PROTEINS

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14 Terms

1
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PROTEINS: WHAT DO THEIR ROLES DEPEND ONE AND EXPAND ON THIS ANSWER?

Roles depend on molectular shape:

  1. FIBROUS: (Alpha helix or Beta pleated, not both)
    Insoluble and very stable molecules.

  2. GLOBULAR: (Alpha, Beta, or both)
    Soluble (because it has hydrophobic and hydrophyllic R-groups)
    Carry out metabolic functions.

2
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FIBROUS AND GLOBULAR PROTEIN EXAMPLES:

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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.

4
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WHAT IS PH?

HOW CAN IT BE CALULCATED?

HOW CAN YOU CONTROL IT?

  • PH: [(aq) H+]

  • PH CHANGE OF ONE CHANGES THE [(aq) H+] BY 10x

  • PH 1 = 0.1moldm-3

  • pH = -Log10(H+)

  • CAN CONTROL IT USING BUFFER SOLUTION.

5
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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.

  • ETC.

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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].

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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.

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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].

9
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ENZYME CONCENTRATION

As long as there is an excess of substate, [enzyme] —> proportionate ROR.

(Diagonal, dp line.)

But if substrate is limiting, increasing enzyme has no effect on ROR, so ROR will stabilise at constant level.

10
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TEST PROTEINS:

  1. BLUE BIURETS + SHAKE.

  2. LILAC.

11
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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.

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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.

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13
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DESC INDUCED FIT AND HOW AN ENZYME ACTS AS A CATALYST?

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14
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DESCRIBE HOW AMINO ACIDS JOIN TO FORM A POLYPEPTIDE SO THERE IS ALWAYS NH2 AT ONE END AND COOH AT THE OTHER.

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