PROTEINS

PROTEINS: WHAT DO THEIR ROLES DEPEND ONE AND EXPAND ON THIS ANSWER?

THEIR ROLES DEPEND ON THEIR MOLECULAR SHAPE:

1. FIBROUS: (ALPHA HELIX) FORM LONG CHAINS RUNNING PARALLEL TO EACH OTHER WITH CROSS-BRIDGES BETWEEN THE CHAINS —> PRODUCES VERY STABLE MOLECULES.

2. GLOBULAR: (BETA PLEATED) CARRY OUT METABOLIC FUNCTIONS.

COLLEGEN STRUCTURE:

1. UNBRANCHED POLYPEPTIDE CHAIN.

2. THE POLYPEPTIDE CHAIN FORMS AN ALPHA HELIX. LOTS OF THE AMINO ACID GLYCINE HELPS CLOSE PACKAGING.

3. THE POLYPEPTIDE CHAIN IS FURTHER TWISTED.

4. THREE POLYPEPTIDE CHAINS ARE WOULD TOGETHER WITH COVALENT BONDS BETWEEN AMINO ACIDS OF ADJACENT CHAINS.

FIBROUS AND GLOBULAR PROTEIN EXAMPLES:

ENZYME

• NEEDED BECAUSE OTHERWISE TEMP IN LIVING CELLS WOULD BE TOO LOW FOR CHEM MOLECULES TO REACT FAST ENOUGH TO SUPPORT LIFE.

• ENZYMES ARE GLOBULAR PROTEINS SO EACH ENZYME (AND ITS ACTIVE SITE) HAS A SPECIFIC TERTIARY STRUCTURE AND SHAPE.

• SPECIFIC TO ONE SUBSTRATE.

TEMP + ENZYMES:

LOW: AT VERY LOW TEMPS, ENZYME MOLECULES ARE INACTIVE BUT UNDAMAGED. THEY MAY MOVE AROUND SLOWLY IN AQUEOUS SOLUTION, SO THEY COLLIDE ONLY RARELY - RATE OF ENZYME ACTION REMAINS LOW.

WARM: INCREASE IN Ek —> MORE SUCCESSFUL COLLISIONS —> MORE ES-COMPLEXES.

HOT: OPTIMUM, MOST COLLSIONS OCCURE, SHAPE IS MOST COMPLIMENTARY.

TOO HOT: DENATURE.

WHAT IS PH?

HOW CAN IT BE CALULCATED?

HOW CAN YOU CONTROL IT?

• PH: CONCENTRATION OG H+ IONS IN SOLUTION.

• PH CHANGE OF ONE CHANGES THE [(aq) H⁺] BY TEN TIMES.

• PH 1 = 0.1moldm^-3

• pH = -Log10(H⁺)

• CAN CONTROL IT USING BUFFER SOLUTION.

PH + ENZYME

• PH CHANGE ALTERS CHARGES ON THE AA WHICH MAKE UP THE AS OF THE ENZYME.

• CAUSES SOME H BONDS AND IONIC BONDS TO BREAK AND REFORM IN DIFFERENT PLACES.

• THESE CHANGES CAN ALTER THE SHAPE OF THE ACTIVE SITE AND THE SUBSTRATE MAY THEREFORE NO LONGER FIT.

• NO ES-COMPLEXES CAN FORM.

• DENATURE.

COMPETITIVE INHIBITOR

1. MOLECULES THAT HAVE A SIMILAR SHAPE TO THAT OF THE SUBSTRATE.

2. ALLOWS IT TO COMPLETE WITH THE SUBSTRATE AS IT BINDS TO THE AS OF THE ENZYME.

3. SINCE AS ARE BLOCKED BY THE CI, THEY CANT BE OCCUPIED BY SUB.

4. FEWER ES-COMPLEXES FORMED AND ROR REDUCED.

   EFFECT DEPENDS ON ITS CONC RELATIVE TO [SUB].

NON- COMPETITIVE INHIBITOR

1. BINDS TO ALLOSTERIC SITE.

2. CHANGES TERTIARY STRUCTURE OF ENZYME → CHANGES SHAPE OF AS.

3. SUB CAN’T BIND.

4. FEWER ES-C.

5. ROR DECREASE.

EFFECT IS NOT DEPENDANT ON ITS CONC RELATIVE TO [SUB].

IT INACTIVATES THE ENZYME.

FEW ENZYMES MAY REMAIN UNEFFECTED AND REACTION MAY PROCEED AT A SLOW RATE.

SUBSTRATE CONCENTRATION

INCREASING ITS CONC:

1. AS [SUBSTRATE] INCREASES, ROR INCREASES. (Bc there are more molecules → higher chance of collisions, more ES-complexes.)

2. UNTIL SATURATION POINT WHERE ACTIVE SITES ARE FULL, ENZYME IS LIMITING.

3. ROR REMAINS CONSTATE AT A [CERTAIN].

ENZYME CONCENTRATION

AS LONG AS THERE IS AN EXCESS OF SUBSTRATE, AN INCREASE IN THE AMOUNT OF ENZYME LEADS TO A PROPORTINATION INCREASE IN ROR.

(STRAIGHT DIAGONAL DP LINE.)

IF SUBSTRATE IS LIMITING HOWEVER, ANY INCREASE IN ENZYME CONC WILL HAVE NO EFFECT ON THE ROR.

(ROR ATP WILL STABILISE AT A CONSTANT LEVEL)

TEST PROTEINS:

1. BLUE BIURETS + SHAKE.

2. LILAC.

The secondary structure of a polypeptide is produced by bonds between amino acids. Describe how.

Two proteins have the same number and type of amino acids but different tertiary structures. Explain why.

A dipeptide consists of two amino acids joined by a peptide bond. Dipeptides may differ in the type of amino acids they contain. Describe two other ways in which all dipeptides are similar and one way in which they might differ.