Download SBI4U Molecular genetics UNIT_AK

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SBI4U Unit Test: Molecular Genetics
Multiple Choice (K/17)
___ 1.Which of the following terms does not describe DNA replication?
a. uni-lateral
c. bi-directional
b. semi-discontinuous
d. semi-conservative
___ 2. The central dogma of molecular biology states that the information in DNA is copied to
_______ by the process of __________, and then to ___________ via____________.
a. protein, transcription, RNA, translation
b. RNA, transcription, protein, translation
c. RNA, translation, protein, transcription
d. protein, translation, RNA, transcription
___ 3. Pairs of homologous chromosomes line up along the equator of the cell in:
a. Prophase I
c. Metaphase I
b. Prophase II
d. Metaphase II
___ 4. The universal start codon also encodes:
a. does not code an A.A.
c. threonine
b. leucine
d. methionine
___ 5.Two alleles for fur colour exist in a population of mice: B (dominant brown) and b
(recessive white). If the allele frequency of b is 0.34, what proportion of the mouse
population will be heterozygous (Bb)?
a. 0.12
c.0.45
b. 0.34
d. 0.66
___ 6.A base-pair substitution can be classified as all of the following, except:
a. nonsense mutation
c. silent mutation
b. frameshift mutation
d. missense mutation
___ 7.Which of the following genes is not under the control of the lac operator?
a. LacI
c. LacZ
b. Lac X
d. LacY
___ 8.Chinchillas have 32 pairs of chromosomes. After a round of mitotic division, how many
chromosomes will each daughter cell contain?
a. 16
c. 64
b. 32
d. 128
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___ 9. Gel electrophoresis is a method for separating DNA fragments according to their size.
When electrophoresed, DNA fragments will migrate at a rate that is:
a. proportional to their size
b. proportional to the logarithm of their size
c. inversely proportional to their size
d. inversely proportional to the logarithm of their size
___ 10.Which of the following is true about the bonding in a DNA molecule?
a. The sugar-phosphate backbone is joined by hydrogen bonds
b. The sugar-phosphate backbone is joined by phosphodiester
bonds
c. The nitrogenous bases are joined by phosphodiester bonds
d. The nitrogen bases are joined by glycosyl bonds
___ 11. The synthesis of Okazaki fragments occurs because:
a. DNA strands run antiparallel, and synthesis can only occur in the 3’  5’ direction.
b. DNA strands run antiparallel, and synthesis can only occur in the 5’  3’
direction.
c. RNA primase as a high affinity for the lagging strand, and primes it often
d. There is only enough free energy to synthesize one strand continuously
___ 12.Which of the following post-transcriptional modifications is carried out in a prokaryotic
cell?
a. 5’ cap of 7-methyl guanosine and 3’ poly-A tail are added
b. introns are spliced out, and exons are ligated
c. both a and b
d. neither a nor b
___ 13. In eukaryotes, what part of the mRNA transcript is recognized by the ribosome so it
knows to assemble?
a. Shine-Dalgarno sequence
c. the 3’ poly-A tail
b. the promoter
d. the 5’ cap
___ 14. During protein synthesis, the ribosome moves along the mRNA in groups of ______
bases, called ______. A tRNA with a complementary _______ delivers the amino acid.
a. three; codons; anticodon
b. three; anticodons; codon
c. four; codons; anticodon
d. four; anticodons; codon
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___ 15.A particular operon is controlled by a repressor. It is not transcribed in the presence of
protein X. When concentrations of protein X decline, transcription of the operon
occurs. Which statement describes the situation best?
a. this is positive regulation, and protein X is the effector
b. this is positive regulation, and protein X is the regulator
c. this is negative regulation, and protein X is the effector
d. this is negative regulation, and protein X is the promoter
___ 16. SalI is a restriction enzyme that recognizes the following sequence, and cuts between
the G and the T:
5’-GTCGAC-3’
3’-CAGCTG-5’
What type of ends are formed by cleavage with SalI?
a. split end
b. 5’ overhang sticky end
c. 3’ overhang sticky end
d. blunt end
___ 17. Which of the following is not an application of RFLP?
a. genetic screening
c. forensic identification
b. bacterial transformation
d. paternity testing
Short answer
18. Look at the following picture of an agarose gel following electrophoresis of the four
samples in a Sanger sequencing reaction. (T/2)
a. What is the sequence of DNA? 5’-GTACATTAAA-3’
b. Label the end with positive electrode (+), and the negative electrode (-).
(-)
(+)
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19. DNA replication has proofreading mechanisms whereas transcription does not. Describe
two reasons why a cell can more easily tolerate an error in transcription. (C/2)
Description of any two of the following:
 Error is not inherited
 Normal proteins should outnumber defective proteins
 Redundancy of the genetic code
 Wobble base pairing
20 a. Compare the processes of DNA replication, Polymerase Chain Reaction, and
transcription: (K/3)
Replication
____ separates
Type of primer
the DNA strands
helicase
RNA
Enzyme that synthesizes
the new strand
DNA Pol III
PCR
Heat
DNA
DNA Pol
(Taq polymerase)
Transcription
RNA polymerase
None
RNA polymerase
b. PCR does not require single stranded binding proteins, or a DNA gyrase. What
purpose does each of these molecules serve in DNA replication, and why are they not
required in PCR? (C/3)



SSB – keeps strands from re-annealing
DNA gyrase – relieves torsional strain caused of strands unwinding
Not required bc heat completely denatures the strands from each other
21. This picture depicts transcription and translation in a cell. Is this cell prokaryotic, or
eukaryotic? How do you know? (A/2)
 Prokaryotic
 coupled transcription-translation
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22.An individual is homozygous at two loci: A/a·B/b.
a. Write out the four possible combinations of alleles (T/1)
AB, ab Ab, aB
b. What proportion of each gamete can be expected if loci A and B are unlinked? (T/1)
25% of each
Suppose the individual described above received the following combinations of alleles
from its parents: A·B from its mother, a·b from its father.
c. Rewrite the allele combinations from (a). What proportion of each gamete can be
expected if loci A and B experience a recombination frequency of 0.2? (T/1)
AB – 40%
ab – 40%
Ab – 10%
aB – 10%
23 a. List the four levels at which gene regulation can act (K/2)
Transcriptional, post-transcriptional, translational, post-translational
b. In gene therapy, antisense oligonucleotides act by bonding to targeted mRNA’s in
order to prevent their products from being made. At what level is this form of gene
regulation acting? (A/1)
Translational
24. Describe either the experiment of either Meselson and Stahl, or Beadle and Tatum: What
did they do, what did they see, and what did they conclude? (C/3)
Check notes – One mark for each point
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25. Considering the following DNA sequence:
3’ – CCGTACCCTGTATAAAAC – 5’
a. Write the complementary DNA strand that would be produced in replication (A/1):
3’ – CCGTACCCTGTATAAAAC – 5’
5’ – GGCATGGGACATATTTTG – 3`
b. Write the complementary mRNA strand that would be produced in transcription (A/1):
3’ – CCGTACCCTGTATAAAAC – 5’
5’ – GGCAUGGGACAUAUUUUG – 3`
c. What is the sequence of amino acids that would result from translation of the mRNA
sequence? (A/1)
GLY – MET – GLY – HIS – ILE - LEU
The Genetic Code (the pocket version):
Codon
AAC
AUG
AUU
CCG
CAU
CUU
Amino acid
Asn
Met
Ile
Pro
His
Leu
Codon
GAC
GGA
GGC
UAA
UCA
UUG
Amino acid
Asp
Gly
Gly
Stop
Ser
Leu
Bonus: (K/1)
What was the name of the scientist who actually performed the experiment that produced
the X-ray crystallography image of DNA?
rosalind franklin
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26. Consider the following mRNA sequence, and the polypeptide produced as a result:
5’ - AUG
Met
mRNA:
polypeptide:
AAC
Asn
CCC
Pro
ACA
Thr
UAA - 3’
stop
A mutation occurs so that the following mRNA is transcribed:
mRNA:
polypeptide:
5’ – AUG UAA
Met
STOP
CCC
CAC
AUA- 3’
a. Is this a base pair substitution, insertion, or deletion? (T/1) insertion
b. Use the genetic code from #25 to translate the polypeptide sequence, and write it in
above (T/1)
c. Is this a nonsense, missense, or a silent mutation? (T/1) nonsense
27. A DNA fragment that is 1000 bp long is analyzed. Its composition is found to be 32%
thymine. How many guanines bases are in the fragment? (T/2)
Guanine – 18%
18% x 1000 = 180
*Note should be 18% x 2000 = 360 since each base pair contains 2 bases, but I forgot to
point this out when taking up the sheet. I accept either answer.
Matching (A/8)
For each of the following items, indicate in which process they are found. Do this by
writing in the letters corresponding to the process next to the component.
Processes:
A. DNA replication
B. Transcription
C. Translation
D. PCR
_D__ 28. DNA primers
_D__ 32. heat denaturation
_C__ 29. small and large ribosomal subunits
_A__ 33. DNA gyrase
_B__ 30. RNA polymerase
_D__ 34. Taq polymerase
_C__ 31. transfer RNA
___ 35. RNA polymerase
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