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Chemistry > Option B: Biochemistry HL > Flashcards

Option B: Biochemistry HL Flashcards

1
Q

How many dissociation constants does a cationic form of 2-amino acids with non-ionizable side-chain have?

A

two: -COOH and -NH3^+
- COOH has relatively higher acidity and dissociates more easily than the protonated amino group, so pKa1 characterizes the equilibrium between the cation and the zwitterion

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2
Q

Acid-base buffer

A

Contains a weak conjugate acid-base pair which can neutralise small amounts of strong acids and bases without significantly changing the pH.

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3
Q

Buffer pH range of an amino acids as acid-base buffer and why?

A

from pH = pKa1 - 1 to pKa1 + 1
and
from pH = pKa2 - 1 to pKa2 + 1

outside of these ranges, the amino acids exist predominately as a single ionic species and loses its ability to maintain a constant pH of the solution

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4
Q

What is the role of protein buffers?

A

They maintain a constant pH of biological fluids, which is essential for the integrity of body tissues and enzyme functions

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5
Q

Allosteric site

A

Many enzymes can temporarily bind to specific molecules via weak non-covalent interaction.
When allosteric site is occupied, the shape of the enzyme changes, which alters the configuration of the main active site, which affects the stability of the enzyme-substrate complex

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6
Q

Non-competitive inhibition

A

When substrate and the inhibitor have different chemical structures and bind to different sites of the enzyme.

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7
Q

Competitive inhibition

A

When substrate and the inhibitor have similar chemical structure and the inhibitor may occupy the main active site fo the enzyme

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8
Q

Michaelis-Menten equation

A

v = (Vmax [S])/(Km + [S])

v: actual reaction rate
Vmax: maximum reaction rate
Km: Michaelis constant, which is equal to the substrate concentration when v=0.5Vmax

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9
Q

Turnover number

A

the maximum number

of substrate molecules that one molecule of enzyme can convert to product per second

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10
Q

What does small km (Michaelis constant) mean?

A

A small Km
indicates high affinity, which means that
enzyme–substrate complex ES is particularly stable and the rate will approach Vmax
even at relatively low substrate concentrations

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11
Q

Protein assay

A

Analytical procedure that detects proteins and the determines their concentrations in solutions

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12
Q

UV-vis spectroscopy

A

Technique that measures the absorption of UV and/or visible light by proteins or their complexes with organic dyes and transition metal ions

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13
Q

Why almost all proteins absorb UV light?

A
  • due to the presence of aromatic rings in phenylalanien, tyrosine and tryptophan residues
  • complexes of proteins with transition metal ions absorb visible light due to d-orbital electron transitions
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14
Q

Absorbance of a sample

A

The logarithmic ration between the intensity of light emitted by the monochromator (Io) and the intensity of light passed through the sample (I)

A = log (Io/I)

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15
Q

Beer-Lambert law

A

A = εcL

where L is the cuvette length and ε is a constant (known as the molar absorptivity or extinction coefficient) that depends on the solvent nature and the temperature of the solution

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16
Q

Nucleic acids

A

Condensation polymers of nucleotides

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17
Q

Nucleotides

A

Products of condenstaion of a nitrogenous base, a pentose sugar (ribose or deoxyribose) and phosphoric acid.

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18
Q

Nitrogenous bases

A

Heterocyclic aromatic amines that contain several nitrogen atoms and act as proton acceptors in aqueous solutions

  • derived from two parent amines: pyrimidine and purine
  • crystalline substance with high melting points due to the presence of multiple polar groups
  • insoluble in water because their molecules are held togehter by strong hydrogen bonds
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19
Q

Pyrimidine

A

cytosine, thymine, uracil

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20
Q

Purine

A

adenine, guanine

21
Q

Diester bridge

A

When nucleotides combine with one another, the phosphate groups form a digester bridge between 3’ and 5’ carbon atoms of adjacent pentode residues

22
Q

At what end of the strand does the synthesis of nucleic acid begin?

A

5’

23
Q

What is the sequencing of human genome useful for?

A
  • treating various diseases
  • designing new forms of medication
  • understanding human ancestry, migration, evolution and adaptation to environmental changes
24
Q

What are the concerns of the man genome project?

A

1) possible discrimination of genetically disadvantaged people
2) gene modification
3) access to genetic information

25
Q

Initiator proteins

A

Group of enzymes that separate the two DNA strands and create short polynucleotide fragments (primers) paired with the separated strands by complementary nitrogenous
bases

26
Q

DNA polymerase

A

Group of enzymes that add more nucleotides to the primers using the existing DNA
strands as templates

27
Q

RNA polymerase

A

During transcription, it reads the DNA sequence and produces an RNA molecule complementary to an existing DNA
strand

28
Q

point mutation

A

Single mismatched nucleotide in the DNA sequence

29
Q

What causes point mutations?

A
  • can occur naturally
  • UV light
  • ionising radiation
  • free radicals
  • mutagens
30
Q

Diester bridge

A

When nucleotides combine with one another, the phosphate groups form a digester bridge between 3’ and 5’ carbon atoms of adjacent pentode residues

31
Q

At what end of the strand does the synthesis of nucleic acid begin?

A

5’

32
Q

What is the sequencing of human genome useful for?

A
  • treating various diseases
  • designing new forms of medication
  • understanding human ancestry, migration, evolution and adaptation to environmental changes
33
Q

What are the concerns of the man genome project?

A

1) possible discrimination of genetically disadvantaged people
2) gene modification
3) access to genetic information

34
Q

Initiator proteins

A

Group of enzymes that separate the two DNA strands and create short polynucleotide fragments (primers) paired with the separated strands by complementary nitrogenous
bases

35
Q

DNA polymerase

A

Group of enzymes that add more nucleotides to the primers using the existing DNA
strands as templates

36
Q

RNA polymerase

A

During transcription, it reads the DNA sequence and produces an RNA molecule complementary to an existing DNA
strand

37
Q

point mutation

A

Single mismatched nucleotide in the DNA sequence

38
Q

What causes point mutations?

A
  • can occur naturally
  • UV light
  • ionising radiation
  • free radicals
  • mutagens
39
Q

Biological pigments

A

Coloured compounds produced in living organisms

Have extensive systems of alternate single and double carbon-carbon bonds.

40
Q

What does the colour of the biological pigment depend on?

A

Molecular structure and on the number of delocalised electrons

41
Q

Antioxidants

A

Protects cells from UV light, peroxides and free radicals, including a highly reactive single oxygen produced by photosynthesis

42
Q

Porphyrins

A

Biological pigments that are complexes of metal ions with large cyclic ligands.

The organic backbone of porphyrins (porphin) contains four nitrogen atoms in a highly conjugated heterocycle

43
Q

Haemoglobin in venous blood

A

Has a slightly lower pH and higher CO2 concentration than arterial blood, protons and CO2 bind to side-chains of amino acids in haemoglobin and act as non-competitive inhibitors

44
Q

Cytochrome c oxidase

A

a large metalloprotein assembly containing four heme prosthetic groups and several ions of other metals includingcopper, magnesium, and zinc.

45
Q

Anthocyanins

A

Biological pigments that give bright colours of flowers, ripe fruits, berries and vegetables

Are water soluble and concentrate in the vacuoles of plant cells

46
Q

Stereoisomers

A

Molecules that have the same sequence of atoms and chemicalbonds but different arrangements of atoms in space

47
Q

Configurational isomers

A

Stereoisomers that cannot be

transformed into one another without breaking a chemical bond

48
Q

What are the two types of configurational isomers?

A

1) cis-/trans- isomers

2) optical isomers

49
Q

CORN rule

A
  • identifies enantiomers of 2-amino acids:

1) CO-R-N: COOH, R. and NH2
2) position the molecule so that the H atom at the C2 carbon faces away from you
3) CORN is spelled clockwise for D-enantiomer and anti-clockwise for an L-enantiomer

Chemistry (25 decks)

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