How do you do Question 7 and Question 8 for Part 2. The last 2 open-ended questions

GENERAL BIOCHEMISTRY 115:404/504 Final Examination, Spring, 2015
NAME: ____________________________

Page 1

I.D. number: ____________________

Part I. Answer all questions on the answer sheet. Give the answer which is most nearly correct.
Unless otherwise indic?ted each question has only one answer. Each question is worth two points.
1.

The DG'? values for the two reactions shown below are given.
Oxaloacetate + acetyl-CoA + H2O ? citrate + CoASH
citrate
synthase

DG'? = ?32.2 kJ/mol

Oxaloacetate + acetate ? citrate
citrate lyase

DG'? = ?1.9 kJ/mol

What is the DG'? for the hydrolysis of acetyl-CoA?
Acetyl-CoA + H2O ? acetate + CoASH + H+
A. ?34.1 kJ/mol
B. ?32.2 kJ/mol
C. ?30.3 kJ/mol

D. +61.9 kJ/mol
E. +34.1 kJ/mol
(from last year?s final exam)

2. When

a mixture of glucose 6-phosphate and fructose 6-phosphate is incubated with the enzyme
phosphohexose isomerase, the final mixture contains twice as much glucose 6-phosphate as fructose 6phosphate. Which one of the following statements is most nearly correct, when applied to the reaction below
(R = 8.315 J/mol?K and T = 298 K)?
Glucose 6-phosphate ? fructose 6-phosphate
A.
B.
C.
D.
E.
3. Which

?G'? is +1.7 kJ/mol.
?G'? is ?1.7 kJ/mol.
?G'? is incalculably large and negative.
?G'? is incalculably large and positive.
?G'? is zero.

(from last year?s final exam)

of the following reactions in glycolysis requires ATP as a substrate?

A.
B.
C.
D.
E.

Hexokinase
Glyceraldehyde-3-phosphate dehydrogenase
Pyruvate kinase
Aldolase
Phosphoglycerate kinase

(from last year?s final exam)

4. Which of the following reactions in glycolysis produces ATP as a product?
A.
B.
C.
D.
E.

Hexokinase
Glyceraldehyde-3-phosphate dehydrogenase
Pyruvate kinase
Aldolase
Phosphoglycerate kinase

(from last year?s final exam)

GENERAL BIOCHEMISTRY 115:404/504 Final Examination, Spring, 2015

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5. The main function of the pentose phosphate pathway is to:
(from the first hour exam)
A.
B.
C.
D.
E.

Supply energy.
Give the cell a back-up capacity should glycolysis fail.
Provide energy and reducing power.
Provide a mechanism for the utilization of the carbon skeletons of excess amino acids.
Provide pentoses and NADPH

6. Place the following steps in lipid oxidation in their proper sequence.
(from the first hour exam)
(a)
Thiolysis
(b)
Reaction of fatty acyl-CoA with carnitine
(c)
Oxidation requiring NAD+
(d)
Hydrolysis of triacyiglycerol by lipase
(e)
Activation of fatty acid by joining to CoA
(f)
Hydration
A. d, e, b, c, f, a
D. a, d, e, b, f, a
7.

B. e, d, c, b, a, f
E. e, d, b, c, a, f

C. d, b, e, c, f, a

How are fatty acid oxidation and synthesis controlled so that futile cycling does not occur?
(from the second hour exam)
A. They occur in different cellular compartments.
B. They employ different electron carriers.
C. The product of the first oxidation reaction inhibits the rate-limiting step of biosynthesis.
D. A and B
E. A, B and C

8. The proton-motive force generated by the electron transfer chain:
(from the second hour exam)
A. includes a pH-gradient component.
B. includes an electrical-potential-gradient component.
C. is used for active transport processes.
D. is used to synthesize ATP.
E. has all of the above characteristics.
9. Which one of the following statements is true?
A.
B.
C.
D.

The brain prefers glucose as an energy source, but can use ketone bodies.
Muscle cannot use fatty acids as an energy source.
In a well-fed human, about equal amounts of energy are stored as glycogen and as triacylglycerol.
Fatty acids cannot be used as an energy source in humans because humans lack the enzymes of
the glyoxylate cycle.
E. Amino acids are a preferable energy source over fatty acids

GENERAL BIOCHEMISTRY 115:404/504 Final Examination, Spring, 2015

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10. The Spiegelman monster
A. defines the approximate minimum size of the replicative unit, given the modern set of enzymes
and other materials.
B. arose by gradual loss of unneeded genes.
C. arose by spontaneous polymerization in the manner of the Eigen or Orgel experiments.
D. eats cookies.
E. A and B
11. Leucine zippers
A. are protein regions which prevent protein-protein interactions when these would interfere with
gene regulation.
B. are protein-protein interaction domains which interact with DNA polymerase.
C. are protein-protein interaction domains which help regulate RNA polymerase activity.
D. are a synthetic material used in certain types of clothing.
E. none of the above.
12. In the synthesis of lagging strand DNA, polymerization of the Okazaki fragment near the replication
fork stops when the RNA primer of the previous strand is reached. The RNA is then removed by which of
the following processes?
A. kinetic proofreading
B. base excision repair
C. nick translation

D. ligation
E. none of the above

13. On the leading and lagging strands DNA polymerase III attaches the next base at the
A.
B.
C.
D.
E.

leading strand 3?-OH; lagging strand 3?-OH
leading strand 3?-OH; lagging strand 5?-OH
leading strand 5?-OH; lagging strand 3?-OH
leading strand 5?-OH; lagging strand 5?-OH
none of the above: it attaches at a phosphate of the growing strand

14. In what order do the following enzymes act in the synthesis of Okazaki fragments?
A.
B.
C.
D.
E.

helicase, primase, polymerase III, polymerase I, ligase
ligase, polymerase I, polymerase III, primase, helicase
primase, polymerase I, helicase, polymerase III, ligase
polymerase I, polymerase III, helicase, ligase, primase
none of the above.

15. The binding of RNA polymerase to DNA is driven largely by electrostatic forces. This implies that
A. the polymerase carries a substantial net negative charge.
B. the polymerase carries a roughly even mixture of lysines, arginines, glutamic acids and aspartic
acids.
C. the polymerase exposes a large hydrophobic binding surface to the DNA.
D. the polymerase carries a substantial net positive charge.
E. none of the above.

GENERAL BIOCHEMISTRY 115:404/504 Final Examination, Spring, 2015

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16. In Crick?s wobble hypothesis
A. the ribosome wobbles by one base on the m-RNA, shifting the reading frame of protein synthesis.
B. the first (5?) base of m-RNA codons can make unusual hydrogen bonds with t-RNA anticodons
explaining codon redundancy.
C. a runner rounding third base doesn?t make it to home plate before being thrown out.
D. the third (3?) base of m-RNA codons can make unusual hydrogen bonds with t-RNA anticodons
explaining codon redundancy.
E. mitochondrial codons replace chromosomal codons.
17. Which of the following is characteristic of the polymerase III holoenzyme?
A.
B.
C.
D.
E.

It is twice the size of a half-enzyme.
It is a stripped down protein with only some of the pol III catalytic activities.
It contains ribonuclease H.
It contains all the catalytic activities of pol III.
none of the above.

18. Splicing mRNA in eukaryotes
A. Removes the introns
B. Removes the exons
C. Always involves folding of the mRNA

D. Always involves proteins
E. Two of the above

19. How are amino acids attached to tRNAs?
A. The N-terminal end is attached to the 2' or 3' hydroxyl of the tRNA
B. The N-terminal end is attached to the 5' hydroxyl of the tRNA
C. The C-terminal end is attached to the 2' or 3' hydroxyl of the tRNA
D. The C-terminal end is attached to the 5' hydroxyl of the tRNA
E. None of the above
20. Which has the lowest error rate?
A. reproduction as used by Dyson
B. modern replication
C. metabolism

D. parasitic infection
E. the average student taking this examination

21. What is the role of the DAM methylase in E. coli?
A.
B.
C.
D.
E.

to label the template strand so that base errors in the newly synthesized strand can be detected.
to excise bases which are inserted erroneously by polymerases.
to dispose of excess methyl groups.
to stimulate helicase action.
the role is unknown; that?s why people curse it by calling it the DAMn methylase.

22. Which of the following is NOT involved in DNA mismatch repair in E. coli
A. DNA glycosylase
B. DNA helicase II
C. DNA ligase
D. DNA polymersase
E. Exonuclease

GENERAL BIOCHEMISTRY 115:404/504 Final Examination, Spring, 2015

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23. The first enzyme to take part in base excision repair is
A. DNA ligase.
B. DNA glycosylase.
C. AP endonuclease.

D. DNA polymerase I.
E. none of the above.

24. What is a promoter?
A. an origin of replication
B. a start site for binding of RNA polymerase
C. a Shine-Dalgarno sequence

D. a splice site
E. someone who organizes boxing matches

25. In a eukaryotic polyribosome, which ribosome will have the longest growing polypeptide chain attached
to it?
A. the ribosome nearest the 3? end of the mRNA.
B. the ribosome nearest the 5? end of the mRNA.
C. a ribosome near the middle of the mRNA, because synthesis is bidirectional.
D. a ribosome with a chaperone complex adjacent to it.
E. it depends on which protein is being made.
26. The binding reaction between the E. coli lac repressor and the lac operon can be represented by the
-10
dissociation reaction RO R + O. Kd, the dissociation equilibrium constant, for this reaction is 10
-4
M. Kd for dissociation of the repressor from other E. coli DNA sequences is 10 M. This means that
A.
B.
C.
D.
E.

The repressor binds a million times more tightly to the operator than to other parts of the DNA.
The repressor binds 10,000 times more tightly to the operator than to other parts of the DNA.
The repressor binds a million times less tightly to the operator than to other parts of the DNA.
The repressor binds 10,000 times less tightly to the operator than to other parts of the DNA.
None of the above.

27. Free fatty acids in the bloodstream are:
A.
B.
C.
D.
E.

Bound to hemoglobin
Present at levels that are independent of epinephrine.
Carried by the protein serum albumin.
Freely soluble in the aqueous phase of the blood.
Nonexistent; the blood does not contain free fatty acids.

28. Fatty acids are activated to acyl-C?As and the acyl group is further transferred to carnitine because:
A.
B.
C.
D.
E.

acyl-C?As easily cross the mitochondrial membrane, but the fatty acids themselves will not.
Fatty acids cannot be oxidized by FAD unless they are in the acyl-carnitine form.
Carnitine is required to oxidize NAD+ to NADH.
Acyl-carnitines readily cross the mitochondrial inner membrane, but acyl-C?As do not,
None of the above is true.

GENERAL BIOCHEMISTRY 115:404/504 Final Examination, Spring, 2015

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29. In amino acid catabolism, the first reaction for many amino acids is a(n):
A.
B.
C.
D.
E.

Decarboxylation requiring thiamine pyrophosphate (TPP).
Reduction requiring pyridoxal phosphate (PLP).
Transamination requiring pyridoxal phosphate.
Hydroxylation requiring NADPH and O2.
Oxidative deamination requiring NAD+.

30. Which of the following conversion requires more than one step?
1.
2.
3.
4.
5.

Alanine pyruvate
Glutamate ? -ketoglutarate
Aspartate oxaloacetate
Proline glutamate
Phenylalanine succinate

A. 1, 2, and 4
D. 1 and 4

B. 1, 3, and 5
E. 4 and 5

C. 2, 4, and 5

31. Which substance is not involved in the production of urea from NH4+ via the urea cycle?
A. Aspartate
D. Malate

B. ATP
C. Ornithine
E. Carbamoyl phosphate

32. If a person?s urine contains unusually high concentrations of urea, which of the following diets has he or
she probably been eating recently?
A. Very high carbohydrate, very low protein
C. Very high fat, very low protein
E. A purely vegetarian diet

B. Very low carbohydrate, very high protein
D. Very high fat, high carbohydrate, no protein

33. The human genetic disease phenylketonuria (PKU) can result from:
A.
B.
C.
D.
E.

Inability to synthesize phenylalanine
Inability to catabolize ketone bodies
Inability to convert phenylalanine to tyrosine
Production of enzymes containing no phenylalanine
Deficiency of protein in the diet

34. In the re-oxidation of QH2 by purified ubiquinone-cytochrome c reductase (Complex (III) from heart
muscle, the overall stoichiometry of the reaction requires 2 mol of cytochrome c per mol of QH2 because
A. cytochrome c is a one-electron carrier, whereas QH2 is a two-electron donor.
B. cytochrome c is a two-electron acceptor, whereas QH2 is a one-electron donor.
C. cytochrome c is water soluble and operates between the inner and outer mitochondrial membranes.
D. heart muscle has a high rate of oxidative metabolism and therefore requires twice as much
cytochrome c as QH2.
E. two molecules of cytochrome c must combine physically before they are biologically active.

GENERAL BIOCHEMISTRY 115:404/504 Final Examination, Spring, 2015

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35. Uncoupling of mitochondrial oxidative phosphorylation
A. allows continued mitochondrial ATP formation but halts O2 consumption.
B. halts all mitochondrial metabolism.
C. halts mitochondrial ATP formation but allows continued O2 consumption.
D. slows down the citric acid cycle
E. slows the conversion of glucose to pyruvate by glycolysis.
36. In photophosphorylation, absorption of light energy in chloroplast ?light reactions? leads to
A.
B.
C.
D.
E.

absorption of CO2 and release of O2.
absorption of O2 and release of CO2.
hydrolysis of ATP and reduction of NADP+.
synthesis of ATP and oxidation of NADPH
use of iron-sulfur proteins.

37. Photosynthetic phosphorylation and oxidative phosphorylation appear to be generally similar processes,
both consisting of ATP synthesis coupled to the transfer of electrons along an electron carrier chain. Which
of the following is NOT true of both processes?
A.
B.
C.
D.
E.

Both contain cytochromes and flavins in their electron carrier chains.
Both processes are associated with membranous elements of the cell.
Both use oxygen as a terminal electron acceptor.
Each represents the major route of ATP synthesis in those cells in which it is found.
Protons are pumped from the inside to the outside of both the mitochondrial and chloroplast
membranes.

38. What is a signature sequence of amino acids or nucleotides?
A.
B.
C.
D.
E.

It is a run of residues common to a set of related species but not to other groups of species.
It is a way to close a gap in a set of sequences.
It is a specific binding site for an enzyme that processes the protein or nucleic acid.
It is a set of residues which occurs only in gram positive and gram negative bacteria.
none of the above

39. What sort of information is present in a rooted evolutionary tree but not in an unrooted tree?
A. A rooted tree shows the evolutionary relationships among species but not the sequence of
evolutionary development.
B. A rooted tree shows both the evolutionary relationships among species and the sequence of
evolutionary development.
C. A rooted tree offers several possible evolutionary routes from ancestral species to extant species
whereas an unrooted tree shows only one possible route.
D. A rooted tree has nodes, whereas an unrooted one does not.
E. A rooted tree has leaves, whereas an unrooted one is dead and has no leaves.

GENERAL BIOCHEMISTRY 115:404/504 Final Examination, Spring, 2015

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40. As discussed in class, the numbers of rooted and unrooted trees are given by the formulas below.

For a set of five OTUs the ratio of the number of possible rooted to unrooted trees is
A. 105

B. 15

C. 0.14

D. 7

E. none of these

41. The components of a distance matrix are
A.
B.
C.
D.

the lengths of the ?arms? in an unrooted tree.
composite OTUs made up of several species.
the time since species diverged from common ancestors.
the numbers of nucleotide or amino acid differences between each pair of species represented in
the matrix.
E. more than one of the above
42. Informative sites in studies of molecular evolution
A. are sites where the paleontological record can be reconciled with the nucleotide or protein
sequence data.
B. are places in Wikipedia whose information can be trusted.
C. allow an investigator to favor one possible evolutionary tree over another.
D. allow an investigator to determine which species is the root of a rooted tree.
E. allow an investigator to group some species into a clade while excluding others.
43. The serological work of Goodman (p. 7 in the class notes) led him
A. to exclude humans from the same clade as other primates.
B. place humans and orangutans in the same clade and chimpanzees and gorillas in another clade.
C. to conclude that humans share no common ancestors at any level of evolution with the African
apes.
D. to conclude that humans are closer in evolutionary terms to chimpanzees than to gorillas.
E. to place humans, chimpanzees and gorillas in single clade.
44. E. coli choose glucose over lactose as a carbon source when both sugars are present by
A. binding a CRP / c-AMP complex to the operator when repressor is not bound.
B. binding lac repressor.
C. binding neither repressor nor CRP, thereby allowing only a low level of lac operon transcription.
D. binding both repressor and CRP.
E. none of the above

GENERAL BIOCHEMISTRY 115:404/504 Final Examination, Spring, 2015

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45. When tryptophan levels are high in E. coli
A. tryptophan binds to the trp repressor, causing it to fall off the DNA. Transcription of the trp
operon then follows.
B. the trp repressor is not made.
C. synthesis of the leader peptide is disrupted.
D. tryptophan binds to the trp repressor thereby changing its conformation and enabling it to block
the operon?s operator.
E. the cells excrete the excess amino acid.
46. What is the total number of possible RNA sequences of length 50?
A. 4*50

B. 450

C. 504

D. 6*504

E. 6*450

47. If there are 61 non-stop codons and all codons are uniformly distributed, what is the minimal length of a
real ORF at alpha=0.01
A. log0.95 0.01
D. log0.95 0.99

B. log0.01 0.95
C. log0.99 0.05
E. none of the above

48. A predictor identifies people as blond (positive) or brunette (negative) on the basis of their genetic
variation. It makes 100 predictions, identifies 5 brunettes as blond, 12 blondes as brunette, and gets the rest
right. How many type I errors does it make?
A. 12

B.

5

C. 17

D.

7

E. 0.05

49. What information about the genomes do we lose by representing it as a string?
A. Read accessibility
D. All of the above

B. Codon frequency
E. None of the above

C. ORF start/stop sites

50. In preparing for this exam
A.
B.
C.
D.
E.

I had nightmares about an attack of the Spiegelman monsters.
I couldn?t have nightmares because I didn?t sleep.
My metabolic network hydrolyzed vast quantities of ATP to little effect.
My metabolic network ground nearly to a halt.
My metabolic network ran at such a high rate that I?m exhausted.

GENERAL BIOCHEMISTRY 115:404/504 Final Examination, Spring, 2015
Name: ________________________

Page 10

I.D. No. ______________________

Part II, Question 1. Twenty points. Please answer the question on this page, using the back should
you need it. SHOW ALL YOUR WORK. An answer with incomplete, uncle?r, or illegible work will be
penalized. IF YOU OMIT THIS QUESTION, WRITE "OMIT" CLEARLY ON THIS PAGE.
In the last reaction of the citric acid cycle, malate is dehydrogenated to make oxaloacetate (OAA), which is
necessary for condensation with acetyl-CoA to begin the next turn of the cycle:
Malate + NAD+ OAA + NADH + H+

?G?? = +30 kJ/mol

(a) Calculate the equilibrium constant for this reaction at 25?C.

R = 8.315 J/mol-deg

(b) Because ?G?? assumes a standard pH of 7, the equilibrium constant in (a) corresponds to
K?eq = [OAA] [NADH] [L-malate] [ NAD+] The measured concentration of L-malate in rat liver mitochondria is about 0.20 mM when the
[NAD+]/[NADH] ratio is 10. Calculate the concentration of OAA in mitochondria at pH 7.

(c) To appreciate the magnitude of the mitochondrial OAA concentration, calculate the number of OAA
molecules in a single rat liver mitochondrion. Assume the mitochondrion is a sphere of diameter 2.0 ?m.
(Mitochondria are not spherical, but their volumes are equal to that of a 2.0 ?m sphere.)
Avogadro?s number = 6.023 x 1023molecules/mol. Volume of a sphere = 4 ? r3 / 3

This is Question 10 from the end of chapter 16. It was also on this term?s first hour exam.

GENERAL BIOCHEMISTRY 115:404/504 Final Examination, Spring, 2015
Name: ________________________

Page 11

I.D. No. ______________________

Part II, Question 2. Twenty points. Please answer the question on this page, using the back should
you need it. SHOW ALL YOUR WORK. An answer with incomplete, unclear, or illegible work will be
pen?lized. IF YOU OMIT THIS QUESTION, WRITE "OMIT" CLEARLY ON THIS PAGE.
The subunit composition of E. coli DNA polymerase III is shown in the table below. The extra rows and
columns in the table may be helpful to you.
a) The average molecular weight of an amino acid residue in a protein is 115 daltons. Calculate the number
of amino acids needed to make one Pol III.
b) If it ?costs? an average of 10 high energy phosphates from ATP to make one amino acid, how many
glucose molecules would be needed to make one Pol III if the cells are grown anaerobically? If they are
grown aerobically?
From last year?s second exam. Also from this term?s second exam.
LABEL THE TWO PARTS OF YOUR ANSWER BELOW with (a) or (b).

Show your work in
detail. If you simply write down memorized answers, you will receive no credit.
Add more rows and/or columns as needed, but be sure your reasoning is made clear.
Subunit
Name
?
?
?
?
?
?
?'
?
?

No. of

Subunit

subunits

mol. Wt.

2

129,000

258000

2

27,500

55000

2

8,600

17200

2

71,700

143400

1

47,500

47500

1

38,700

38700

1

36,900

36900

1

16,600

16600

1

15,200

15200

4

40,600

162400

Total =
No. of res = Mol. Wt /

Mol. Wt.

No. of
residues

790900
115

6877

(a)

ATP cost =

10

per res. =

68774

Anerobic:

2

ATP/Glc =

34387

Aerobic

30

ATP/Glc =

2292

Answer OK

Aerobic

32

ATP/Glc =

2149

Answer OK

Aerobic

38

ATP/Glc =

1810

Answer OK

(b) & below

GENERAL BIOCHEMISTRY 115:404/504 Final Examination, Spring, 2015
Name: ________________________

Page 12

I.D. No. ______________________

Part II, Question 3. Twenty points. Please answer the question on this page, using the back should
you need it. SHOW ALL YOUR WORK. An ?nswer with incomplete, unclear, or illegible work will be
penalized. IF YOU OMIT THIS QUESTION, WRITE "OMIT" CLEARLY ON THIS PAGE.
The equation shown represents oxygenic photosynthesis:
6 CO2 + 6 H2O + 12 NADPH + 18 ATP C6H12O6 + 12 NADP+ + 18 ADP + 18 Pi + 6 O2
(glucose)
Note that C6H12O6 + 6 O2 6 CO2 + 6 H2O for which ?G?? = -2864 kJ/mol,
for NADP+/NADPH ?E?? = -0.324 v, F = 96.5 kJ/v, ?G?? = -nF?E??
and for ADP + Pi ATP + H2O ?G?? = +30.5 kJ/mol.
The energy required to drive this reaction is provided by light (photons). Assume photons of wavelength
? = 700 nm are used, and the energy of one photon is given by E = hc / ?.
h = Planck?s constant = 6.626 x 10-34 J s
c = velocity of light = 3.00 x 108 meters/s
Avogadro?s number = 6.023 x 1023
Calculate the percentage of the light energy used to form the required amount of NADPH and ATP to yield
one mole of glucose. Note that this is a two-photosystem form of photosynthesis.
This question is from last year?s final exam.
?2
To make one mole of NADPH: DG o ' = x 96.5 x ( ?0.324) = kJ/mol
62.5
For 12 mol: 12 x 62.5 = 750 kJ.
=
=
To make 18 mol ATP: DG o ' 18 x 30.5 549 kJ
Total = 750 + 549 = 1299 kJ
To form O 2 i.e. 2 H 2O ? 4 H + + O 2 + 4e ? :
Each electron requires 2 photons:
6 waters x 4 electrons/water x 2 photons/electron = 48 photons.
For 1 mol (one Einstein) of photons at 70 nm:
E = h c / l x Avogadro's Number
6.626 x10?34 J s / photon x 3×108 m/s x 6.023×1023 photons/Einstein
700 x10?9 m
= 171×103 J/mol = 171 kJ/mol

E=

(Remember, nm must be converted to meters; hence 700×10?9 )
For 48 photons: 48 x 171 = 8208 kJ/mol in the 700 nm light.
% consumed = 100 x

1299
= 15.8 % (not very efficient!)
8208

GENERAL BIOCHEMISTRY 115:404/504 Final Examination, Spring, 2015
Name: ________________________

Page 13

I.D. No. ______________________

Part II, Question 4. Twenty points. Please answer the question on this page, using the back should
you need it. SHOW ALL YOUR WORK. An answer with incomplete, uncle?r, or illegible work will be
penalized. IF YOU OMIT THIS QUESTION, WRITE "OMIT" CLEARLY ON THIS PAGE.
The active form of the lac repressor (R) binds to the operator (O) with a dissociation constant (Kd) of
10 M for the reaction RO <?> R + O. About 10 molecules of repressor per E. coli cell suffice to keep
the lac operon turned off in the absence of inducer.
-13

a) If the average E. coli cell has an intracellular volume of 0.3 x 10
intracellular concentration of repressor.

-12

mL, calculate the approximate

b) If the average cell contains two copies of the lac operon, calculate the approximate intracellular
concentration of lac operators.
c) Write an equilibrium constant expression for the dissociation reaction.
d) Calculate the intracellular concentration of free operator, i.e. operator to which no repressor is bound.
Hint: remember that [R]free + [R]bound = [R]total, and th…