SBI4U: Molecular Genetics Unit Review

DNA and DNA replication1. a. Describe the experiments of:

Joachim HammerlingAlfred Hershey and Martha ChaseErwin ChargaffFranklin, 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 widec. 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?

Prophase Metaphase Anaphase Telophase/Cyto.Meiosis I DNA condenses. Each

chromosome consists of a pair of sister chromatids.

Nuclear membrane disappears.

Homologous pairs line up along metaphase plate.

Spindle fibres contract, pulling one homologue from each pair to opposite sides of the cell.

New nuclear envelope forms around each set of chromosomes.

Cytokinesis occurs. Product: Two haploid cells; each chromosome consists of two sister chromatids

Meiosis II DNA condenses. Each chromosome consists of a pair of sister chromatids.

Nuclear membrane disappears.

Chromosomes line up along metaphase plate.

Spindle fibres contract, pulling one sister chromatid from each chromosome to opposite sides of the cell.

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 pairsb. 32c. 16

DNA replication & repair15. 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

strands DNA polymerase III – Adds on

deoxyribonucleotides to form a new DNA strand. DNA gyrase – Relieves tension caused by DNA

strands unwinding around each other. Does this by cutting and re-ligating the DNA strands.

DNA polymerase I – Removes RNA primers and replaces them with deoxyribonucleotides. Also proofreads the newly-formed strand. Excises any erronenously-incorporated nucleotides and replaces them with correct ones.

single-stranded binding proteins – Binds to the DNA ligase – Forms phosphodiester bonds to join

unpaired bases once helicase unwinds the strands. Prevents re-annealing of strands

the sugar phosphate backbones of DNA strand fragments.

RNA primase - Produces small RNA primers to provide a free 3’ end for elongation to occur

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 #15/16b. 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 Synthesis20. 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.

SUPPLEMENTMutant Arginine Ornithine Citrulline 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 precursorproduct of arg-2 catalyzed the formation of citrulline from ornithineproduct 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 addedMutant A B C Dx - + - -y + + - -

z + + + -

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 protein y protein x

D A B C

22. What is the central dogma of molecular biology? What is transcription? Translation?

CENTRAL DOGMA: DNA RNA proteinDNA 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. transcriptionb. translation

Initiation Elongation TerminationTranscription RNA polymerase binds to

the promoter on DNA and unwinds the strands

RNA polymerase creates a complementary mRNA strand. It adds on ribonucleotides in the 5’ 3’ direction

RNA Pol recognizes a terminator sequence. Translation ceases, and the mRNA is released

Translation Assembly of the initiation complex at the promoter

tRNA delivers the appropriate amino acids to the ribosome for addition to the growing polypeptide chain

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 replication TranscriptionDNA is unwound ...by helicase

... SSBs prevent re-annealing... 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

... the new strand is RNA

... produced by RNA polymerase

... must be primed first

...both strands serve as templates, at the same time

... 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 - STOP5’-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 expression39. 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 Mutations46. 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 organization50. 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. Eukaryotes52. 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 genetics54. 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 15%A-B 3%B-C 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 Aq = 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 genotypeq2 = 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.

Biotechnology68. 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, HindIIIBlunt: 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-chargedb. 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.