Option B: Biochemistry HL Flashcards

(49 cards)

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?

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
Initiator proteins
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
DNA polymerase
Group of enzymes that add more nucleotides to the primers using the existing DNA strands as templates
27
RNA polymerase
During transcription, it reads the DNA sequence and produces an RNA molecule complementary to an existing DNA strand
28
point mutation
Single mismatched nucleotide in the DNA sequence
29
What causes point mutations?
- can occur naturally - UV light - ionising radiation - free radicals - mutagens
30
Diester bridge
When nucleotides combine with one another, the phosphate groups form a digester bridge between 3’ and 5’ carbon atoms of adjacent pentode residues
31
At what end of the strand does the synthesis of nucleic acid begin?
5’
32
What is the sequencing of human genome useful for?
- treating various diseases - designing new forms of medication - understanding human ancestry, migration, evolution and adaptation to environmental changes
33
What are the concerns of the man genome project?
1) possible discrimination of genetically disadvantaged people 2) gene modification 3) access to genetic information
34
Initiator proteins
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
DNA polymerase
Group of enzymes that add more nucleotides to the primers using the existing DNA strands as templates
36
RNA polymerase
During transcription, it reads the DNA sequence and produces an RNA molecule complementary to an existing DNA strand
37
point mutation
Single mismatched nucleotide in the DNA sequence
38
What causes point mutations?
- can occur naturally - UV light - ionising radiation - free radicals - mutagens
39
Biological pigments
Coloured compounds produced in living organisms Have extensive systems of alternate single and double carbon-carbon bonds.
40
What does the colour of the biological pigment depend on?
Molecular structure and on the number of delocalised electrons
41
Antioxidants
Protects cells from UV light, peroxides and free radicals, including a highly reactive single oxygen produced by photosynthesis
42
Porphyrins
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
Haemoglobin in venous blood
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
Cytochrome c oxidase
a large metalloprotein assembly containing four heme prosthetic groups and several ions of other metals includingcopper, magnesium, and zinc.
45
Anthocyanins
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
Stereoisomers
Molecules that have the same sequence of atoms and chemicalbonds but different arrangements of atoms in space
47
Configurational isomers
Stereoisomers that cannot be | transformed into one another without breaking a chemical bond
48
What are the two types of configurational isomers?
1) cis-/trans- isomers | 2) optical isomers
49
CORN rule
- 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