Biochemistry Test 3

Name:  _____________________                                                                              April 19, 2017

Possible:  101 pts

1.   (4 pts)  Write the rate equation for the formation of product for the following elementary step.

2A ® B

Rate (or velocity) = k [A]2       -2 for v = d[B]/dt

2.   (3 pts)  In general, if the concentration of enzyme is doubled, the reaction velocity will tend to

a)      double.          b)  stay the same.            c)  decrease by a factor of 2.

3.   (2 pts)  An enzyme becomes saturated when

a)      The product concentration is very low.

b)      The substrate concentration is very low.

c)      The substrate concentration is very high.

d)      The enzyme concentration is very low.

e)      The enzyme concentration is very high.

4.   (8 pts)  On the plot below indicate the locations of Vmax and KM.  Also, estimate values of Vmax ___________ 30 µM/s     2pts

      and KM _______________ 5 µM     2pts  (-2 pts for 15—took value from y-axis)

P07_094 pts for indicating locations

5.   (3 pts)  In comparing two enzymes, the most effective enzyme at catalyzing a reaction will have

a)      The largest Vmax and the largest KM.

b)      The largest Vmax and the smallest KM.(-1)

c)      The largest kcat and the largest KM.

d)      The largest kcat and the smallest KM.

e)      The smallest Vmax and the largest KM.

f)       The smallest Vmax and the smallest KM.

g)      The smallest kcat and the largest KM.

h)      The smallest kcat and the smallest KM.

This Page:  20 pts

6.   (3 pts)  Which two kinds of inhibitors bind to a site on an enzyme that is not the active site?

a)      A suicide inhibitor

b)      A competitive inhibitor

c)      A mixed inhibitor

d)      A noncompetitive inhibitor

7.   (2 pts)  Which kind of inhibitor can be overcome by using a large concentration of substrate?

a)      A suicide inhibitor

b)      An irreversible inhibitor

c)      A competitive inhibitor

d)      A mixed inhibitor

e)      A noncompetitive inhibitor

8.   (2 pts)  To the right is shown the reaction catalyzed by an enzyme.

That reaction is inhibited by the compound below.

      Which of the following is a transition state analog?

a)      Glyceraldehyde-3-phosphate

b)      Enediolate

c)      Dihydroxyacetone phosphate

d)      Phosphoglycohydroxamate

e)      Acetone (not shown)

This Page:  7 pts

 

9.   (6 pts)  A reaction is carried out in the presence and absence of an inhibitor, and initial velocities are plotted versus substrate concentration.  The results are shown in the graph.  What type of inhibitor is this?  Explain.

P07_28

This is a mixed inhibitor.  Both Vmax and KM are changed.

10. (7 pts)  Sketch a Fischer projection of glucose.  (That’s the open-chain, linear form.)

1 pt for 6 carbons

1 pt for aldehyde on C1

5 pts for OHs on C2-C6

11. (1 pt)  Glucose is 

a)      a d sugar                                     b)  an l sugar

12. (1 pt)  Is glucose a reducing sugar? 

a)      Yes                                             b)  No

13. (3 pts)  Glucose is


a)      an aldotetrose

b)      an aldopentose

c)      an aldohexose

d)      a ketotetrose

e)      a ketopentose

f)       a ketohexose


This Page: 18 pts


 

14. (3 pt)  When the beta anomer of glucose forms a cyclohexane structure, how many of its substituents are in equatorial positions?


a)      Just one

b)      Two

c)      Three

d)      Four (-1 pt)

e)      All of the substituents


15. (1 pt)  Glucose usually has a

a)      pyranose ring structure                                         b)  furanose ring structure

16. (2 pts)  Suppose the –OH of a sugar is reduced.  This sugar would then be


a)      a deoxy sugar

b)      a keto sugar

c)      a ribose

d)      a uronic acid

e)      an epimer


17. (4 pts)  Pyrosequencing is the current method of determining the sequence of nucleotides in DNA.  The following figure shows the light given off in a pyrosequencer for a sample of DNA.  What is the sequence of the first 9 nucleotides in this sample?

  AGGGGTGGC

This Page:  10 pts


 

18. (10 pts)  Draw a Haworth projection (that’s the one that shows rings) of sucrose.

a glycosidic bond

 
box-10-a-2.jpg                                                 00002B9CArt                            BB1CF510:

19. (3 pts)  Label a glycosidic bond in the sketch above.

20. (10 pts)      Sketch the structure of a single strand of the nucleic acid represented by UG.

This is for AGTC.  Since U is present, it represents RNA, so the sugars should have an OH on C2’.  And, here’s the structure of uracil:

21. (2 pts)  Bases are paired in double stranded DNA.  Which base is paired with cytosine? 

a)      adenine       b)  guanine      c)  cytosine      d)  thymine      e)  uracil

22. (3 pts)  Which one of the following statements best describes the thermal behavior of DNA?

a)      The temperature at the midpoint of strand separation is called Tm.

b)      The Tm is influenced by the G:C content of the strands.

c)      Separated DNA strands can be renatured by lowering the temperature slowly.

d)      all of the above

e)      none of the above

This Page:  28 pts

23. (8 pts)  Convert the DNA nucleotide sequence, UCAAAUGGGUGA into a sequence of amino acids.  (Biology students:  assume the sequence is on the coding (nontemplate) strand.)

SerAsnGly

 

 

 

 

 

 

 

 

24. (8 pts) Circle the restriction enzyme in the following table that can cleave the following double strand of DNA.  Sketch what the strands look like after being cut.

5’–AGTTGAAGGTCCCTC–3’

3’–TCAACTTCCAGGGAG–5’

AsuI:

5’–AGTTGAAG     GTCCCTC–3’

3’–TCAACTTCCAG     GGAG–5’

 

Enzyme

Recognition/cleavage site

AluI

AG | CT

MspI

C | CGG

AsuI

G | GNCC

EcoRI

G | AATTC

EcoRV

GAT | ATC

PstI

CTGCA | G

SauI

CC | TNAGG

NotI

GC | GGCCGC

      (N represents any nucleotide)

25. (2 pts) Does the enzyme used in the previous problem produce stick ends? (Must be consistent with answer above.)

                                          Yes                  No                   Maybe (or, sometimes)

This Page:  18 pts