lecture 8B

Question 1

2. Which of the following statement is not correct regarding Topoisomerase II molecules?

(a) Are capable of relaxing supercoiled DNA that is not circular.
(b) Can induce supercoiling in a relaxed DNA molecule.
(c) Change the linking number in units of 2.
(d) Break a single DNA strand to change the linking number.

Answer 1

(d) Break a single DNA strand to change the linking number

Question 2

3. Which of the following statements best describes DNA polymerase III functions in replication?

(a) chain extension in the 3’-5’ direction and 3’-5’ proofreading
(b) chain extension in the 5’-3’ direction and 3’-5’ proofreading
(c) chain extension in the 3’-5’ direction and 5’-3’ proofreading
(d) chain extension in the 5’-3’ direction and 5’-3’ proofreading

Answer 2

(b) chain extension in the 5’-3’ direction and 3’-5’ proofreading

Question 3

4. Which of the following DNA repair mechanisms will work to remove uracil from a DNA strand?

(a) photolyase
(b) base excision repair
(c) homologous recombination
(d) all of the above.

Answer 3

b) base excision repair

Question 4

5. Which of the following is false concerning 2, 3 – biphosphoglycerate (BPG)?

a) It binds in a pocket, present in the T form of hemoglobin, in the center of the hemoglobin tetramer
b) It is normally found associated with the hemoglobin molecules that are extracted from red blood cells
c) It binds with lower affinity to fetal hemoglobin than to adult hemoglobin
d) It stabilizes the oxy form of hemoglobin, increasing the oxygen affinity of hemoglobin

Answer 4

d) It stabilizes the oxy form of hemoglobin, increasing the oxygen affinity of hemoglobin

Question 5

6. The correct abbreviated name for the molecule presented below is:

a) dGDP
b) dT
c) dCMP
d) ATP

Answer 5

c) dCMP

Question 6

7. In a mixture of the five proteins listed below, which should elute second in size-exclusion (gel filtration) chromatography?

a) Cytochome c, Mr = 13, 000
b) Immunoglobulin G, Mr = 145, 000
c) Ribonuclease A, Mr = 13, 700
d) RNA polymerase, Mr = 450, 000
e) Serum albumin, Mr = 68, 500

Answer 6

b) Immunoglobulin G, Mr = 145, 000

Question 7

8. The structures below represent

a) UV-induced formation of thymine dimers
b) Formation of carbon linkages involving carbons 5 and 6 or 4 and 6 in adjacent thymine bases
c) UV-induced formation of pyrimidine dimmers
d) All of the above

Answer 7

d) all of the above

Question 8

9. By adding SDS (sodium dodecyl sulfate) during the electrophoresis of proteins, it is possible to:

a) Determine the amino acid composition of the protein
b) Preserve a protein’s native structure and biological activity
c) Determine a protein’s isoelectric point
d) Separate proteins exclusively on the basis of molecular weight
e) Determine an enzyme’s specific activity

Answer 8

d) Separate proteins exclusively on the basis of molecular weight

Question 9

10. The Klenow fragment of the E. coli DNA polymerase:

a) is a truncated version of the intact DNA polymerase III
b) has DNA polymerization and 5’-3’ exonuclease domains
c) has DNA polymerization and 3’-5’ endonuclease domains
d) contains 3 domains with different functions
e) may be used to proof-read mismatched or defective nucleotides in the 3’ end of the Okazaki fragments

Answer 9

e) may be used to proof-read mismatched or defective nucleotides in the 3’ end of the Okazaki fragments

Question 10

11. Immunoprecipitation is a biochemical technique that may be used to:

a) study DNA gel electrophoresis
b) isolate proteins that interact with one another in cells
c) separate proteins in CsCl gradients
d) grow protein crystals
e) study supercoiling of B-form DNA

Answer 10

b) isolate proteins that interact with one another in cells

Question 11

12. An Okazaki fragment is a:

a) small segment of DNA that is an intermediate in the synthesis of the lagging strand
b) fragment of RNA that is a subunit of the 30S ribosome
c) segment of mRNA synthesized by RNA primase
d) piece of DNA that is synthesized in the 3’-5’ direction
e) fragment of DNA resulting from endonuclease action

Answer 11

a) small segment of DNA that is an intermediate in the synthesis of the lagging strand

Question 12

13. In homologous recombination in E. coli, the protein that moves along a double-stranded DNA, unwinding the strands ahead of it and rewinding those behind it, is:

a) RecA protein
b) RuvC protein (resolvase)
c) chi
d) DNA ligase
e) RecBCD enzyme

Answer 12

RecA protein

Question 13

14. If the structure of a fully relaxed, closed-circular DNA molecule is changed so that the specific linking difference (σ) is -0.05:

a) the number of helical turns is increased by 5%
b) the number of bases is increased by 5%
c) the number of helical turns is decreased by 5%
d) the number of bases is decreased by 5%
e) none of the above is true

Answer 13

c) the number of helical turns is decreased by 5%

Question 14

15. The Meselson-Stahl experiment established:

a) that newly synthesized DNA in E. coli has a different base composition than does the pre-existing DNA
b) the role of DNA polymerase in DNA synthesis
c) that DNA synthesis in E. coli proceeds by a conservative mechanism
d) that DNA synthesis in E. coli proceeds by a semi-conservative mechanism
e) that DNA synthesis requires dNTPs

Answer 14

d) that DNA synthesis in E. coli proceeds by a semi-conservative mechanism

Question 15

These rare enol tautomers occasionally become incorporated into a growing DNA strand but rapidly shift back to their normal, more prevalent form, resulting in a mismatch. How are the majority of tautomeric mispairing errors corrected? (1 mark)

Answer 15

(c) 3’-5’ editing by DNA polymerase I

Question 16

Assume the enol tautomer of guanine is misincorporated into a growing DNA strand and is not repaired. Which base will this tautomer pair with? Show the resulting transition after two rounds of replication. The incoming tautomer should be shown on the right. (2 marks)

Answer 16

(d) Genol will pair with thymine (T).

Question 17

2. (5 marks) Draw out a simple schematic for DNA polymerase I synthesis at the lagging strand and explain the three functions this enzyme has in DNA replication. Be sure to label each nucleic acid strand and note their 3’ and 5’ ends, and show the direction of DNA synthesis.

Answer 17

DNA pol 1 functions:

1. Removes the RNA primer one ribonucleotide at a time, 5’-3’ exonuclease
2. Adds deoxyribonucleotides to the 3’ end of the adjacent Okazaki fragment (lagging stand)

These 2 processes together are called nick translation

3. Excises mismatched or defective nucleotides in the 3’ end of the Okazali fragments: 3’-5’ exonuclease. This process is calling “proof-reading”

Question 18

Describe the concept of hyperchromic shift as it is related to the denaturation (melting) of double stranded (ds) DNA.

Answer 18

Early slides in Lecture 16 or pp. 287-288 in the 5th Edition of Lehninger

Question 19

b) Describe qualitatively how the Tm of a ds DNA depends upon its nucleotide composition

Answer 19

In general, the higher the proportion of G and C, the higher the melting temperature, Tm. More thermal energy is required to break the three hydrogen bonds holding GC base pairs than to break the two hydrogen bonds holding AT base pairs.

Question 20

a) Define, in the context of DNA structure, “topoisomers”.

Answer 20

Topoisomers are different forms of the same DNA molecule that differ only in a topological property such as their linking number.

Question 21

b) If two DNA topoisomers – one supercoiled, one relaxed - are resolved on DNA agarose gel electrophoresis, which one will advance further in the agarose matrix? Why?

Answer 21

The supercoiled topoisomer will advance further because it is more compact; as such, it will experience less friction within the agarose matrix as it migrates towards the + electrode in the electric field.

Question 22

Which of the following statements about the Michaelis constant, KM, is correct?
A. The KM is equal to the concentration of substrate when all enzyme binding sites are filled.
B. The larger the KM, the higher the affinity of the enzyme for substrate.
C. The half-maximum velocity (Vmax/2) of a reaction is reached when the substrate concentration is
equal to the KM.
D. The KM for a given enzyme is independent of nature of the substrate.

Answer 22

. The half-maximum velocity (Vmax/2) of a reaction is reached when the substrate concentration is
equal to the KM

Question 23

6. The function of the oxyanion hole in chymotrypsin is to …
   A. attack the carbonyl carbon of the scissile peptide bond.
   B. align the substrate in the active site by binding a hydrophobic side chain.
   C. activate the catalytic serine.
   D. transiently stabilize the negative charge on the tetrahedral intermediate.

Answer 23

D. transiently stabilize the negative charge on the tetrahedral intermediate.

Question 24

7. The presence of 2,3-bisphosphoglycerate (BPG), low pH, and CO2 have the following effect on
   hemoglobin oxygen affinity:
   A. They all stabilize the T state, increasing the affinity of hemoglobin for oxygen.
   B. They all stabilize the T state, reducing the affinity of hemoglobin for oxygen.
   C. They all stabilize the R state, increasing the affinity of hemoglobin for oxygen.
   D. They all stabilize the R state, reducing the affinity of hemoglobin for oxygen.

Answer 24

B. They all stabilize the T state, reducing the affinity of hemoglobin for oxygen

Question 25

8. Choose the best ending to the statement: “Hemoglobin is better suited than myoglobin for oxygen
   transport and delivery to the tissues because it _________________________________”.
   A. has a higher affinity for oxygen.
   B. is tetrameric and thus binds to 4 oxygen molecules instead of 1.
   C. binds to oxygen cooperatively.
   D. can be regulated by binding to BPG.

Answer 25

C. binds to oxygen cooperatively

Question 26

Which of the following is the correct complementary DNA strand for the sequence
 5’-CAAGCTTCATTCG-3’
A. 3’-GUUCGAACUAAGC-5’
B. 3’-CAAGCTTCATTCG-5’
C. 5’-CGAAUGAAGCUUG-3’
D. 5’-CGAATGAAGCTTG-3’

Answer 26

D). 5’-CGAATGAAGCTTG-3

Question 27

10. Which of the following statements is not correct regarding Topoisomerase II molecules?
    A. They break a single strand of the DNA duplex to change the linking number.
    B. They utilize ATP in their reaction.
    C. They can underwind relaxed DNA molecules.
    D. They change the linking number in units of 2.

Answer 27

B. They utilize ATP in their reaction.

Question 28

11. Which of the following descriptions is not accurate for enzymes involved in DNA replication?
    A. helicases unwind the DNA duplex
    B. topoisomerases relax the DNA that becomes supercoiled from unwinding
    C. primases degrade the RNA primers on the Okazaki fragments
    D. ligases seal the nicks between Okazaki fragments

Answer 28

C. primases degrade the RNA primers on the Okazaki fragments

Question 29

. Name two protein assays discussed in class that utilize antibodies. (1 mark)
\

Answer 29

Western blots (immunoblots), ELISA, immunofluorescence, (affinity purification – not an assay
but acceptable)

Question 30

15. A chymotrypsin mutant has a His57→Thr mutation. Briefly and precisely explain how this
    mutation will affect enzyme activity, including whether it will have a greater impact on KM or Kcat.
    (4 marks)

Answer 30

The uncharged Asn57 will not be able to act as a base catalyst to abstract a proton away from
Ser195, so the alkoxide ion, which is a stronger nucleophile, will not form. This should not affect
KM (i.e. binding affinity) significantly but will impact Kcat (the enzyme will be inactive)

Question 31

Select T (true) or F (false) for the following statements regarding hemoglobin and myoglobin. (4
marks)
(T or F) Hemoglobin and myoglobin have similar affinities for oxygen in tissues.
(T or F) Myoglobin binds tightly to O2 except at very low PO2, as seen in actively respiring cells.
(T or F) Fetal hemoglobin has a His→Ser change that increases its affinity for the heme group and
hence for oxygen, allowing it to extract oxygen from maternal hemoglobin.
(T or F) A mutation in the hemoglobin gene that changes Glu6→Asp would likely cause a disease
similar to that of sickle cell anemia.

Answer 31

F
F
T
F

Question 32

18. A relaxed DNA plasmid has a linking number of 1000. How many base pairs does it have,
    assuming 10.5 bp/turn?
    How many rounds of Topoisomerase
    I catalysis would be required to reduce its Lk to 992? Explain. None

Answer 32

1000 turns x 10.5 bp/turn = 10,500 bp

None – Topo I can only relax
supercoiled DNA – it cannot induce supercoiling (underwinding/overwinding, that would require energy from ATP hydrolysis. (3 marks)

Question 33

9. Select T (true) or F (false) for the following statements regarding DNA replication. (4 marks)
   (T or F) DNA polymerase I binds to the end of an Okazaki fragment and simultaneously degrades
   its RNA primer, replacing each NTP with a dNTP.
   (T or F) DNA polymerases I and III both have 3’-5’ exonuclease activity (proof-reading).
   (T or F) DNA polymerases I and III both have 5’-3’ exonuclease activity.
   (T or F) The β-clamp makes DNA polymerase I highly processive

Answer 33

F
T
F
F

Question 34

21. UV light can induce formation of _____ in DNA strands. These are repaired in
    bacteria by the enzyme ____ and in humans by the nucleotide excision repair pathway.
    Methylation (alkylation) of guanine is repaired by the enzyme ____ for 0.5 marks). (3 marks)

Answer 34

1. Thymine dimers
2. photolyase
3. O^6-alkylguanine alkyltransferase

Question 35

The indirect methods of DNA repair – base excision repair, nucleotide excision repair and
mismatch repair - all use a similar strategy that requires the following 3 enzymes (or enzyme classes):

Answer 35

nuclease, polymerase and ligase.

Question 36

Describe in your own words how Okazaki fragments become joined to produce a continuous
DNA lagging strand. Use a diagram if necessary.

Answer 36

DNA pol I simultaneously added dNMPs to the 3’ end of the Okazaki fragments (5’-3’
polymerization) and digests the RNA primer (5’-3’ exonuclease) of the adjacent Okazaki
fragment. By extending the end of one fragment and simultaneously degrading the end of the
adjacent fragment, the gap or “nick between the two fragments moves or “translates” (nick
translation). Once the RNA primer has been fully excised and DNA synthesis is complete the
Okazaki fragments are ligated together by DNA ligase.

Question 37

2. Briefly describe 2 mechanisms that insure that correct nucleotides are added during DNA
   replication. What is the process call whereby DNA polymerases identify a mismatched base and
   excise the offending nucleotide? Which of the DNA polymerases has this function? What is the
   direction of nuclease activity related to this function?

Answer 37

(i) complementarity between the bases on the template strand and the incoming nucleotide bases
(ii) shape/H-bonding complementarity between the DNA polymerase active site and the bases (in
the minor groove)
Missmatched nucleotides are recognized and excised - this is called proof-reading and is done by
both DNA pol I and III – proof-reading/exonuclease activity is in the 3’-5’ direction
Note that only DNA pol I has 5’-3’ exonuclease activity – both polymerases need to proof-read
and excise mismated nucleotides but noly DNA pol I also has to degrade RNA primers on
adjacent Okazaki fragments.

Question 38

3. What is the function of the b-clamp on DNA polymerase?

Answer 38

The clamp holds the DNA on the core polymerase of DNA pol III, increasing the processivity of
this enzyme.

Question 39

4. Comment on the processivity of DNA pol I vs. DNA pol III. What do you think causes DNA
   pol I to terminate polymerization?

Answer 39

DNA pol III is highly processive, polymerizing ~500,000 nt before it falls off. DNA pol I is much less processive, polymerizing <200 nt at a time. It likely falls off when it reaches the 5’
end of the last Okazaki fragment synthesized (i.e. the RNA primer end).

Question 40

5. What are transitions and tranversions? Why do you think transitions are more common than
   tranversions?

Answer 40

Transitions are mutations that result in a purine-to-purine or pyrimidine-to-pyrimidine change
(eg. C:G to C:A or A:T to A:C) whereas transversions result in a purine-to-pyrimidine or pyrimidine-to-purine change (eg. C:G to C:T or A:T to A:G. Tranversions are more dramatic
structural changes and are thus more likely to be recognized and corrected by DNA repair
2
enzymes and by DNA polymerases during replication. Note that damage to bases that potentially
result in transitions and transversions are often transient (eg. tautomerization, which we didn’t
discuss in class, or damage that is repaired before replication).

Question 41

6. Why does deamination of adenine result in an A-T to a G-C transition and what mechanism is
   used to repair this damage?

Answer 41

Deamination of adenine produces hypoxanthine (H), which base pairs with cytosine instead of
thymine (both are pyrimidines, hence a transition). The abnormal base is repaired by base
excision repair (the base is first removed, then the rest of the nucleotide and some nucleotides
adjacent to it are removed by an endonuclease, then DNA pol I replaces the nucleotide and DNA ligase ligates it).

A:T -> (deamination) -> H:T -> (replication) -> H:C -> (replication) G:C

Question 42

7. Why does alkylation of guanine results in a G-C to an A-T transition and what mechanism is
   used to repair this damage?

Answer 42

Alkylation of guanine produces O6
-methylguanine (mG), which base pairs with thymine instead
of cytosine. mG is directly repaired by a O^6-alkylguanine alkyltransferase (which removed the methyl group). It could also be repaired by base excision repair. This is a form of direct repair.
G:C -> (alkylation) -> mG:C -> (replication) -> mG:T -> (replication) A:T

Question 43

8. How does ultraviolet light damage DNA and how is this repaired in E. coli?

Answer 43

UV light induces covalent bond formation between adjacent thymines, which will stall DNA replication and transcription. In bacteria, thymine dimers (pyrimidine dimers) are repaired by a photolyase, which uses visible light energy to hydrolyzes the bonds between the thymines. This is a form of direct repair. (In mammals the dimers are repaired by nucleotide excision repair. Humans defective in nucleotide excision repair pathways suffer from xeroderma pigmentosum.)

Question 44

9. List 2 mechanisms for (a) direct DNA repair, and (b) indirect DNA repair

Answer 44

(a) DNA repair enzymes O^6-alkylguanine alkyltransferase and photolyase
(b) Base excision repair, nucleotide excision repair, mismatch repair

Question 45

10. List the 3 steps that are common to every type of indirect repair: base excision repair,
    nucleotide excision repair and mismatch repair.

Answer 45

They all use (1) nucleases (exo, endo) to excise the damaged base or nucleotide (usually excise a
segment of nucleotides around the damaged site), (2) polymerases to fill in the missing
nucleotides using the undamaged complementary DNA strand as a template, and (3) a DNA
ligase to ligate the final newly added nucleotide to the remaining DNA strand.

Question 46

11. What is the role of MutH in mismatch repair and why is this important? Mismatch repair is
    only effective when it occurs soon after DNA replication. Why?

Answer 46

MutH identifies the parent (correct) strand because it is methylated – newly synthesized daughter
strands are not methylated right away. By identifying the correct strand, MutH then cuts the
newly synthesized unmethylated strand near the region of the mis-incorporated nucleotide, identified by the MutS/MutL complex, and an exonuclease then excises that unmethylated DNA
strans through the mismatch. (It is then filled in by DNA pol III, ligated.) This must occur before
the daughter strand also becomes methylated.

Question 47

12. List 3 cellular processes that utilize DNA recombination.

Answer 47

repair of double strand breaks
- repair of replication forks stalled at the site of DNA damage (to one strand)
- recombination of antibody H and L genes to produce a diverse immune repertoire with more
than a million distinct antibodies with unique antigen specificities (by VJD recombination)
- integration/excision of viral genomes.

Question 48

13. Explain how the non-recombinant and recombinant heteroduplexes shown in (l) of the Holliday model slides differ.

Answer 48

The non-recombinant heteroduplexes have only a segment of a single DNA strand in each heteroduplex replaced, whereas the recombinant heteroduplexes have their duplex ends replaced (i.e. both strands of the duplex). Imagine these ends represent the ends of a chromosome. The
recombinant heteroduplex is the result of a full cross-over event, where the genes an that end of
the chromosome become swapped between the two homologous duplexes

Question 49

Number each of the key steps in homologous recombination in order of occurrence:

• Resolution (cleavage) of the Holliday junction
• Formation of Holliday junction
• Strand invasion
• Branch migration
• Alignment of homologous duplexes
• Isomerization of the Holliday junction

Answer 49

Resolution (cleavage) of the Holliday junction (6)
Formation of Holliday junction (3)
Strand invasion (2)
Branch migration (4)
Alignment of homologous duplexes (1)
Isomerization of the Holliday junction (5)

Question 50

15. What is a Chi sequence and what is its role in double strand break repair? What is the protein
    E. coli that mediates homologous recombination in E. coli?

Answer 50

The Chi sequence is distributed throughout the E. coli genome. When a double strand break
occurs, RecBCD digests both strand of the duplex DNA until it hits a Chi sequence, at which
point it stalls and 3’-5’ exonuclease activity is halted but its 5’-3’ exonuclease activity continues,
resulting in a ssDNA with a 3’ end. RecA binds to this ssDNA and mediates strand invasion of a
homologous duplex, initiating recombination in E. coli.

Question 51

16. Name 3 enzymes we have encountered that utilize an active site tyrosine. What do all these
    enzyme mechanisms have in common?

Answer 51

DNA ligase, Topo I, Topo II, Cre recombinase – tyrosine performs a nucleophilic attack on a phosphate in the DNA backbone, forms a covalent intermediate

Question 52

DNA polymerase I is an example of an enzyme that is involved in base-excision repair and direct repair

Answer 52

base-excision repair and

NUCLEOTIDE-EXCISION repair.]

Question 53

What is an example of an intercalating agent that can be introduced to the agarose gel in order to visual the DNA
fragments? Where would the intercalating agent bind to in these DNA fragments? (2 marks)

Answer 53

Ethidium bromide

Intercalate between adjacent base pairs in the DNA double helix.

Question 54

Plasmid X contains AmpR
, which is a commonly-used selectable gene marker. What protein product does AmpR
code
for and what is the function of this protein product? (2 marks)

Answer 54

AmpR
represents the ampicillin resistance gene which encodes gene for -lactamase (1 mark), an enzyme
that degrades -lactam antibiotics like ampicillin

Question 55

Chymotrypsin belongs to a group of proteolytic enzymes called the “serine proteases.” Besides serine-195, what are
the other two amino acid residues in the active site that make up the catalytic triad in chymotrypsin? During the
generation of the acyl-enzyme intermediate, what is the main function of each of the two residues? (2 marks in total)

Answer 55

His57
Asp102
His57 = acts as base catalyst, abstracting proton from hydroxyl group of Ser195 (0.5 mark) Asp102 = stabilizes the resulting positively charged His57 (0.5 mark)

Question 56

Which of the following three amino acid sequences will be cleaved twice by chymotrypsin? Please circle your answer
(only one answer) (1 mark):
 Thr-Pro-Gly-Met-Ala
 Trp-Pro-Gly-Lys-Ala
 Tyr-Pro-Gly-Met-Ala

Answer 56

 Tyr-Pro-Gly-Met-Ala

Question 57

The oxygen–hemoglobin dissociation curve plots the proportion of hemoglobin in its saturated form on the vertical
axis against the prevailing oxygen tension on the horizontal axis. What is the proper name of the curve for adult
hemoglobin? How does the shape of this curve arise? (2 marks in total)

Answer 57

Adult hemoglobin = sigmoidal (1 mark). The sigmoidal shape arises from cooperative binding between hemoglobin and oxygen molecules

Question 58

Certain physiological factors can shift the oxygen-hemoglobin dissociation curve either left or right. Briefly explain
what a left Bohr shift represents? Briefly explain what a right Bohr shift represents? (1 mark for explanation for left
Bohr shift + 1 mark for explanation for right Bohr shift = 2 marks in total)

Answer 58

Left Bohr shift = oxygen binds tightly to hemoglobin (or hemoglobin does not release oxygen readily) (1 mark)
 Right Bohr shift = oxygen is released by hemoglobin more readily (or hemoglobin releases oxygen readily) (1
mark)

Question 59

In the table, determine if each of the factors will cause a right or left Bohr shift. (0.5 mark each x 4 = 2 marks in total)
ANSWER:
FACTORS BOHR SHIFT (left or right)
Hypocapnia LEFT (0.5 mark)
Decrease in 2,3-bisphosphoglycerate LEFT (0.5 mark)
Metabolic acidosis RIGHT (0.5 mark)
Increased exercise RIGHT (0.5 mark)

Answer 59

Hypocapnia LEFT (0.5 mark)
Decrease in 2,3-bisphosphoglycerate LEFT (0.5 mark)
Metabolic acidosis RIGHT (0.5 mark)
Increased exercise RIGHT (0.5 mark)

Question 60

4. A closed circular bacterial plasmid has 4725 base pairs and a topological linking number of 423. Use 10.5 base pairs
   per turn in answering the following questions:
   a. What would the writhe number (Wr) be for this molecule in order for it to optimize its base pairing? Show your
   work too. (1 mark)
   b. What is its superhelical density? (1 mark)
   c. Is the molecule underwound or overwound? (1 mark)
   d. How could you increase the linking number to 433? (1 mark)

Answer 60

The molecule would have 4725 bp/10.5 base pairs per turn or 450 turns to optimize its base pairing. Since Lk = Tw +
Wr, 423 = 450 +Wr, and Wr = -27. (1 mark)
 σ = Lk/Lko = -0.06 = -6%. (1 mark)
 Since the writhe number is right-handed, the molecule is negatively supercoiled or UNDERWOUND. (1 mark)
 Treat the DNA to ten cycles of topoisomerase I, which will increase Lk by 1 each time and partially relax the
supercoiling. (1 mark)

Question 61

16. Name 3 enzymes we have encountered that utilize an active site tyrosine. What do all these
    enzyme mechanisms have in common?

Answer 61

DNA ligase, Topo I, Topo II

tyrosine performs a nucleophilic attack on a phosphate in the DNA backbone, forms a covalent intermediate