SBI4U – Molecular Genetics Unit Test

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Name:
Date:
TEACHER’S NOTES – SBI4U Molecular Genetics Unit Test
(Summative Test)
MINISTRY EXPECTATIONS
BIG IDEA
DNA contains all the genetic information for any living organism.
Proteins control a wide variety of cellular processes.
OVERALL EXPECTATIONS
D2. Investigate through laboratory activities, the structures of cell components and their roles in
processes that occur within the cell
D3. Demonstrate an understanding of concepts related to molecular genetics, and how genetic
modifications is applied in industry and agriculture
SPECIFIC EXPECTATIONS
D2.1 Use appropriate terminology related to molecular genetics
D2.2 Analyze a simulated strand of DNA to determine the genetic code and base pairing of DNA
D3.1 Explain the current model of DNA replication, and describe the different repair
mechanisms that can correct mistakes in DNA sequencing
D3.2 Compare the structures and functions of RNA and DNA, and explain their roles in the
process of protein synthesis
D3.3 Explain the steps involved in the process of protein synthesis and how genetic expression is
controlled in prokaryotes and eukaryotes by regulatory proteins
D3.4 Explain how mutagens, such as radiation and chemicals, can cause mutations by changing
the genetic material in the cells
D3.5 Describe some examples of genetic modification, and explain how it is applied in industry
and agriculture
D3.6 Describe the functions of some of the cell components used in biotechnology
D3.7 Describe, on the basis of research, some of the historical scientific contributions that have
advanced our understanding of molecular genetics
SOURCES
Nelson Biology 12: University Preparation
http://seniorbiology8.wikispaces.com/Genetic+Processes
NOTES
This unit test is a summative assessment (assessment OF learning) of what the students
have learned throughout the unit. It is to be completed after all appropriate expectations
(according to the Grade 12 Science Ontario Curriculum) have been covered in the
molecular genetics unit.
Calculators are not required, but students may use them if they wish.
A codon chart is provided for students, attached at end of test.
Students with special needs may require more time to complete the unit test.
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Unit Test Analysis
Test
Question
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1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
Question Type
Multiple choice
Diagram
Numerical
Problem
Essay
Short Answer
Data Analysis
Essay
Essay
Essay
Coded
Ministry
Expectations
D3.3
D3.3
D3.1
D3.1
D3.7
D3.1, D3.2
D2.2
D3.4
D2.1
D3.2
D2.1
D3.2
D3.7
D3.5
D3.4
D2.1
Achievement Chart Category
Knowledge and
Understanding
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Thinking and
Investigation
Communication Application
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D3.7
D2.1, D3.7
D2.2, D3.4
D3.3, D3.5
D2.1, D3.1
D2.1, D3.6
D2.1, D3.3
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Date:
SBI4U – Molecular Genetics Unit Test
K/U
/15
T/I
/13
C
/10
A
/14
TOTAL
/52
PART A: Knowledge/Understanding [15]
Select ONE answer from the following questions.
1. After the primary transcript (mRNA) has been formed, what mediates the cutting of
introns?
a) 5’ cap
b) Exons
c) Introns
d) Poly – A tail
e) Spliceosomes
2. Why are archaea more like eukaryotes than prokaryotes?
a) Because plasmids are associated with their DNA
b) Because histones are associated with their DNA
c) Because nucleoids are associated with their DNA
d) Because chromosomes are associated with their DNA
3. Which enzymes are responsible for most DNA repair?
a) DNA polymerase I and II
b) DNA polymerase II and III
c) DNA polymerase I and II
d) DNA ligase
4. What is the function of DNA polymerase III:
a) To unzip the double stranded DNA chromosome
b) To help uncoil the DNA due to being ‘supercoiled’
c) To extend the new DNA strand by adding new nucleotides onto the growing
strand
d) To cut out the RNA primers and replace them with DNA nucleotides
5. What did Edward Chargaff’s experiments reveal about DNA?
a) DNA is the genetic material
b) Bases occur in definite ratios in double-stranded DNA
c) There is a material that can transform bacteria
d) X-rays can be used to find the shape of DNA
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6. Why is RNA structurally different from DNA?
a) RNA is generally single stranded
b) RNA is generally double stranded
c) RNA is always read in the 5’ to 3’ direction
d) RNA is always read in the 3’ to 5’ direction
7. What is the correct tRNA anti-codon sequence for the 5’-ACT-3’ DNA sequence?
a) 5’-UGA-3’
b) 3’-UGA-5’
c) 5’-ACU-3’
d) 3’-ACU-5’
8. Why is the flu virus considered to be dangerous?
a) There is no vaccine for it
b) It uses reverse transcriptase to insert its DNA into its host
c) It has a low mutation rate, which makes a vaccine hard to develop
d) It has a high mutation rate, which makes a vaccine hard to develop
9. The process by which mRNA is converted into a protein sequence is called:
a) Elongation
b) Initiation
c) Transcription
d) Translation
10. In which direction is the mRNA read by the ribosome:
a) 3’ to 5’
b) 5’ to 3’
c) 1’ to 3’
11. What does a codon code for:
a) Amino acid
b) Carbohydrates
c) Cholesterol
d) Nucleotides
12. There are two pockets/active sites inside the ribosome, what are they called:
a) A site and B site
b) B site and C site
c) P site and A site
d) P site and B site
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13. What does the Sanger method involve?
a) Using radioactive RNA primers as chain promoters to determine DNA sequences
b) Using radioactive deoxynucleotides as chain elongators to determine DNA
sequences
c) Using dideoxynucleotides as chain terminators to determine DNA sequences
d) Using dideoxynucleotides as chain elongators to determine DNA sequences
14. What is the first step in the polymerase chain reaction (PCR)?
a) Annealing the DNA primers
b) Building a complementary DNA strand between the primers
c) Placing the solutions in an ice-water bath
d) Separating the strands of DNA
15. Which kind of mutation has the least impact on an organism:
a) Deletion
b) Insertion
c) Nonsense
d) Silent
PART B: Application [14]
16. Label any 10 different things in the diagram below of replicating DNA. [10]
17. A double stranded DNA sequence is composed of 14 % Guanine. Calculate the
percentage of the other 3 bases A, T, and C. [4]
Given: 14% G
Required: % A, T, and C
Analysis: According to the Chargaff rule, in a double stranded DNA sequence:
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%G=%C
%A=%T
%G=14
Solution:
%A +%T +%G +%C =100
% A +%T +14+14=100
Therefore, we have
%A+%T=72
%A=%T=72/2=36
Paraphrase:
Therefore, the percentages of the following bases are:
A=36%
T=36%
C=14%
PART C: Thinking/Inquiry [13]
18. What if Meselson and Stahl’s observations had looked like the tubes below? What
would this mean for their conclusion and why? [3]
Parents.
First Generation
Second Generation
This would mean that the DNA replication is taking place in a conservative way (versus
semiconservative replication) and that’s why they have light 14N bands in the first
generation along with the heavy 15N bands as well. If the DNA is replicating
conservatively, one of the original heavy 15N copies is saved for one of the daughter cells
and the other 14N light copy would be used for another daughter cell in the first
generation. Even in the second generation when one of the daughter cells with the heavy
15
N DNA replicates, it produces one heavy 15N DNA and one light 14N DNA. When the
other daughter cell from the first generation replicates, it produces the daughter cells both
with light 14N DNA. Hence, in the second generation there are three daughter cells with
light 14N DNA.
19. Use the DNA sequence 5’-ATCTCCTGGTAA-3’ to demonstrate the effect of each
type of mutation on the resulting polypeptide sequence. Include sequences for DNA,
mRNA, and polypeptide. [6]
a) Missense mutation
DNA  5’-ATGTCCTGGTAA-3’
3’-TACAGGACCATT-5’
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mRNA  5’-AUGUCCUGGUAA-3’
polypeptide  ---Met-Ser-Trp---
b) Frameshift mutation
DNA  5’-ATCTCCGTGGTAA-3’
3’-TAGAGGCACCATT-5’
mRNA  5’-AUCUCCGUGGUAA-3’
polypeptide  ---Ile-Ser-Val-Val--20. You are a scientist who has discovered a bacterium that has a biochemical pathway of
one reactant, three intermediate substrates and one product, five chemicals all
together, which we will call A, B, C, D, and E, which may not occur in this order.
You expose some of the bacteria to radiation to create defective mutants. You find
four nutritional mutants that need one of the five substances added to them in order to
get them to grow. The data is shown in the table below. Determine the order of the
substrate molecules (A-E) in the biochemical pathway, and the order of the enzymes
(E1-E4) that catalyze these reactions. [4]
Enzyme
destroyed
1
2
3
4
Growth
media with A
+
+
-
Growth
media with B
-
E4
D
Growth
media with C
+
-
E1
E
E3
A
Growth
media with D
+
+
+
+
Growth
media with E
+
+
+
-
E2
C
B
PART D: Communication [10]
Choose ONE of the following questions to answer. You may answer on the back of
this page.
21. Describe how errors, such as base-pair mismatches, are dealt with during DNA
replication. [10]
DNA polymerase enzymes have the ability to proofread and correct errors during
DNA replication. There are two ways to deal with errors during DNA replication.
One, repair mechanisms fix damage and base-pair mismatches while DNA is
being replicated. If an incorrect base is added, the two bases cannot form their
proper hydrogen bonds and the strand is unstable. This prevents DNA polymerase
III, which builds new DNA strands from nucleotides, from moving forward. The
DNA polymerase enzyme will reverse and act as a deoxyribonuclease to remove
the mispaired nucleotide. Second, a DNA replication event can be completed with
rare mismatching errors present. Repair enzymes read the strands for errors that
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may have been missed by DNA polymerase III. These repair mechanisms consist
of complexes of proteins and enzymes, including DNA polymerase I and II. DNA
polymerase II is a slow enzyme and works to repair DNA damage that occurs
between replication events. When a mismatch is located, the repair mechanism
determines which of the two helices contains the incorrect base, and an enzyme
removes a portion of the strand surrounding the mismatch. The resultant gap is
filled in by a DNA polymerase and the nicks are sealed by DNA ligase.
22. Describe the chronological steps involved in genetic recombination/DNA cloning and
how a target gene is identified. [10]
1. Determine desired DNA fragment (gene of interest) for isolation.
2. Cut DNA into fragments using restriction enzymes. One fragment should
include target gene.
3. Select artificial engineered plasmid with antibiotic resistance (e.g. Ampicillinresistant) and multiple cloning site.
4. Cut the circular plasmid, using same restriction enzymes used to cut the gene of
interest. This allows both DNA and plasmid to anneal.
5. Place DNA fragments and plasmids into the same solution and allow annealing.
6. Transform recombinant DNA molecule into bacterial cells (e.g. E. coli) to allow
replication within cells.
7. Grow bacteria containing plasmids on a culture medium containing antibiotic
resistance.
8. Use special filter paper laid on the bacterial plate to pick up cells from each
bacterial colony. This is a replica of the plate.
9. Filter paper is treated to lyse bacteria, and dsDNA is denatured to ssDNA,
which remains stuck to the filter paper in the same position as the colony it
came from.
10. A radioactive-labelled probe that is complementary to the target gene is
incubated with the ssDNA on the filter paper. The probe hybridizes only with
the target gene, and excess probe is washed off.
11. The filter is placed against a film that is exposed wherever the radioactive probe
has hybridized. The position of any radioactive spot is correlated to the original
plate.
12. The colony is isolated and used to produce large quantities of the target gene.
23. Describe the lac operon that prokaryotes use for controlling gene expression. [10]
The lac operon is a set of three genes that code for the proteins involved in the
metabolism of lactose, an energy source for prokaryotes that is acquired from the
environment. The operon consists of a promoter (site where DNA transcription
begins), an operator (a sequence of bases that control transcription), and the
coding regions for the various enzymes that actually metabolize the lactose, lacZ,
lacY, and lacA. Upstream from the operon is a gene that codes for a repressor
protein. This protein, lacI, takes cues from the environment (i.e. the concentration
of lactose within a cell) and regulates the production of the lactose-metabolizing
proteins. The genes for lacI are always transcribed, so this repressor is always
present within a cell.
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In the absense of lactose, lacI is active and binds to the operator, preventing RNA
polymerase from transcribing the metabolizing proteins. However, in the presence
of lactose in the cell, lactose molecules bind to lacI preventing it from binding to
the operator. This allows RNA polymerase to bind to the promoter and begin
transcription of the lactose metabolizing proteins. The enzymes that metabolize
lactose are synthesized and start to break down the lactose in the cell. When the
concentration of the lactose in the cell decreases, the amount of inactivated lacI
decreases and is eventually reactivated to binds to the operator, preventing
transcription.
Codon Chart
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