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Exam 4 with Answers | Biochemistry 2006 | CHEM 431, Exams of Biochemistry

Material Type: Exam; Professor: Campbell; Class: BIOCHEMISTRY; Subject: Chemistry; University: Jackson State University; Term: Unknown 1989;

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Name__________________________________________
CHEM 431/531 - Biochemistry I/Exam IV/December 04, 2006
Each Question is Worth 7 Points. Answer ONLY a total of 18 Questions. Maximum Points Possible =
126 Points.
CREDIT WILL BE GIVEN TO THE FIRST “18” QUESTIONS ANSWERED. ADDITIONAL
QUESTIONS ANSWERED WILL NOT BE GRADED!!!!!!!!!!!!
CHapter 10: regulatory strategies
1. Why was it surprising to find that CTP inhibits ATCase?
The substrates for ATCase are carbamoyl phosphate and aspartate. These molecules do not
resemble CTP. Thus, it was clear that the CTP must not bind to the active site, but to a distinct
regulatory site
2. The scheme S T U V W X Y represents a hypothetical pathway for the
metabolic synthesis of compound Y. The pathway is regulated by feedback inhibition. Indicate
where the inhibition is most likely to occur and what the likely inhibitor is.
S T U V W X Y (most likely inhibitor)

– – – – – – – – – – – – – – –
(most likely inhibited step)
3. How does the sequential model differ from the concerted model for allosteric enzymes?
The concerted model does not allow for anything other than an “all-or-none” complete tense- or
relaxed-form protein. In contrast, the sequential model allows for a mixed type of protein,
containing some tense and some relaxed subunits. The form is in response to the ligand binding
by a particular subunit.
4. Why is covalent modification advantageous when compared to proteolytic activation?
Covalent modification is usually a reversible process.
5. Explain how a biochemist might discover that a certain enzyme is allosterically regulated.
The enzyme would show kinetics that do not fit the Michaelis-Menten equation; the plot of V vs. [S] would
be sigmoidal, not hyperbolic. The enzyme kinetics would be affected by molecules other than the
substrate(s).
6. Phosphorylation is an extremely effective tool for catalytic control. Explain the reasons.
A phosphoryl group adds negative charges, allowing new electrostatic interactions and new
hydrogen-bond formation. The free energy charge of phosphorylation is large, which can affect the
conformational equilibrium of different states. Using ATP means that the reaction is linked to the
energy status of the cell. Phosphorylation is rapid and reversible and can result in amplified effects.
These factors affect structural, thermodynamic, regulatory, and kinetic properties.
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Download Exam 4 with Answers | Biochemistry 2006 | CHEM 431 and more Exams Biochemistry in PDF only on Docsity!

Name__________________________________________

CHEM 431/531 - Biochemistry I/Exam IV/December 04, 2006

Each Question is Worth 7 Points. Answer ONLY a total of 18 Questions. Maximum Points Possible =

126 Points.

CREDIT WILL BE GIVEN TO THE FIRST “18” QUESTIONS ANSWERED. ADDITIONAL

QUESTIONS ANSWERED WILL NOT BE GRADED!!!!!!!!!!!!

CHapter 10: regulatory strategies

  1. Why was it surprising to find that CTP inhibits ATCase?

The substrates for ATCase are carbamoyl phosphate and aspartate. These molecules do not

resemble CTP. Thus, it was clear that the CTP must not bind to the active site, but to a distinct

regulatory site

  1. The scheme S  T  U  V  W  X  Y represents a hypothetical pathway for the

metabolic synthesis of compound Y. The pathway is regulated by feedback inhibition. Indicate

where the inhibition is most likely to occur and what the likely inhibitor is.

S  T  U  V  W  X  Y (most likely inhibitor)

(most likely inhibited step)

  1. How does the sequential model differ from the concerted model for allosteric enzymes?

The concerted model does not allow for anything other than an “all-or-none” complete tense- or

relaxed-form protein. In contrast, the sequential model allows for a mixed type of protein,

containing some tense and some relaxed subunits. The form is in response to the ligand binding

by a particular subunit.

  1. Why is covalent modification advantageous when compared to proteolytic activation?

Covalent modification is usually a reversible process.

  1. Explain how a biochemist might discover that a certain enzyme is allosterically regulated.

The enzyme would show kinetics that do not fit the Michaelis-Menten equation; the plot of V vs. [S] would

be sigmoidal, not hyperbolic. The enzyme kinetics would be affected by molecules other than the

substrate(s).

  1. Phosphorylation is an extremely effective tool for catalytic control. Explain the reasons.

A phosphoryl group adds negative charges, allowing new electrostatic interactions and new

hydrogen-bond formation. The free energy charge of phosphorylation is large, which can affect the

conformational equilibrium of different states. Using ATP means that the reaction is linked to the

energy status of the cell. Phosphorylation is rapid and reversible and can result in amplified effects.

These factors affect structural, thermodynamic, regulatory, and kinetic properties.

  1. What do cigarette smoking and elastase have in common?

 1-Antitrypsin is a neutrophil protein that inhibits elastase. Some individuals with a genetic  

disease causing deficiency of functional1-antitrypsin have scarred lungs due to overactive

elastase. One side effect of smoking is oxidation of a met in the1-antitrypsin, and it no

longer functions efficiently as an inhibitor. Thus, both a genetic disorder and smoking can

lead to emphysema.

CHAPTER 11: CARBOHYDRATES

  1. List two reasons carbohydrates are considered important molecules.

(i) Carbohydrates serve several important functions as fuels, metabolic intermediates,

and energy stores. (ii) They are the basis of most of the organic matter on our planet.

(iii) Carbohydrates serve as the structural framework or building blocks for DNA, RNA,

and polysaccharides. (iv)They are also linked to other molecules, such as proteins and

lipids, and play important roles in signaling and structure.

  1. Explain in molecular terms why humans cannot use cellulose as a nutrient, but goats and

cattle can.

The ruminant animals have in their rumens microorganisms that produce the enzyme

cellulase, which splits the (14) linkages in cellulose, releasing glucose. Humans do not) linkages in cellulose, releasing glucose. Humans do not

produce an enzyme with this activity; the human digestive enzyme-amylase can split only

(14) linkages in cellulose, releasing glucose. Humans do not) linkages (such as those in glycogen and starch).

  1. What are lectins? What are some biological processes which involve lectins?

Lectins are proteins that bind to specific oligosaccharides. They interact with specific cell-

surface glycoproteins thus mediating cell-cell recognition and adhesion. Several microbial

toxins and viral capsid proteins, which interact with cell surface receptors, are lectins.

  1. The structure of lactose is shown below. Identify the type of linkage in lactose?

O

OH

OH

CH 2

OH

OH

O

OH

OH

CH 2

OH

OH O

 1 - 4 glycosidic linkage

D-glucose ( anomer)

D-galactose

  1. Name two molecules that form the polar head group of a phospholipids.

Examples of head groups include serine, ethanolamine, choline, glycerol, and inositol.

embedded in the membrane, with a hydrophobic channel submerged about halfway through

the bilayer. The arachidonic acid is a product of membrane lipid hydrolysis, and enters the

protein channel from within the membrane, successfully avoiding any interaction with

aqueous environments.

  1. What kinds of forces or bonds anchor an integral membrane protein in a biological membrane?

The forces that hold integral membrane proteins in the membrane are hydrophobic interactions between

hydrophobic domains of the protein and the fatty acyl chains of the bilayer interior.

CHAPTER 13: MEMBRANE CHANNELS AND PUMPS

19. What is simple diffusion? Provide an example.

In simple diffusion, molecules pass through a membrane from areas of higher to lower

concentration. Only lipophilic molecules can easily pass through the membrane. An

example would be a steroid hormone molecule such as vitamin A.

20 How does active transport differ from passive?

In active transport, molecules must by pumped against a concentration gradient.

This movement requires the active input of energy.

  1. Many pumps are members of the P-type ATPases. If you discovered a new

enzyme with similar function, what reaction product would help convince

you that your enzyme was a member of this family?

The members of this family transfer the phosphate from the ATP to a

specific aspartyl side chain in the enzyme.

  1. Compare and contrast symport and antiport. Which term best describes

the transport system mediated by the Na

K

ATPase?

Symport and antiport are both types of cotransport systems in which two

solutes move through the membrane simultaneously. In symport, both

move in the same direction; in antiport, one solute goes in one direction,

the other in the opposite direction. The Na

K

ATPase of the plasma

membrane is an antiport system. It moves K

in and Na

out in a ratio of

2 K

per 3 Na

. Neither of the two ions can be transported unless the

other is present, which is characteristic of cotransport systems.

  1. Describe the functional domains of the sarcoplasmic reticulum Ca

2+

ATPase.

The protein has an integral membrane domain and a cytosolic head with

three separate domains. One of the head domains is responsible for

binding ATP, another accepts the phosphate group, and another appears to

serve as the actuator. The membrane-spanning domain is the site of

calcium ion binding.

  1. Describe the proposed mechanism of the lactose permease symporter. 1. The cycle begins with the two halves oriented with the opening to the binding pocket

facing outside the cell. A proton from outside the cell binds to a Glu residue in the

permease.

_2. The protonated permease binds lactose from outside the cell.

  1. The structure everts to a form with the binding pocket facing inside the cell._

4) linkages in cellulose, releasing glucose. Humans do not. The permease releases the lactose to the inside of the cell.

_5. The permease releases the proton to the inside of the cell.

  1. The permease everts to the original form and completes the cycle._