Download SBI4U: Molecular Genetics Unit Review

SBI4U: Molecular Genetics Unit Review
DNA and DNA replication
1. a. Describe the experiments of:
Joachim Hammerling
Alfred Hershey and Martha Chase
Erwin Chargaff
Franklin, Wilkins, Watson & Crick
b. In what way did they contribute to our current knowledge of DNA?
IN YOUR BOOK.
2. Describe the directionality of a single strand of DNA (5’ vs. 3’)


The 5’ end of a strand of DNA is the one to which the 5’ carbon of deoxyribose is pointed. It
carries the phosphate group.
The 3’ end of the strand carries a Hydrogen
3. Describe the structure of DNA. Include the structural components, as well as the important
bonds.





Two nucleotide strands, bonded together and running antiparallel, such that one strand
runs 5’ to 3’, and the other strand runs 3’ to 5’. The two strands together form a helical
shape.
Each nucleotide is composed of: deoxyribose sugar (pentose); negatively-charged phosphate
group on the 5`carbon of the sugar, and a nitrogenous base attached to the 1`carbon.
The base can be either a purine (adenine or guanine) or a pyrimidine (thymine or cytosine)
The backbone of the double helix is composed of alternating sugars and phosphate groups.
Each nucleotide is joined via a condensation reaction with a phosphodiester bond.
The two antiparallel nucleotide strands are joined to each other by hydrogen bonds. In an
H-bonding pair, adenine always bonds with thymine, and guanine always bonds with
cytosine (purine + pyrimidine).
4. When we talk about DNA sequence, why do we call it the “base sequence”?
The only component that differs between nucleotides is the nitrogenous base it contains; hence
the base confers to the nucleotide its identity.
5. What is the difference between the A-T bond and the G-C bond?
The A-T bond is formed by two H-bonds, whereas the G-C bond is formed by three. Because of
this, the G-C bond is stronger than the A-T bond.
6. Write the complementary DNA strand:
5’- A A T G G C T -3’
3’ –T T A C C G A -5’
7. a. How many base-pairs (nucleotides) are in one turn of the DNA helix? How long is one turn
of the helix?
b. How wide is the double helix?
c. If a DNA sequence contains 1500 base pairs, how many turns are in the helix? How long is
the DNA?
a. 10 nucleotides in one turn, 3.4 nm.
b. 2 nm wide
c. 1500 bp/10 (bp/turn) = 150 turns
150 turns x 3.4 (nm/turn) = 510 nm
Nuclear division: Mitosis and Meiosis
8. Is cell division the same thing as mitosis? Meiosis?
No, it is the same as neither. Mitosis and meiosis involve division of the nucleus. Strictly
speaking cell division is only cytokinesis – the division of the membrane, cytoplasm, and
organelles.
9. Describe the cell cycle of a somatic cell.
Interphase – Cell grows, performs regular functions. Eventually DNA replicates. This is the
longest stage of the cell cycle.
Mitosis – The nucleus divides so that each new nucleus contains identical information.
Cytokinesis – The rest of the cell divides to produce two new identical daughter cells.
10. How do mitosis and meiosis differ in the number and composition of their end products?
Where does mitosis occur? Meiosis?




Mitosis products contain all the same genetic information as its parent cell. Two daughter
cells are produced.
Meiosis products contain only half the genetic information of its parent cell. Four daughter
cells are produced; in males this means four spermatozoa are produced, but in females only
one ovum is produced. Three of the four haploid cells becomes a polar body, due to uneven
distribution of cytoplasm and organelles during meiotic division.
Mitosis: somatic cells
Meiosis: germ line cells; gives rise to gametes
11. Name the stages of mitosis. Describe each stage.


Prophase – Chromatin condenses to form chromosomes. The centrioles migrate to the
poles of the cell, forming spindle fibres. The nuclear membrane breaks down.
Metaphase – The chromosomes line up at the equator of the cell (the metaphase plate)


Anaphase – The spindle fibres contract, pulling the sister chromatids apart at the
centromeres
Telophase – The chromatids move towards the polar ends of the cell. The nuclear
membrane reforms around each set of chromatids.
12. Meiosis involves two nuclear divisions instead of just one. Why?
The products are gametes that are utilized in sexual reproduction – they must be haploid.
13. Describe and compare the stages of Meiosis I vs. Meiosis II. What are the products of each?
Meiosis I
Prophase
DNA condenses. Each
chromosome consists
of a pair of sister
chromatids.
Metaphase
Homologous pairs
line up along
metaphase plate.
Nuclear membrane
disappears.
Meiosis II
DNA condenses. Each
chromosome consists
of a pair of sister
chromatids.
Nuclear membrane
disappears.
Chromosomes
line up along
metaphase plate.
Anaphase
Spindle fibres
contract, pulling
one homologue
from each pair to
opposite sides of
the cell.
Spindle fibres
contract, pulling
one sister
chromatid from
each
chromosome to
opposite sides of
the cell.
Telophase/Cyto.
New nuclear
envelope forms
around each set
of chromosomes.
Cytokinesis
occurs.
Product: Two
haploid cells;
each
chromosome
consists of two
sister chromatids
New nuclear
envelope forms
around each set
of chromosomes.
Cytokinesis
occurs.
Product: Two
haploid cells per
division (four per
parent cell); each
chromosome
consists of one
chromatid
Some might prefer to draw what is occurring at each stage. I recommend colour-coding the
chromosomes, and drawing 2-3 chromosomes as they move through the stages of Meiosis.
14. An American badger has 32 chromosomes total in each of its somatic cells:
a. How many pairs of homologous chromosomes are in each somatic cell?
b. How many total chromosomes will the products of mitotic division contain?
c. How many chromosomes will an American badger’s gamete contain?
a. 32 chromosomes/2 (chromosomes/pair) = 16 pairs
b. 32
c. 16
DNA replication & repair
15. Name and describe the three models of DNA replication.



Conservative - The two parental strands stay together, and somehow produce another
daughter helix with completely new strands.
Semi-conservative - Two parental strands separate and each serves as a template for a new
progeny strand. Progeny strand contains one parental, one new strand.
Dispersive - DNA becomes fragmented so that new and old DNA coexist in the same strand after
replication.
16. Describe the Meselson-Stahl experiment: setup, results, conclusions.



Setup: Grew E. coli on medium with heavy nitrogen (15N). All E. coli DNA contained 15N, and so
was denser than DNA containing regular 14N. Introduced this E. coli to a medium containing
regular 14N and allowed replication to occur. Following replication, DNA was extracted,
purified, and then separated according to density.
Results: Following one round of replication, only one density band was obtained; this band was
intermediate to both the heavy band and the light band. After another round, two bands were
obtained: one intermediate, and one light.
Conclusion: DNA replication is semi-conservative. The intermediate band obtained after one
round is the result of each DNA molecule containing one heavy band and one light. After the
second round of replication, half of the molecules is composed of one heavy and one light,
where other half is entirely composed of light bands.
17. Name and briefly describe the three stages in which DNA replication occurs.





Initiation: DNA helicase breaks the H-bonds between strands, and “unzips” them. RNA primase
anneals short complementary RNA strands to the unzipped DNA template strands.
Elongation: DNA Polymerase III adds on free deoxyribonucleotides in the 5’  3’ direction.
Termination: DNA Polymerase III reaches another replication fork, or reaches the end of its
strand.
18. Describe the roles of the following enzymes/proteins in DNA replication:
DNA helicase - Breaks H-bonds between DNA
 DNA polymerase III – Adds on
strands
deoxyribonucleotides to form a new DNA strand.
DNA gyrase – Relieves tension caused by DNA
 DNA polymerase I – Removes RNA primers and
strands unwinding around each other. Does this
replaces them with deoxyribonucleotides. Also
by cutting and re-ligating the DNA strands.
proofreads the newly-formed strand. Excises any


single-stranded binding proteins – Binds to the
unpaired bases once helicase unwinds the
strands. Prevents re-annealing of strands
RNA primase - Produces small RNA primers to
provide a free 3’ end for elongation to occur


erronenously-incorporated nucleotides and
replaces them with correct ones.
DNA ligase – Forms phosphodiester bonds to join
the sugar phosphate backbones of DNA strand
fragments.
DNA telomerase – In prokaryotes, and in
eukaryotic germ line cells, prevents the shortening
of telomeres during successive rounds of
replication.
19. Explain what is meant by the following statements:
a. DNA replication is semi-conservative – See #15/16
b. DNA replication is semi-discontinuous – One strand is replicated continuously (the leading
strand), while the other is replicated discontinuously, in fragments (the lagging strand). The
discontinuously-synthesized strand is must constantly be primed by RNA primase as the
replication fork moves. This occurs because DNA synthesis can only occur in the 5’ 3’
direction, and the two template strands run antiparallel to each other.
c. DNA replication is bi-directional – Synthesis proceeds outwards in both directions from a
replication fork.
Protein Synthesis
20. Describe the Beadle and Tatum experiment: setup, results, analysis, conclusions.
Setup: Bombarded N. crassa with radiation to produce mutants. Identified three classes of
mutants that were deficient in their ability to synthesize arginine. Mapped the mutations to
three separate loci (genes) in the N. crassa genome. Added in compounds related to arginine in
order to see their effects on the mutants’ ability to grow.
Results: Compounds added were arginine, ornithine, citrulline.
SUPPLEMENT
Arginine
Ornithine
Citrulline
Mutant
+
+
+
arg-1
+
+
arg-2
+
arg-3
Analysis: The metabolic pathway for synthesis of arginine is:
precursor  ornithine  citrulline  arginine
the product of the arg-1 gene catalyzed the formation of ornithine from precursor
product of arg-2 catalyzed the formation of citrulline from ornithine
product of arg-3 catalyzed formation of arginine from ornithine.
Conclusions: Each gene controls the production of one polypeptide
21. The data below represents the response of three mutants, defective for genes x, y, and z
respectively, to the addition of four compounds: A, B, C and D.
Compound added
Mutant
x
y
z
A
B
C
+
+
D
+
+
+
+
-
a. Assume all four compounds exist within the same biosynthetic pathway. In what order
are they found?
b. Where in the biosynthetic pathway are the products of genes x, y, and z located?
protein z
D

protein y
A

protein x
B

C
22. What is the central dogma of molecular biology? What is transcription? Translation?
CENTRAL DOGMA: DNA  RNA  protein
DNA directs the production of proteins by first being transcribed into an mRNA molecule whose
sequence is dependent on the sequence of DNA. The mRNA is then “read” or translated by
ribosomes in the cytoplasm in order to produce a polypeptide.
23. How does RNA differ from DNA?
 Sugar: Ribose instead of deoxyribose. 2’ carbon has a hydroxyl group.
 Bases: Contains uracil where DNA would contain thymine. Hence in RNA Uracil
complementary base-pairs with adenine in another nucleic acid strand (be it RNA or DNA)
 Location: DNA is located only in the nucleus (eukaryotes) where RNA is free to move
between the nucleus and the cytoplasm.
24. Describe what happens in intiation, elongation, and termination of:
a. transcription
b. translation
Transcription
Translation
Initiation
RNA polymerase binds to
the promoter on DNA and
unwinds the strands
Assembly of the initiation
complex at the promoter
Elongation
RNA polymerase creates a
complementary mRNA
strand. It adds on
ribonucleotides in the 5’ 
3’ direction
tRNA delivers the
appropriate amino acids to
the ribosome for addition
to the growing polypeptide
chain
Termination
RNA Pol recognizes a
terminator sequence.
Translation ceases, and the
mRNA is released
Ribosome recognizes a
stop codon. Transcription
ceases, and the
polypeptide is released
25. Why is the promoter region high in A-T base pairs as opposed to G-C?
A-T has only 2 H-bonds: easier to break the strands apart.
26. What post-transcriptional modifications occur to an mRNA before it leaves the nucleus?



splicing –by spliceosomes; the removal of non-coding regions (introns) in order to produce a
final transcript containing only expressed regions (exons)
5’ cap - 7-methyl guanosine added to 5’ end of primary transcript
3’ tail – chain of 200 adenine nucleotides added to the 3’ end. Capping and tailing protects
the mRNA from degradation in the nucleus
27. What is the function of reverse transcriptase?
For an RNA molecule, makes a complementary DNA strand. It then uses that DNA strand to
produce a double-stranded DNA molecule.
28. What is a retrovirus? Describe how HIV works. How does AIDS differ from HIV?





A retrovirus is one whose genome is contained in RNA rather than DNA.
HIV is a retrovirus that contains reverse transcriptase within its structure.
It selectively targets T-cells in a host’s immune system, and infects them. Once inside the
host cell, reverse transcriptase makes a dsDNA copy of the viral RNA genome.
The dsDNA incorporates itself into the host’s genome, where it can then direct synthesis of
more viral particles.
HIV is Human Immunodeficiency Virus – the actual virus that infects the host.
AIDS is Acquired Immunodeficiency Syndrome – It is the group of symptoms that is the
cumulative result of host immunosuppression by the HIV virus.
29. What are the three classes of RNA discussed in class?



mRNA – messenger RNA; the product of transcription of a DNA gene. Carries the instructions for
constructing a polypeptide at the ribosome.
tRNA – transfer RNA; delivers amino acids to the ribosome during translation
rRNA – structural component of ribosomes. Produced in the nucleolus.
30. Why are errors in transcription not as tightly-regulated as errors in replication?
Replication – Only one copy of DNA per cell. Must ensure fidelity from one generation to the
next.
Transcription – Many copies of mRNA per cell → correct copies should outnumber, and
compensate for, incorrect copies. The redundancy of the genetic code should also help make up
for errors.
31. The processes of transcription and DNA replication are actually quite similar. How are they
similar? Different?
Common feature
DNA is unwound
DNA replication
...by helicase
... SSBs prevent re-annealing
Transcription
... by RNA polymerase
... DNA re-anneals as soon as
RNA polymerase moves away
A new complementary strand
is formed, using DNA as the
template
...the new strand is DNA
... produced by DNA
polymerase
... must be primed first
...both strands serve as
templates, at the same time
... the new strand is RNA
... produced by RNA
polymerase
... does not require priming
... either strand can serve as
the template, depending on
the protein
32. How many bases are in a codon? Why can a codon not consist of only 2 bases?


Three bases in a codon
Two bases per codon would only allow for 42=16 possible combinations of nucleotides. Since
there are 20 amino acids, this would not be sufficient to code for all twenty.
33. A polypeptide sequence is composed of 45 amino acids. How many bases are read by the
ribosome to produce this polypeptide?
 45 amino acids x 3 (bases/amino acid) = 135 bases – this of course assumes you do not count the
stop codon, which would add another 3 bases on to your total
34. Describe the formation of the initiation complex
 The start codon (AUG) is recognized.
 The following components bind: small ribosomal subunit; met-tRNA; large ribosomal subunit
35. Describe the formation of a polypeptide at the ribosome
 met-tRNA is at the P site of the ribosome after formation of initiation complex
 The ribosome reads the next codon, and the tRNA bearing the complementary anticodon enters
the A (acceptor site), delivering the correct AA.
 The ribosome translocates by one codon, moving the met-tRNA into the E site, and the next
tRNA into the P site. A peptide bond forms between these two AA’s. The tRNA that once
carried methionine exits the ribosome.
 The next tRNA enters at the A site, carrying its A.A.
 The ribosome translocates, and the cycle continues until a stop codon is reached. At this point,
a release factor arrives and dismantles the ribosome-mRNA complex.
36. Identify the three possible reading frames for the sequence of mRNA below. For each reading
frame, what polypeptide will be synthesized?
5’-AUGGCUAAGCUGUGA-3’
5’-AUG/GCU/AAG/CUG/UGA-3’: Met – Ala – Lys – Leu - STOP
5’-A/UGG/CUA/AGC/UGU/GA-3’: ...Trp – Leu – Ser – Cys - ...
5’-AU/GGC/UAA/GCU/GUG/A-3’: .. Gly - STOP
37. a.One arm of a tRNA bears an anticodon that is complementary to the codon on the mRNA.
How does the sequence of successive anticodons compare to the sequence of DNA from which
the mRNA was transcribed?
Identical to the template DNA sequence, but containing uracil instead of thymine.
b. Will the sequence of anticodons, if strung together, be exactly the same as the sequence of
template DNA?
Only in a prokaryote. In eukaryotes, splicing of introns may occur.
38. What is wobble-base pairing? How is it adaptive?
The term used to describe the fact that the third base of the tRNA anticodon may not be
complementary to the third base on the codon. Since the genetic code is redundant, with
codons for the same A.A. differing usually at the last base, it allows the correct amino acid to be
delivered to the ribosome despite the fact that the anticodon and codon may not be 100%
complementary.
Regulation of gene expression
39. At what four levels can control of gene expression occur?
transcriptional, post-transcriptional, translational, post-translational.
40. What is an operon? What are the regulatory sequences of an operon? What type of
regulation is being exhibited?



An operon is a cluster of related genes that are located together on a chromosome. They
are under the control of one set of regulatory sequences.
Regulatory sequences: The Promoter (recruits RNA Polymerase for transcription to occur),
and the Operator sequence (controls whether RNA Polymerase can bind or not)
Transcriptional regulation
41. What is a regulator? What are two types of regulator?


Regulator – A protein that binds to the operator sequence in order to affect the rate of
transcription.
Two types: Repressor (inhibits transcription); Activator (promotes transcription)
42. How does a repressor work?

Binds to the operator to physically prevent RNA Polymerase from being able to initiate
transcription
43. What is an effector? What are the two types of effector and how do they act on a repressor?


An effector is a molecule that acts on the regulator to either enhance or diminish the
regulator’s activity
Two types: Co-repressor (enhances the activity of a repressor by allowing it to bind);
Activator (diminishes the repressor’s ability to bind, thereby activating transcription)
44. How is the lac operon regulated? Is this negative or positive regulation?
 LacI repressor protein is normally bound to the operator
 If lactose concentrations are high, lactose will act as an activator by allosterically binding to
the repressor. This binding changes the conformation of the repressor so that it can no
longer bind to the operator
 Transcription occurs.
 Negative regulation – In the presence of effector, the repressor is removed.
45. How is the trp operon regulated? Negative or positive?
 Repressor is not normally bound – Transcription will occur.
 If tryptophan levels are high, tryptophan will act as a co-repressor. It will bind to the trp
repressor and change its conformation so that the repressor can now bind to the operator.
 Transcription ceases.
 Positive regulation – In the presence of effector, the repressor is added.
Genetic Mutations
46. What are the three types of point mutations? Which have the potential to be the most
detrimental to a cell?


Substitution, Insertion, Deletion
Insertion and Deletion have the highest potential for being detrimental because they can
cause shifts in the reading frame (frameshift mutations)
47. What are the five types of chromosomal mutations we discussed? How does gene duplication
occur and what might be the result?



Insertion, Deletion, Inversion, Translocation, Gene duplication
Gene duplication occurs due to misalignment of homologous chromosomes during crossing
over in Meiosis I.
This can give rise to “gene families” – clusters of genes that arose by duplication, but have
been modified over time to have slightly different sequences and functions
48. Distinguish between a silent mutation, missense mutation, and nonsense mutation. Which
has the potential to be the most detrimental?



Silent mutation – Has no effect on protein structure; can be a bp-substition such that the
new sequence codes for the same amino acid, or it can be a mutation within an intron.
Missense – A different amino acid is encoded. Might not be too bad if the new Amino acid
has similar properties to the old one.
Nonsense – A codon is converted to a stop codon. Most potentially dangerous.
49. Name and describe the two causes of mutation.
 Innate (spontaneous) mutation – due to an error by replication machinery
 Environmental – due to exposure to mutagenic agents
Levels of DNA organization
50. Describe the levels of DNA organization




Double stranded-DNA
Nucleosome – dsDNA coiled around histone proteins
Chromatin – Coiled strands of nucleosomes (beads on a string)
Chromosome – Highly condensed fibres of chromatin, formed by supercoiling
51. How much of the genome is coding vs. non-coding. What practical purpose can non-coding
regions serve?



5% coding, 95% noncoding
Non-coding regions contain long stretches of repeating elements. These can be used as
genetic markers for individualization
Telomeres at ends of chromosomes are also repetitive – serve to protect chromosomes
from loss of information during replication
Prokaryotes vs. Eukaryotes
52. How does the structure of the prokaryotic genome compare with the eukaryotic one?



Much smaller
Circular
No introns
53. Transcription and translation occur differently in a prokaryote. Outline the differences:




Eukaryotes have nuclei. The mRNA transcript must be completed and processed before it
can exit the nucleus for translation.
Prokaryotes have no nuclei; translation can therefore occur simultaneously with
transcription. This is called coupled transcription-translation.
Prokaryotic ribosomes recognize the purine-rich Shine-Dalgarno sequence in order to
initiate translation, where eukaryotic ribosomes recognize the 5’ cap of mRNA
Eukaryotic ribosomes are bigger.
..Remember: Eukaryotic EVERYTHING is bigger!!!
Population genetics
54. What is allele frequency? How is it calculated?
 The proportion of a specific allele in a gene pool, for a given population.
 Allele frequency = (# of allele)/(all possible alleles for that locus in the gene pool)
55. Mendel`s law of Independent assortment is sometimes violated. Describe what this means,
and the term given to such a violation.
 Mendel’s law of independent assortment states that during gamete formation, alleles for
different genes will segregate independently of each other.
o


E.g. if you received A and B from your mother, and a and b from your father, your
gametes are free to contain a combination of A/a and B/b that is “non-parental”.
o Gametes can contain combo AB, ab, or Ab, aB – the alleles are not associated with
another and so will segregate independent of each other.
Violation of this law would constitute inheritance of these alleles as a “package”. A is more
likely to segregate with B (AB) than it is to segregate with b (Ab).
This type of violation is termed “gene linkage”.
56. What is recombination? How is it accomplished for loci that are unlinked? Linked?
 Recombination is the process of forming gametes that have different combinations of alleles
than those received from the individual’s parents.
 For unlinked loci, this is accomplished by independent assortment. The gametes have equal
probability of containing a parental combination (AB or ab, from above) as of containing
non-parental(recombinant) combinations (aB or Ab).
 For linked loci, this is accomplished only by crossing-over during meiosis I. This is because
the loci are on the same chromosome, so the only way to separate the parental alleles from
each other is to physically take the piece of chromosomes and switch them.
57. What is recombination frequency, and how is it calculated?


Recombination frequency is the probability of the occurrence of one recombination event
between two loci.
RF= (# of non-parental allele combinations)/(total # of gametes formed)
58. Linkage mapping is a linear representation of a chromosome that places loci based on their
recombination frequency. What does 1 map unit on a linkage map represent?

1 m.u. represents a RF of 1%
59. Construct a linkage map for loci A, B, and C based on the following RF’s:
A-C
A-B
B-C
15%
3%
18%
B 3 mu A
15 mu________C
60. State the two equations used to describe Hardy-Weinberg equilibrium. Be sure to define the
variables.
For a population with alleles A and a at a particular locus,
p = allele frequency of A
q = allele frequency of a
and p + q = 1
The genotypic frequencies can be related as follows:
p2 + 2pq + q2 = 1
where p2 = frequency of AA genotype
2pq = frequency of Aa genotype
q2 = frequency of aa genotype
61. For a particular locus there exist two alleles, B and b. If the frequency of the genotype BB is
0.64 for a population, what are the allele frequencies for B and b? What are the genotype
frequencies for Bb and bb?
62. Hardy-Weinberg equilibrium states that allele frequencies do not change within a population,
from one generation to the next. However, for this type of equilibrium to occur several
conditions must be met. State these conditions.





large population
equal mating opportunities
no mutations occur
no migration occurs
no natural selection
63. How do these conditions serve as an outline for the forces that drive evolution?


By definition, evolution is defined as a change in allele frequency over generations – hence a
state of Hardy-Weinberg disequilibrium.
The list of conditions for H-W equilibrium to occur represents a list of factors that can drive
evolution
64. What is genetic drift and why does it affect small populations more?


Genetic drift refers to any change in allele frequency that occurs as a result of random
events (chance) (as opposed to natural selection or sexual selection)
Small populations have smaller gene pools. As such their allele frequencies are more
susceptible to forces of random change than a larger one that contains more individuals and
more alleles.
65. What is a bottleneck effect, and when might it occur?
 A bottleneck effect is a dramatic reduction in population size. It results in a significant
genetic drift.
 After a bottleneck event, the sample of alleles left in the population might significantly differ
from those in the population before the event

It can occur after a natural disaster
66. How does the founder effect differ from gene flow?


The founder effect is a type of genetic drift that occurs when individuals from a population
leave to establish a new population. The allele frequencies of the new population are
dependent on the AFs in the small sample of individuals in the founder group.
Gene flow occurs when individuals from one population move to a new area, and join an
existing population. These individuals bring with them their alleles.
67. A neutral mutation is not necessarily the same thing is a silent mutation. What is the
difference
 A silent mutation is one that has no consequence on amino acid sequence.
 A neutral mutation is one that has no consequence on an individual`s reproductive fitness.
It may very well be a silent mutation, but it may also be a nonsense or a missense mutation.
o Contrast with harmful or beneficial mutations, which do affect an individual`s
fitness. Either of these can also be nonsense or missense mutations.
Biotechnology
68. Describe the roles of the following tools or techniques in forming and expressing a
recombinant DNA product:






Restriction endonuclease: Used to cut DNA (“molecular scissors”). Functions as a crude
immune system within a bacterium. RE’s that produce sticky ends (ends having overhangs) can
be used to digest DNA from two sources so that they can be ligated together
Ligase – Catalyzes the formation of phosphodiester bonds between the sugar-phosphate
backbones of two DNA fragments
Methylase – Enzymes isolated from bacteria; add methyl groups onto the recognition sites for
RE’s in order to prevent cleavage. Can be used in biotechnology to “protect” certain restriction
sites from being cleaved
Gel electrophoresis – A method of separating DNA fragments, based on the fragment size. A
sample containing DNA fragments of various lengths is loaded into wells at one end of a slab of
agarose gel that is immersed in buffer solution. A current is applied across the gel so that the
DNA fragments will migrate towards the opposite end (positive electrode). Migration rate is
dependent on fragment size, so that the rate of migration is inversely proportional to the
logarithm of the fragment size.
Vector (plasmid) – A vector is a vehicle by which foreign DNA is inserted into a bacterial cell.
Plasmids are small, circular pieces of DNA that are found in bacteria. They replicate
independently of the bacterial genome, and can be used as vectors for bacterial transformation.
A gene of interest is inserted into a plasmid, and then the plasmid is inserted into a bacterium.
Plasmid vectors are engineered to contain restriction sites for several common RE’s, an origin of
replication, a selectable “reporter” gene, and an inducible operon.
Transformation – Refers to the process of inserting a foreign DNA into a bacterial cell. This is
commonly accomplished by treating the bacterial cell with CaCl2 to stabilize the membrane,
then introducing the foreign DNA and subjecting the system to a heat shock. The heat shock
opens up pores in the membrane and creates a draft that sweeps the plasmid inside. Other
methods of transformation include electroporation, and the use of gene guns (this is a common
method with plant cell transformation).
69. Give an example of a RE that cleaves leaving a sticky end. Give an example of one that leaves
a blunt end. Which of the two would be more useful, and why?
Sticky: EcoRI, HindIII
Blunt: SmaI, AluI
Sticky is more useful in molecular biology because it is easier to ligate two DNA fragments
having sticky ends. The overhangs on each strand can H-bond to the complementary bases on
the opposite strand, whereas the bases in blunt ends cannot H-bond with each other.
70. What properties of DNA are utilized in gel electrophoresis?
a. DNA is negatively-charged
b. The mass-to-charge ratio of a DNA molecule is consistent (all parts of the fragment carry
the same charge concentration)
71. What are the features that are engineered into a plasmid vector?




Origin of replication
Multiple cloning site
Selectable reporter gene
Inducible operon
72. Name and describe three methods for transformation.
See Q#68
73. How is transformation checked? How is insertion of the desired gene checked?


Transformation: Reporter gene, e.g. one that confers antibiotic resistance, is engineered
into the plasmid. Plate bacteria on medium containing antibiotic to check for plasmid
uptake.
Insertion of gene: Extract DNA from antibiotic-resistant colonies. Electrophorese + probe
for gene of interest.
74. How is expression of a foreign gene induced in a bacterial cell?


Operator and promoter for an inducible operon (usually lac) are engineered in frame with
the restriction site and inserted gene.
Add an analogue of lactose (IPTG) to the culture so it can induce expression of the gene.
75. Describe the process of polymerase chain reaction.




Reaction components: template DNA, DNA primers, DNA polymerase, free
deoxyribonucleotides
Heat is used to separate the strands
Temperature is lowered so that primers can anneal
DNA Polymerase (Taq polymerase) produces a new strand
76. For what is RFLP analysis useful? What is the principle behind RFLP analysis?


Useful for individualizing samples
Principle: Individuals are polymorphic for restriction sites within their genomes. After
digesting genomic DNA with RE’s and determining the size of the restriction fragment
lengths, individuals can be distinguished from one another due to the banding patterns that
are obtained.
77. Why is PCR a more attractive technique for individualizing samples than RFLP?
It only requires a small amount of DNA.
78. Describe the process of Sanger dideoxy sequencing.







PCR-based technique requiring four separate amplification reactions (1 per nucleotide base)
Each reaction tube: template for sequencing, DNA Pol, radioactively-labelled DNA primers,
dNTPs and one of four types dNTP analogue (ddNTP = dideoxyribonucleotide triphosphates)
– either dATP, dGTP. DTTP, dCTP
ddNTPs lack a 3’ OH group  elongation cannot continue after incorporating a dNTP
Carry out successive amplifications
ddNTPs will be incorporated at random where their dNTP equivalents would be, eventually
generating fragments of various lengths that correspond to the location of that particular
base in the sequence.
Electrophorese all four sequencing reactions alongside each other. Fragments will differ in
length by one base pair.
“Read” the sequence from the shortest fragment to the longest.
79. What is gene therapy?
Any method for treating a disease that involves alteration of the genetic sequence/gene
expression.
80. The use of antisense oligonucleotides in gene therapy is a form of gene regulation. Describe
how it works.
The sequence of a gene targeted for silencing must first be known. Generate short fragments of
RNA that are complementary to the mRNA’s of the targeted genes. Add these to the cells  they
will complementary base pair with the mRNA’s, thus preventing transcription of the protein.