Practice Midterm 1 CHM 362

7/23/2019 Practice Midterm 1 CHM 362

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Not for distribution!

Student Name: Student Number:

UNIVERSITY OF TORONTO MISSISSAUGA

DEPARTMENT OF CHEMICAL AND PHYSICAL SCIENCES

Dr. Jan F. Matejovic

CHM 362H5S Term Test 1

Monday, July 28, 2014.

Duration 50 min

Only the TI-30XIISor the CASIO fx-260solarnon-programmable calculators are permitted. All

others will be confiscated.

Notes to students:1.There are 4 questions on 10 pages. Attempt allquestions.2.Part marks will be given for partly correct answers. Show your work.3.Answer all questions on the test paper. Use the backs of pages if necessary, but indicate if you do so.4.A table of constants is on page 2 and formulae are on the last page.5.Write in PEN ONLY. White-out is not permitted.6.Total marks = 80Not for distribution!


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Page 2

R = 8.314 Jmol-1K-1

kB= 1.38 x 10-23JK-1

h = 6.63 x 10-34Jsec

0 C = 273.15 K



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Page 3

1. High energy phosphate compounds (20 marks)

a. Using point form and diagrams, explain why the G for the hydrolysis of glucose-1-phosphate would be higher than that for fructose-1-phosphate. (There is more space than

you require on this page!!) (10 marks)

G(hydrolysis)= -21.0 kJ/mol G(hydrolysis)= -16.0 kJ/mol



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Page 4

b. Would the G for the hydrolysis of acyl phosphate be higher or lower than that for

glucose-1-phosphate? Explain using diagrams and point form. (There is more space than

you require on this page!!) (10 marks)

OPO32-

O



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Page 5

2. Consider the reaction of glycolysis. (20 marks)

a. The molecule shown below is a competitive inhibitor of a certain enzyme in the

glycolysis pathway. Draw the structure and name the glycolysis intermediate this

molecule is supposed to mimic. What is the name of the enzyme that will be inhibited by

this molecule? Why is this particularly reaction in glycolysis so important? (5 marks)

2-O3P

NH

NH

PO32-

O

OH

BONUS (2 marks) Sketch a Lineweaver-Burk plot for the uninhibited and inhibited reaction.



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Page 6

b. One common experimental tool utilized to study metabolism is to genetically mutate an

enzyme thought to be on a metabolic pathway, rendering that enzyme inactive. Suppose

the seventh enzyme in glycolysis was mutated into an inactive form and then all

enzymes, necessary cofactors, coenzymes and cosubstrates were placed into liposomes

(unilamellar vesicles). The vesicles were then exposed to a finite quantity of glucose.

Vesicles were drawn off at three time-points and the contents extracted and analysed for

glycolysis intermediates. The concentration of these metabolites was then plotted for

these three time points. Below is an example from the textbook on what such a plot

would look like under steady-state conditions.

For this question, you must use your knowledge of the G values for the reactions of

glycolysis!!

On the provided grids, plot approximations of what the glycolysis intermediate

concentrations would be at three time-points:

i. Moments after the addition of glucose

ii. Steady-state conditions a short while after the addition of glucose

iii. After all the glucose is consumed (remember that there were not saturatingamounts of glucose present).

In point form, briefly outline why you chose to show the concentrations the way you did

for each plot. (15 marks)



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Page 7

(Note: E1 = First enzyme in glycolysis; I1 = first intermediate in glycolysis; etc.)

Moments after the addition of glucose

i. Steady-state conditions a short while after the addition of glucose

Explanation:

Steady-state conditions a short while after the addition of glucose

Explanation:

E

1

E

2

E

3

E

4

E

5

E

6

E

7E

8

E

9

E

10

I

1

I

2

I

3

I

4

I

5

I

6

I

7

I

8

I

9

P

y

r

u

v

a

t

e

Mutated

Concentrationof

Metaboliteinter!ediate

E

1

E

2

E

3

E

4

E

5

E

6

E

7

E

8

E

9

E

10

I

1

I

2

I

3

I

4

I

5

I

6

I

7

I

8

I

9

P

y

r

u

v

a

t

e

Mutated

Concentrationof




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Page 8

After all the glucose is consumed(remember that there were not saturating amounts of glucose present)

Explanation:

E

1

E

2

E

3

E

4

E

5

E

6

E

7

E

8

E

9

E

10

I

1

I

2

I

3

I

4

I

5

I

6

I

7

I

8

I

9

P

y

r

u

v

a

t

e

Mutated

Concentrationof




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Page 9

3. Do only one of the following two mechanisms. If you do both, only the first one will be graded.

(20 marks)

OPTION 1

Glyceraldehyde-3-phosphate dehydrogenase reaction

There are two substrates for this reaction and you have available the following:

Enz149CysS- EnzB: N+

NH2

O

R



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Page 10

OPTION 2

Pyruvate dehydrogenase (E1)/dihydrolipoyl transacetylase (E2) conversion of

hydroxyethyl-TPP to acetyl-CoA and with conversion of oxidized lypoyllysine to fully

reduced lipoyllysine

The substrate for this reaction is CoA-SH. The structures of hydroxyethyl-TPP and oxidized

lypoyllysine are given below:

S

N+

R

R' R''

CH3

OH

H

S S

Lipoamide

Also available is: EnzB: EnzBH+



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Page 11

4. Consider the citric acid cycle (20 marks)

a. Suppose mitochondria were isolated, liberated of all TCA intermediate and then bathed in

H218O such that any radiolabelled oxygen would be incorporated into TCA intermediates

for only one round of the cycle. (Note, all necessary cofactors and coenzymes were still

present.) You have this page and the next for your answer, if you require it.

i. If mitochondria were then exposed to citrate, allowing the Krebs Cycle to

commence (note: there is no acetyl-CoA available), would any radiolabelled

water be incorporated? If so, where would the radiolabel appear? Use diagrams

and brief point forms for explanation. Would the cycle get stuck at a certain point

or continue on?

ii. If the mitochondria were then bathed in non-radiolabelled water and acetyl-CoA

was added, how many rounds of the TCA cycle (if any) would it take for the

radiolabelled oxygen to be lost (if any). Use diagrams and brief point forms for

explanation.



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Page 12



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Page 13

Formulae you may find useful:

G= -RTlnKeq

G = G+ RTlnQ

1

v=K

m

Vmax

1+I[ ]

KI

1

S[ ]

+

1

Vmax

Go= H

oTS

o

Question Mark

1 /20

2 /20

3 /20

4 /20

Total Marks /80


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Practice Midterm 1 CHM 362