
represent the change made to the test during the class period
While mom has a short fragment that could predispose her to diabetes of this type, she
may be protected by the larger fragment band. Dad has diabetes, most likely due to a
homozygous situation where he has inherited two alleles both with a shorter number of
repeats. The daughter only inherited bands of shorter lengths, those of which would
represent the number of repeats necessary to bring on diabetes
(kb) Ladder
Mom
Dad
Daughter
850
700
550
400
350
200
150
100
22.
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)
If this happened, any tRNA anticodon would be accepted on the ribosome with
any amino acid no matter the codon on the mRNA
Thus, the whole amino acid sequence would be messed up, faulty protein made,
resulting in death for the bacterium
b) Not every antibiotic that can affect ribosome operation can be given to humans
for treatment. Give a reason as to why. (2 marks)
Humans have their own ribosomes and need to make key proteins for life too so
a potential antibiotic might not just affect the bacterium but the human too
(possibly killing them)
For Question 23 and 24, choose ONLY ONE of the following.
23.
a) Only a small section of the mammalian genome is made of genes . The rest has
been called “Junk DNA” in the past. How do popular terms such as this help or hinder
society’s understanding of the need for further research? Explain. (2 marks)
Terms like this hinder society as it gives no incentive to research the topic further.
Quick conclusions like this could stop scientific research even though a relevant
function for JUNK DNA could be present (and as we found out is)
b) What is a possible advantage and disadvantage of an organism having large
quantities of JUNK DNA? (2 marks)
Different answers accepted
Advantage – helps to regulate the genes in terms of when they are turned on and off
- Telomeric DNA at chromosome ends help protect valuable genes
Disadvantage – more JUNK DNA means more DNA to replicate – thus more energy
expended by organism
OR
24.
The Canadian government is considering cutting back funding for the DNA barcoding
project. You are a lobbyist in support of this project and have given time to speak to the
MP in charge of scientific research. Provide an argument to the government for keeping
the project in Canada. (4 marks)
Different answers were accepted; Looking for 4 different ideas or 2 different ideas that
were very well explained.
Amino Acids
The one letter abbreviation for each of the twenty amino acids is listed below:
Alanine
A
Glutamine
Q Leucine
L
Serine
S
Arginine
R
Glutamic Acid E
Lysine
K
Threonine
T
Asparagine
N Glycine
G
Methionine
M Tryptophan W
Aspartic Acid D
Histindine
H
Phenylalanine F
Tryosine
Y
Cysteine
Isoleucine
I
Proline
Valine
V
C
P