MCBG Exam 1 Flashcards

1
Q

What is the central dogma of genetics?

A

Genes containing codons for amino acids are TRANSCRIBED into mRNA containing codon sequences of amino acids which are then TRANSLATED into proteins with specific amino acid sequences

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2
Q

How many genes are in the human genome? How many proteins do these genes code for?

A

About 20,000 genes code for nearly 1 million proteins

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3
Q

What phenomena allow for a million proteins to be coded from only 20,000 genes?

A

Alternate splicing

Postranslational modifications: cleaving, acetylation, methylation … etc.

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4
Q

What is the difference between common and derived amino acids?

A

Common amino acids have codons and corresponding tRNA molecules

Derived amino acids do not have codons and are formed enzymatically from common amino acids.

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5
Q

What is the common structure of an amino acid?

A

An alpha carbon has an ammonium group, a carboxylate group, and a variable side chain as well as a hydrogen atom bound to it

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6
Q

List the monoamino, monocarboxylic amino acids (i.e. the amino acids with no acidic, basic or ringed side groups)

A

Glycine, Alanine, Valine, Leucine and Isoleucine

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7
Q

List the amino acids with ring structures

A

Proline (*heterocyclic ring*)

Phenylalanine (benzene ring)

Tyrosine (phenol ring)

Tryptophan (indole ring)

Histidine (imidazole ring)

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8
Q

List the aromatic amino acids

A

Phenylalanine

Tyrosine

Tryptophan

Histidine

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9
Q

Which amino acids contain Sulfur?

A

Methionine and Cysteine

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10
Q

Which amino acids contain alcoholic side chains (hydroxy group)?

A

Serine, Threonine and Tyrosine

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11
Q

What is considered to be the 21st amino acid and what makes it unique?

A

Selenocysteine is the 21st amino acid

Its codon is redundant with a stop codon, so it was not discovered until much later than the other 20 amino acids.

Has same structure as cysteine, but with SeH instead of SH

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12
Q

What are the carboxamide amino acids?

A

Asparagine and Glutamine

Both contain CONH2groups

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13
Q

What are the dicarboxylic amino acids?

A

Aspartate and Glutamate

both have COOH group (in addition to the C-terminal COOH)

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14
Q

Which amino acids are the diamino amino acids?

A

Lysine, Arginine and Histidine

All have active H+ attached to N group in their side chains

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15
Q

How does hydrophobicity/hydrophilicity affect protein folding?

A

Hydrophobic residues (Aromatic and Alkly amino acids) will be found inside folded proteins.

Hydrophilic residues (ammonium and carboxylate amino acids) will be found outside folded proteins.

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16
Q

Which amino acid are the hyrdophobicities and hydrophilicities measured with respect to?

A

Glycine

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17
Q

Describe the formation of a peptide bond.

A

The carboxylate (COO-) from one amino acid combines with the ammonium (NH3+) from another amino acid

H2O is released as bond is formed between the carboxylate carbon and ammonium nitrogen

This takes place at the ribosome

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18
Q

Conventionally, in which direction proteins are numbered ?

A

N-terminus to C-terminus

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19
Q

Describe how cystine is formed.

A

2 cysteine residues within the same peptide chain can spontaneously oxidate to form disulfide bridges

This is an example of a derived amino acid

These bridges stabilize the folded conformation of proteins

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20
Q

What is the cutoff in length between a peptide and a protein

A

Peptides <50 Amino Acids

Proteins >50 Amino Acids

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21
Q

What is the definition of a Bronsted-Lowry acid and base?

A

B-L Acid: Proton donor

B-L Base: Proton acceptor

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22
Q

If acid HA dissociates into H+ and A-, what is the equation for the equilibrium constant?

A

equilibrium constant = Ka = [H+] [A-] / [HA]

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23
Q

What is the Henderson-Hasselbalch equation?

A

The difference between pH and pKa is equal to the log of the ratio of conjugate base to conjugate acid

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24
Q

What are the values for log(1), log(10), amd log(100)?

A

log(1) = 0

log(10) = 1

log(100) = 2

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25
Q

An enzyme is only active when histidine is in its basic form. At physiologic pH, what percentage of the enzyme is active?

A

pKa of histidine = 6

pH-pKa = 1

log(ratio) = 1 —> ratio=10/1

If 10 parts are basic, and 1 part is acidic, then the enzyme is 90% active

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26
Q

What is the approximate pKa range for the N-terminal residue (and also lysine)?

A

7.6 - 10.6

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27
Q

What is the approximate pKa range for the C-terminal, glutamate and aspartate?

A

3-5.5

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28
Q

What is the approximate pKa range for arginine?

A

11.5-12.5

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29
Q

What is the approximate pKa of cysteine?

A

8-9

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30
Q

What is the approximate pKa range for histidine?

A

6-7

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31
Q

What is the approximate pKa range of Tyrosine?

A

9.5 - 10.5

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32
Q

What is a zwitterion?

A

The form of an amino acid in which the net charge is equal to zero. This form exists at the isoelectric point of the amino acid.

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33
Q

On the titration curve of an amino acid, what values are found in regions where the pH is not changing very quickly (with respect to equivalents of base)?

A

The pKa values of the active protons in the molecule are found at the regions in the titration curve where pH is relatively constant

***NOTE THE AXES****

if pH is y-axis then the pKa’s have low slope

if pH is x-axis then the pKa’s have high slope

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34
Q

How is the isoelectric point calculated?

A

Identify the range of pH where the charge on the amino acid is equal to zero.

Take the average of the two boundaries of this range to get the pI

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35
Q

What is the charge on a protein if the pH is greater than the pI?

A

pH > pI, then protein charge is negative

(more basic solutions have lower H+ concentration, more negative charge)

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36
Q

What determines the rate of migration of proteins in agarose isoelectric focusing?

A

The difference between the pH of the gel and the pI of the proteins

Larger differences will result in faster migrations covering greater distance

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37
Q

How does 2D electrophoresis work?

A

Also known as isoelectric focusing

Can separate thousands of proteins by separating based on pI, on one axis, and molecular weight, on the other axis

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38
Q

How can you identify the pI from a titration curve of an amino acid?

A

The pI is found at the inflection point between the cationic and anionic form of an amino acid

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39
Q

Are disulfide bonds considered primary or secondary structures?

A

Disulfide bonds are primary structure because they are determined by the presence of cysteine residues within the sequence

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40
Q

True or False: The side chains of a peptide chain determine the secondary structure of a protein.

A

False: Secondary structure is only determined by the polypeptide backbone, not the sidechains. Sidechains dictate the tertiary structure

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41
Q

Which level of structure is dictated by non-covalent interactions?

A

Quaternary structure

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42
Q

In peptide chains, which atoms are connected by the Φ bonds? And the Ψ bonds?

A

Φ bonds connect the alpha carbon with the amino group (C-N)

Ψ bonds connect the alpha carbon with the carboxylate carbon (C-COOH)

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43
Q

What is the only amino acid that does not freely rotate about the Φ bond?

A

Proline because of the heterocyclic ring formed by the sidechain

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44
Q

What prevents free rotation about peptide bonds?

A

Resonance stabilization from the C=O gives peptide bonds 50% double bond character, thus preventing rotation

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45
Q

What happens when Φ and Ψ bond angles are the same throughout a protein structure?

A

Regular conformations such as alpha helices and beta sheets form

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46
Q

What two parameters dictate the shape of a helix?

A

n = number of residues per turn

pitch = distance between repeating turns of the helix along a line parallel to the axis of the helix

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47
Q

Describe the alpha helix shape structure

A

A coiled helix with n = 3.6 residues per turn

Every amino acid is hydrogen-bonded to the AA 4 above and 4 below

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48
Q

Describe the beta sheet structure

A

2 or more strands from the same protein interact via hydrogen bonds

Can be parallel (all same N–>C direction) or anti-parallel (alternating N–>C direction)

Adjacent residues have side chains extending in opposite directions: up, down, up, down …etc.

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49
Q

What is a structural motif?

A

Common arrangements of secondary structural elements

example: alpha-turn-alpha

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50
Q

What is a domain?

A

A functional part of a protein molecule with conserved shape (hydrophobic interior, hydrophilic exterior)

Can fold independently from other domains

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51
Q

What is a protein fold?

A

The way the secondary structure elements of the structure are arranged relative to each other in space

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52
Q

What is the difference between a protein family and a protein superfamily?

A

Sequence homology. Protein superfamilies have less homology (>30%) than protein families (>50%). Superfamilies are evolutionarily related proteins, but not as close in sequence as families

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53
Q

What is an ortholog?

A

A protein from the same family that performs a similar function in different organisms

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54
Q

What is a paralog?

A

Proteins from the same family that operate in the same organism.

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55
Q

What is a superfold?

A

A domain fold that is found in more than one protein that are unrelated by sequence or evolution.

Example: Globin folds

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56
Q

Explain the concept of protein interactomes

A

Intracellular proteins present in complexes that are interconnected forming networks.

Some proteins may be shared between complexes, which are therefor related to one-another via the shared protein

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57
Q

Describe the dynamic nature of proteins

A

They are constantly rapidly changing positions by wiggling and rotating about an average position

The ‘images’ we see are average shapes and positions of atoms within the proteins

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58
Q

What are the differences between fibrous and globular proteins?

A

Fibrous proteins have larger MW, more repetitive AA sequences, and are less soluble

They play a structural (rather than functional) role

example: collagen

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59
Q

Describe the structure of collagen.

A

3 CHAINZ

30% of amino acids are glycine, 10% proline and 10% hydroxyproline

2 Repeated sequences: Gly-Pro-X and Gly-X-HyPro

Helical shape: polyprolene with n=3

glycine residues ‘stick out’ allowing room for chains to come together

INTER-chain H-bonds form

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60
Q

What is the predominant form of hemoglobin in human adults?

A

HbA1

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61
Q

What is the difference between the chain composition of HbA1 and that of HbA2?

A

HbA1 has 2 alpha and 2 beta subunits

HbA2 has 2 alpha and 2 delta subunits

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62
Q

What Hb chains are found in human embryos?

A

zeta, epsilon and gamma

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63
Q

At what point in time does the concentration of Hb beta chains become greater than the concentration of Hb gamma chains?

A

The gamma subunits make up fetal hemoglobin. Beta subunits replace the gamma subunits shortly after birth to form the more common HbA1

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64
Q

What is the name of the ring structure found within each hemoglobin subunit? Describe its structure.

A

Porphyrin ring

This structure is composed of 4 pyrrole subunits (5 membered ring with 4C, 1N) interconnected by methenes. An iron atom binds to the nitrogen of the pyrrole groups

All atoms are in the same plane (except iron can sometimes be out of plane)

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65
Q

What are the two commonly found oxidation states of iron?

A

Fe2+ is the ferrous form that results in a functional heme group

Fe3+ is the ferric form that shoul be immediately reduced back to ferrous iron

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66
Q

Describe how the heme group is attached to the hemoglobin protein chain.

A

The iron atom binds the proximal histidine F8

The opposite side of the iron (at which O2 binds) is stabilized by the distal histidine E7

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67
Q

How is the structure of myoglobin different from that of hemoglobin?

A

Myoglobin has no quaternary structure unlike hemoglobin (which has 4 subunits)

Myoglobin is structurally homologous to one subunit of hemoglobin

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68
Q

In the dissociation of oxymyoglobin, what does a smaller Keq signify?

A

A smaller Keq (for dissociation) would signify better binding and more oxymyoglobin present

Keq = [Mb][O2]/[MbO2]

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69
Q

Define P50

A

The partial pressure of oxygen at which 50% of the binding sites have O2 bound

This is a constant value

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70
Q

How is fractional saturation (Y) defined?

A

Y = number of binding sites occupied / total number of binding sites in solution

Y = [MbO2]/([MbO2] + [Mb])

Y = pO2/(P50+pO2)

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71
Q

How does the binding of oxygen to hemoglobin differ to the binding in myoglobin?

A

Oxygen binds cooperatively to hemoglobin, or in other words: the presence of one bound oxygen increases the Hb affinity for the next oxygen to bind to the molecule

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72
Q

Of the 4 Hb subunits, which has the largest K (and thus poorest O2 affinity)?

A

K1 > K2 >K3 >K4 due to cooperative bonding

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73
Q

What are the two conformations of hemoglobin?

A

T and R

T is the tight formation of deoxyhemoglobin

R is the relaxed formation of oxyhemoglobin

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74
Q

Describe the shape of the hemoglobin dissociation curve.

A

The hemoglobin dissociation curve is sigmoidal due to the cooperative (allosteric) binding of oxygen to hemoglobin.

The steep slope represents how ver small changes in pO2 can allow oxygen to bind or dissociate

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75
Q

What is the Hill equation and what is it used to measure?

A

The Hill equation quantifies the cooperativity of a binding interaction

log(Y/1-Y) = nHlog(pO2) - constant

where Y is the fractional saturation

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76
Q

How can a plot of log(Y/1-Y) vs log(pO2) be interpreted?

A

The slope signifies the binding cooperativity (Hill coefficient)

For example, the plot for Hb has a steeper slope than the plot for Mb

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77
Q

What are the units of the cooperativity index (Rx)?

A

Fold change

For example, Rx of Hb is 4.8, meaning that a 5 fold change in pO2 will raise the saturation from 10% to 90%

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78
Q

Based on the pO2 in the lungs and in the tissue cappilaries, how can Hb deposit O2 in the tissues?

A

There is approximately a 5-fold increase in pO2 between the lungs and capillary tissues, which basically allows Hb to completely associate and dissociate with O2

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79
Q

In which Hb conformation is the iron atom out of the plane of the porphyrin?

A

The iron is out of the plane when Hb subunits are in their T formation due to steric hindrance between the proximal histidine and the porphyrin ring.

When oxygen binds, this hinderance can be overcome and iron is pulled into the plane

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80
Q

What is the Bohr effect?

A

Hemoglobins binding affinity for oxygen is inversely related to acidity

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81
Q

Describe the mechanism for the Bohr effect.

A

When Hb changes from T to R conformation, particular acid groups have lower than normal pKa’s, which leads to the release of protons

If [H+] increases, then the equilibrium is shifted towards deoxyhemoglobin

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82
Q

How does a decrease in pH affect the saturation curve of hemoglobin?

A

Lower pH results in a decreased O2 affinity and therefore a higher P50 value

RIGHT SHIFT of the curve

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83
Q

Describe the ionic stabilization of the T conformation of hemoglobin.

A

An ion pair forms between the positively charged histidine and the negatively charged aspartatic acid residue.

This results in a pKa of 7.7 for the histidine, which is less acidic (compared to 7.3 in the R formation)

Good feedback mechanism

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84
Q

What role does diphosphoglycerate (DPG) play in hemoglobin?

A

H+DPG •Hb + 4O2 <–> Hb(O2)4 + DPG + nH+

A 3 carbon isomer that binds to deoxyhemoglobin which promotes the release of oxygen from the hemoglobin molecules via allosteric interaction (cooperative release of O2)

Forces above equation to the LEFT

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85
Q

How does an increased concentration of DPG affect the saturation curve of hemoglobin?

A

Increasing [DPG] causes a right shift of the curve

Hb’s affinity for O2 will be decreased in the presence of DPG

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86
Q

What is the main way that red blood cells transport carbon dioxide (metabolic waste) from the tissues to the lungs?

A

CO2 diffuses into RBC’s and then interact with carbonic anhydrase (CA)

CA combines CO2 with H2O to form H2CO3, which immediately dissociates into bicarbonate (HCO3-) and H+

The HCO3- is soluble in the plasma and travels in blood to the lungs

The H+ (decreased pH) forces the Hb/O2 equilibrium to release more oxygen

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87
Q

What is carbamino-Hb?

A

Carbondioxide can interact with the N-terminal group of Hb in order to be transported from the tissue to the lungs

This interaction releases a proton which promotes the release of O2 from Hb

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88
Q

What are the parameters that characterize reaction rate expressions?

A

Velocity of reaction

Reaction order

Rate constant

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89
Q

What are the key properties of enzymes?

A

Enzymes are catalysts that are not consumed by reactions.

They affect the rate by which equilibrium is achieved BUT not the equalibrium constant between products and reactants (Keq)

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90
Q

What is free energy and how is it related to the reaction constant?

A

ΔG is the difference in free energy between the products and reactants.

Negative ΔG values correspond with spontaneous reactions

ΔG = - RT ln(Keq) <—A higher Keq will lead to a more negative ΔG

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91
Q

What is a transition state? How does the energy of the transition state correspond with products and reactants?

A

The transition state is a high energy intermediate between the products and reactants. The amount of energy difference between reactants and the transition state is called the activation energy. The product will spontaneously form from the transition state.

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92
Q

How is activation energy affected by the presence of an enzyme?

A

Activation energy is lowered by enzymes, which increases the rate constant

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93
Q

How is ΔG affected by the presence of an enzyme.

A

ΔG is not affected by enzymes, only affected by the product and reactant energy difference

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94
Q

What is the reaction order for:

A —k—> P

What is the rate expression of this reaction?

A

First order reaction

v = k[A] = -d[A]/dt = d[P]/dt

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95
Q

What is the reaction order of:

A + B —k—> P

What is the rate expression?

A

Second order reaction

v = k[A][B] = -d[A]/dt = -d[B]/dt = d[P]/dt

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96
Q

What assumptions are made during the derivation of the Michaelis-Menten equation?

A

1) Substrate concentration is much higher (excessive) than enzyme concentration
2) No back reaction occurs because the product is held at a low concentration
3) Steady-state assumption: d[ES]/dt = 0, the enzyme-substrate complex concentration remains constant because it is formed and broken at the same rate

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97
Q

What is the experimental meaning of Km, Vmax and kcat?

A

Km is the substrate concentration at which v = Vmax/2

Vmax is the maximum velocity of the reaction, reached when [S]>>>>Km

kcat is the macroscopic rate constant of bound substrate transformed to product

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98
Q

How is Vmax related to Eo?

A

Vmax = kcat [Eo]

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99
Q

What is the Michaelis-Mentin equation?

A
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100
Q

If [S] >>> Km, what is the velocity of the reaction?

A

v = Vmax

When [S] is very high, velocity is at its highest

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101
Q

What are the axes of a Lineweaver-Burk plot?

A

x-axis: 1/[S]

y-axis: 1/vo

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102
Q

What are the x and y intercepts of a Lineweaver-Burk plot?

A

x-intercept = -1/Km

y-intercept = 1/Vmax (NOTE: Vmax, not Vmax/2)

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103
Q

What does the slope of a Lineweaver-Burk plot signify?

A

Slope = Km/Vmax

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104
Q

How does a competitive inhibitor alter Km and Vmax? How does this look on an LB plot?

A

Km increases, but Vmax remains the same

On an LB plot, the x-intercept is closer to zero, while the y-intercept remains the same, resulting in a steeper slope (slope = km/vmax)

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105
Q

How can competitive inhibition be overcome?

A

By adding additional substrate, the inhibitors can be competed out of the enzyme active sites. This works because both the inhibitor and substrate bind to the same site of the enzyme.

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106
Q

How does a non-competitive inhibitor alter Km and Vmax? How does this look on a LB plot?

A

Km remains the same, but Vmax decreases

On an LB plot, the x-intercept remains the same, while the Y intercept increases, resulting in an increased slope (slope = Km/Vmax)

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107
Q

What are the 2 substrate, 2 product enzyme reactions?

A

1) Sequential: Order of binding matters, A + E —>EA; B + EA —> EAB —> EPQ —> Q + EP —> Q + P + E
2) Random: Order of substrate binding does not matter, A and B form P and Q
3) Ping-pong: First substrate alters enzyme in order to be able to catalyze second substrate

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108
Q

What are the characteristics of a rate determining enzyme?

A
  1. Multisubunit protein with a high molecular weight
  2. Allosteric regulation: often feedback from reaction products
  3. Low Vmax (SLOW)
  4. Rapidly turned-over (i.e. short half life)
  5. Expression highly regulated at genetic level

* In a metabolic pathway, usually one of the first 2 enzymes is the rate controlling enzyme*

109
Q

What is the difference between the concerted model of cooperativity and the sequential model?

A

The concerted model states that subunits of an enzyme are either in T (inactive, greater affinity for inhibitor) OR R (active, greater affinity for activator and substrates) conformation

The sequential model states that a substrate can bind to any subunit and each individual subunit can exist in either the T or R conformation. Subunit interactions contribute to the cooperativity in binding

110
Q

How does excess ATP affect Aspartate transcarbamylase?

A

Excess ATP binds to the regulatory subunit of ATCase and releaves the inhibition by competing with CTP for binding sites. This activates the enzyme and increases reaction velocity.

If CTP is in excess, the opposite occurs and the enzyme is deactivated

111
Q

What is a homotropic effect?

A

An allosteric effector that binds to a protein and changes the binding affinity of the same type of molecule

Example: O2 binding in hemoglobin

112
Q

What is a heterotropic effect?

A

An allosteric effector that is a different molecule than the molecule whose binding constant is affected

Example: DPG affecting O2 in hemoglobin

113
Q

In the cell cycle, when is chromatin in its most extended form? Most condensed form?

A

Chromatin is most extended during interphase (most of cell cycle)

Chromatin is highly condensed in mitosis

114
Q

What phase of the cell cycle is DNA copied?

A

S phase of interphase

115
Q

What are the types of DNA sequences required to produce a functional eukaryotic chromosome?

A

Telomere, origin of replication and centromere

116
Q

What is the function of a telomere?

A

Telomeres are repetitive sequences found at the end of chromosomes

They enable the ends of chromosomes to be replicated

They cap the chromosome to protect it from being degraded by the cell

117
Q

What is the function of the origin of replication?

A

The site at which DNA duplication begins

These are necessary to ensure that a complete copy of DNA is made

118
Q

True or false: Eukaryotic chromosomes can contain more than one origin of replication

A

True: multiple origins of replication ensure that the entire chromosome can be replicated rapidly

119
Q

What is the funciton of a centromere?

A

Attachment point to mitotic spindle, allowing fro each duplicated chromosome to be apportioned to each daughter cell.

Centromeres link sister chromatids together via cohesin molecules.

120
Q

What are the components of a gene?

A

Exons: the coding regions for proteins

Introns: non-coding material, often regulatory roles, removed via splicing before protein synthesis

Regulatory regions

121
Q

What is a gene?

A

DNA sequences that produce a functional RNA molecule which encodes a protein or forms a structural/regulatory RNA

122
Q

What percentage of genomic DNA has unique sequences?

A

50% uniqe sequences, 50% repetitive sequences

123
Q

What percentage of DNA is found within exons?

A

Only 1.5%

124
Q

What are the different types of repeated sequences found in DNA?

A

Transposons including:

Interspersed nuclear elements (Long = LINEs, Short = SINEs)

Retroviral-like elements

DNA-only transposon ‘fossils’

Simple sequence repeats that code for telomeres, centromeres, etc.

125
Q

What are nucleosomes?

A

The basic unit of eukaryotic chromosome structure composed

Includes core histone protein and about 200 base pairs of DNA (~147 base pairs wrap around histone and the rest link to adjacent histone)

126
Q

How many times does DNA wrap around each histone?

A

1.67 times

127
Q

What are the components of the core histone protein?

A

The histone core is octameric, made up of 2 of 4 different proteins

2 H2A, 2 H2B, 2 H3 and 2 H4

128
Q

Based on the charge of DNA, what amino acids would you guess to be prevalent in the histone protein molecules?

A

DNA is negatively charged

Histones have a high concentration of positively charged amino acids (lysine and arginine) which help tight binding between the histone core and the DNA

129
Q

What form of DNA is most active: condensed or decondensed?

A

Decondensed DNA is more active than highly condensed DNA

130
Q

What is the major role that histone tails play in chromatin organization?

A

The N-terminal histone tails extend ‘away’ from the histone complex and are major sites for covalent chemical modifications that control many aspects of chromatin structure

131
Q

Describe the secondary,tertiary and quaternary structure of histone proteins

A

3 alpha helices (secondary) make up the characteristic histone fold (tertiary)

H2A and H2B heterodimerize into 2 heterodimers, and 2H3 + 2H4 form a tetramer (quaternary) to form the histone octamer

132
Q

Describe the dynamic nature of nucleosomes wrapping and unwrapping.

A

Nucleosomes are wrapped (DNA unavailable for binding) for 250 milliseconds, and then a short window of opportunity exists in which the nucleosome unwraps for 10-50 milliseconds (DNA available for binding)

Sequence specific DNA-binding proteins can bind during that brief time

133
Q

What role do ATP-dependent chromatin remodeling complexes play?

A

They facilitate movement of DNA around the nucleosome allowing access to DNA

These proteins use energy from ATP hydrolysis to loosen the DNA wrapping around the histones leading to a conversion from condensed chromatin to decondensed chromatin

They can also replace or exchange histone proteins within nucleosomes with the help of histone chaperones

134
Q

What are the two different models to explain the 30-nm diameter chromatin fiber? Which model seems more accurate?

A

Zig zag model

Integrated solenoid model

Evidence exists for both models, they may both exist in different situations

135
Q

What components of histone proteins help with chromatin compaction?

A

N-terminal tails (H4 tail mainly) of adjacent histones interact with one another

An additional protein, histone H1 binds where DNA enters/exits the histone making it bind a little tighter

136
Q

What is a chromatin scaffold?

A

Proteins that help organize the folded 30-nm fiber into looped domains

Chromatin can be decondensed while remaining anchored to these proteins in order to increase expression of genes in the loop

137
Q

Describe the model for how condensin may help chromosome to condense.

A

A dimer between Smc2 and Smc4 interacts with each loop of DNA

The condensin dimers can interact with eachother forming an 8 dimer ring

The 8-dimer rings can then stack into condensed chromatin

138
Q

Describe the idea of nuclear neighborhoods.

A

Different spatial locations of genes results in different gene expression.

Near the nuclear envelope is the gene silencing neighborhood, and the gene expression increases in more central locations.

Genes can move to different neighborhoods in order to modify expression

139
Q

Describe the difference between genetic inheritenance and epigenetic inheritance based on chromatin structures

A

In genetic inheritance, the DNA is mutated and a gene is permanently off. After somatic cell multiplication (mitosis) or germ cell production (meiosis) the gene will still be off.

In epigenetic inheritance, chromatin changes cause the gene to be silenced, but DNA sequence is unaltered. After somatic cell multiplication, the genes are still off because chromatin structure is copied from the parent cell. After germ cell production, the gene is once again on.

140
Q

True or false: Euchromatin is resistant to gene expression.

A

FALSE

Euchromatin is the loosely wound and highly expressed chromatin structure

Heterochromatin is resistant to gene expression. Genes relocated to heterochromatin are silenced.

141
Q

What role do barrier DNA sequences play?

A

They are sequences that separate heterochromatin from euchromatin. Without a barrier, the heterochromatin can spread into neighboring euchromatin and silence the gene expression within normally expressed genes.

142
Q

What are the various ways that barrier proteins can stop the spread of heterochromatin?

A

Barrier proteins can attach to the barrier DNA sequences and stop heterochromatin from spreading by:

physically blocking the spread

protecting euchromatin

enzymatically blocking the spread

143
Q

What are the different covalent modifications common in the regulation of chromosome structure?

A

Methylation, phosphorylation, acetylation and ubiquitylation

144
Q

What are the two modifications of lysine that occur frequently within histone tails?

A

Lysine can be methylated 1, 2 or 3 times thus changing the structure and determining which proteins can bind to the lysine

Lysine can also be acetylated, which removes the positive charge that gives lysine a high affinity for DNA

These two modifications are competing reactions.

145
Q

What are reader and writer proteins?

A

Modifications of histone tails attract specific proteins that assemble into a code-reader complex.

They recruit other components of a protein complex that have catalytic activity

These complexes can change gene expression levels or lead to a specific biological function

146
Q

How do histones within the centromere differ from histones elsewhere?

A

They contain a centromere specific histone H3 protein which can interact with binding proteins and the kinetochore assemply to attach to microtubules

147
Q

How much shorter is a mitotic chromosome than it’s extended length?

A

10,000 fold shorter

148
Q

Describe how epigenetic drugs targeting histone modifications may be helpful for fighting cancer.

A

Modfications resulting in increased acetylation may prevent tumor cells from proliferating

example: histone deacetylase inhibitor Vorinostat leads to increased acetylation in tumor cells

149
Q

What are the Watson-Crick base pairs?

A

C pairs with G via 3 H-bonds

A pairs with T via 2 H-bonds

150
Q

If the template strand of DNA is in the following orientation:

5’ —————————————–3’

What is the orientation of the primer strand?

A

Template: 5’ —————————————–3’

Primer: 3’ —————————————–5’

Template and primer strands are anti-parallel

151
Q

During DNA replication, what part of the DNA nucleotide performs a nucleophilic attack to add an additional nucleotide?

A

The 3’ hydroxyl group at the end of the DNA strand performs a nucleophilic attack on the innermost phosphate of the pyrophosphate of the next nucleotide.

152
Q

What group is found at the 5’ end of a DNA strand? What about at the 3’ end?

A

5’ end: phosphate

3’ end: hydroxyl

153
Q

What are the four characteristics that govern nucleic acid synthesis?

A
  1. A pre-existing nucleic acid strand
  2. Nucleic acids strand grow only in 5’–>3’ direction
  3. Polymerases synthesize nucleic acids
  4. Duplex DNA synthesis requires a special growing fork because of antiparallel strands
154
Q

What group is required for DNA polymerase to add base pairs to DNA chains?

A

DNA polymerase requires the 3’ -OH group

155
Q

Where does DNA polymerase receive its energy from?

A

During the nucleophilic attack by the 3’OH on a DNTP, a pyrophosphate molecule is released

This generates energy that drives the reaction and the growth of the DNA chain

156
Q

How does DNA synthesis differ in the leading and lagging strands?

A

The synthesis of DNA at the leading strand is continuous in the 5’ —> 3’ direction

The synthesis of DNA on the lagging strand is discontinuous, with Okazaki fragments forming in the 5’ —> 3’ direction

157
Q

Why does DNA polymerization only occur in the 5’—>3’ direction?

A

The proofreading ability of DNA polymerase is compromised if new bases are added to the 5’ end instead of the 3’ end.

The addition of a DNTP to the 3’ end is possible due to the breakage of the high energy phosphate bond. This high energy bond cleavage would not occur if bases were added to the 5’ end.

158
Q

True or false: DNA replication is unidirectional

A

False

DNA replication is bi-directional.

From an origin of replication, DNA replication occurs in both directions extending outward from growing forks.

Synthesis on each side occurs on the leading strand and the lagging strand.

159
Q

What are the six mechanisms that are utilized during DNA replication?

A
  1. Initiation
  2. Unwinding
  3. Priming
  4. Unidirectional fork movement
  5. Untangling
  6. Termination
160
Q

What base-pairs are present in high amounts within origins of replication?

A

Origins of replication are A-T rich

Only 2 H-bonds hold A/T together, so they are easier to separate.

161
Q

What is the function of DNA helicase?

A

DNA helicase is an allosteric motor protein that unzips DNA enabling the entry of RNA primase to place primers on the DNA chains

162
Q

What is the function of single-strand binding proteins (SSBs)?

A

They prevent reannealing and formation of hairpins within single chains of DNA

SSBs are cooperative proteins that straighten out the DNA backbone but leaving the bases still accessible for DNA polymerase

163
Q

True or false: DNA primase places and RNA primer on the DNA chain that is required for DNA synthesis to begin.

A

True.

DNA synthesis requires an RNA primer, which is placed by DNA primase

This RNA primer is about 10 base pairs long in humans

164
Q

What prevents DNA polymerase from ‘falling off’ the DNA strand during DNA synthesis?

A

A sliding clamp protein (PCNA) permits processive DNA synthesis and keeps the DNA polymerase on target as it moves along the DNA chain.

165
Q

Why are the errors often found in RNA primers not problematic?

A

RNA primers are removed (by an RNase) and replaced by DNA (by DNA polymerase)

166
Q

How are nicks in the phosphate backbone sealed?

A

DNA ligase synthesizes phosphodiester bonds linking the Okazaki fragments together into one complete DNA chain

167
Q

Other than building DNA chains, what is another major function of DNA polymerase? Describe how this works.

A

DNA polymerase also proofreads the DNA as it is synthesized.

DNA polymerase has 3’—->5’ exonuclease activity to ‘chew back’ to create a proper base-paired 3’OH end on the primer strand

168
Q

How is DNA polymerase able to achieve such high-fidelity ( only 1 error in 109 base pairs) DNA synthesis?

A

The 5’—>3’ polymerization has about 1 error every 105 bp

The proofreading capability increases the fidelity about 100 fold, and strand-directed mismatch repair increases fidelity another 100 fold resulting in 1 error in every 109 base pairs

169
Q

Describe how DNA polymerase is able to rapidly move from the end of one Okazaki fragment to the beginning of the next.

A

A protein assembly of DNA polymerase, the sliding clamp protein, DNA helicase and DNA primase is found at the fork.

The lagging strand loops around so that the end of one Okazaki fragment is adjacent to the beginning of the next.

DNA polymerase is reloaded to the beginning of the loop upon finishing an Okazaki fragment (SEE BLUE ARROW BELOW)

170
Q

What is DNA supercoiling?

A

Supercoiling is when the DNA helix in front of DNA polymerase becomes tangled due to strain requiring rapid rotation. Every 10 base pairs synthesized equals one required rotation of the DNA chain. This is problematic because there is not enough energy available for rotation.

171
Q

Explain how DNA topoisomerase I relieves supercoiling.

A

The tyrosine at the active site of DNA topoisomerase I covalently attaches to DNA phosphate, thus introducing a nick into the DNA backbone.

This allows the DNA strand more flexibility and allows for rotation to relieve the accumulated strain.

The energy is conserved in the tyrosine-phosphate bond, so the reaction is reversible.

172
Q

Explain how DNA topoisomerase II is able to untangle DNA.

A

DNA topoisomerase II untangles DNA after it is replicated.

DNA double helices that are interlocked are able to ‘pass through’ one another when DNA topoisomerase II makes a reversible covalent attachment to opposite DNA strands.

It creates a ‘gate’ through which one chain can pass through the other and then restores the intact helix

173
Q

What is the effect of DNA topoisomerase II inhibitors on cancer?

A

DNA topoisomerase II inhibitors are chemotherapeutic drugs

They increase the stability of topoII-DNA complexes causing high levels of DNA breaks

Too many breaks result in cell death

174
Q

True or false: Eukaryotic chromosomes have many replication origins

A

True

Replication origins are located throughout eukaryotic chromosomes

175
Q

How does the cell ensure that each origin is activated only once per cell cycle?

A

Proteins called origin recognition complexes (ORCs) bind to replication origins and recruit other proteins during the G1 phase

These proteins are phosphorylated during S phase, which marks them as activated and prevents them from being used again during that cell cycle.

176
Q

What happens to the histones during DNA replication?

A

Nucleosomes are removed from the parental chromatin and redistributed to daughter strands

Epigenetic modifications can be inherited. After replication, half of the histones will have the modifications of the parent strand, and half will be new, unmodified, histones

Parental histone modifications are then re-established by reader-writer complexes

177
Q

How are centromere-specific nucleosomes conserved between parent and daughter DNA strands?

A

Which histone proteins are added into newly synthesized DNA is determined by the existing histones from the parent strand in both daughter strands.

If existing histones have the CENP-A H3 variant, then the CENP-A H3 variant will be used to fill in missing histones in the area

178
Q

How does eukaryotic chromosomal DNA replication differ from prokaryotic DNA replication?

A
  1. Compartmentalized in the nucleus
  2. Begins at multiple origins, activated throughout S-phase in a preceise and regulated manner
  3. Nucleosomal proteins must be duplicated along with DNA to maintain proper chromosomal organization
179
Q

What is the function of telomerase?

A

Telomerase is a riboprotein that provides an RNA template for synthesizing the G-rich telomere sequence found at the ends of chromosomes

180
Q

Which end (5’ or 3’) does telomerase extend?

A

Telomerase extends the 3’ ends of DNA in order to make room for the completion of lagging strands by DNA polymerase.

181
Q

What cells in the human body express telomerase?

A

Only germ cells and stem cells

182
Q

How are the ends of telomeres protected from degradation?

A

They fold into T-loop structures that prevent degradation

183
Q

What is the function of telomeres?

A

They protect single-stranded chromosomal ends from recombination, fusion and from being recognized as damaged DNA

If they are shortened, they can trigger replicative senescence in human cells (will no longer replicate)

184
Q

How does trinucleotide repeat expansion occur in diseases like Huntingtons?

A

Within repetitive sequences, the strands can separate momentarilly and then rejoin at slightly different spot.

Replication will proceed as normal, but with a loop of repeats that gets duplicated due to misalignment between the strands.

185
Q

What are the consequences of increasing numbers of trinucleotide repeats in Huntington’s disease?

A

<28 CAG repeats = Normal

28-35 CAG repeats = Intermediate

36-40 CAG repeats = Reduced penetrance

> 40 CAG repeats = Full Penetrance

SO: Mo’ repeats, mo’ problems

186
Q

What are diseases of anticipation?

A

In trinucleotide repeat diseases, over generations, more repeats can be expected resulting in earlier disease onset.

187
Q

What do restriction enzymes do?

A

They cleave DNA at specific sequences (which are often palindromic)

This is very useful for reproducibly generating DNA fragments in lab

188
Q

What are the two different types of cuts that restriction enzymes can make?

A

Blunt cut: straght across base pairs

Cohesive/sticky end cuts: cut that leaves an overhang on both strands

189
Q

The following DNA strand is produced with a restriction enzyme. What bases must be present on a different strand for the strands to anneal together?

A

The 5’ overhang of the attaching strand will have:

5’-AATT-3’

190
Q

How is recombinant DNA created?

A

A restriction enzyme cuts DNA opening up sticky ends

A DNA fragment from another source is cut with the same enzyme, giving it the same sequence of sticky ends

The fragments stick together by base pairing

DNA ligase seals the cut sites to produce a recombinant DNA molecule

191
Q

How can bacterial plasmids be used to amplify DNA fragments?

A

With restriction enzymes, a DNA fragment can be added into a circular DNA called a bacterial plasmid

This plasmid can be taken up by a bacterial cell via Transformation

The bacterial cells can be cultured to multiply and replicate the DNA sequence of interest

192
Q

What property does gel electrophoresis separate DNA molecules based on?

A

Size

193
Q

In a gel electrophoresis, does DNA move towards the anode or the cathode?

A

DNA moves toward the positively charged anode

194
Q

Explain the physical basis for why molecules of different lengths migrate through gels at different speeds.

A

Longer DNA fragments have a harder time moving through the gel because they get caught up in the cross linked matrix of the gel

195
Q

Describe the 3 different types of gels used for electrophoresis and the different lengths that of DNA that they are used to analyze.

A

Agarose gel: can separate DNA between 200 and 50,000 base pairs

Polyacrylamide: can separate DNA shorter than 200 base pairs

Pulsed-field agarose gel: can separate DNA longer than 50,000 base pairs

196
Q

What is hybridization?

A

When complementary nucleic acid strands bind to each other

If DNA strands are denatured with heat or high pH, under the right conditions they can spontaneously reform into their original double helix confirmation

197
Q

What is Southern blotting?

A

A lab technique used to detect a specific DNA molecule wihtin a complex mixture

First, DNA is separated by size using electrophoresis

DNA is transfered to a nitrocellulose paper, and probes are then added

Uses single stranded DNA probes that will hybridize with their target sequence if the target sequence is present.

198
Q

What is FISH?

A

Fluorescent in situ hybridization

An advanced technique used to locate specific genes on chromosomes

Useful for identifying over expressed DNA often associated with cancer

199
Q

What is PCR?

A

Polymerase Chain Reaction: a technique for multiplying a specific piece of DNA

200
Q

What is necessary in order for a PCR reaction to occur?

A

DNA (existing DNA to be copied)

Nucleotides (to be used in new DNA synthesis)

DNA polymerase (enzyme to catalyze synthesis)

PCR primers (flank the region of interest)

Buffers / Mg2+, Ca2+

201
Q

Describe the steps in a PCR reaction

A
  1. Denature DNA sample
  2. Primers bind to DNA strands (Annealing)
  3. Polymerase synthesizes new DNA strands (Extension)
  4. Separate DNA strands (Denaturation)
  5. Repeat 2-4 25 to 40 times
202
Q

Describe the number of copies made by an increasing number of PCR cycles.

A

PCR is an exponential growth

2 —> 6 —-> 14 —->30 …. etc.

By the 20th cycle over a million copies will be made

203
Q

How can microsatellites be analyzed for “genetic fingerprinting”?

A

Microsatellites are small tandem repeats that occur in non=coding regions

Using PCR, the number of copies of microsatellites can be determined

The number of copies varies greatly between individuals

Legally, 13 microsatellites are needed to prove that a sample is from a specific person

204
Q

Why is cDNA useful for the analysis of RNA?

A

RNA is single stranded and relatively unstable, so it is not able to be directly replicated by PCR

Reverse transcriptase is used to produce complementary DNA (cDNA) from an RNA template

The cDNA strand can then be used as a template to produce a double stranded DNA copy of the original mRNA that can be amplified using PCR

205
Q

What are microarrays used for?

A

They measure the expression of many genes at the same time using hybridization techniques

206
Q

Describe the microarray procedure.

A

mRNA is isolated from a tissue sample

cDNA is produced from mRNA using labeled nucleotides

The cDNA is applied to the microarray chip, which has wells with various single-stranded DNA fragments each representing a different gene

cDNA hybridizes in wells with complementary bases

Wells with hybridization will fluoresce, indicating that the particular gene is expressed in the tissue sample

207
Q

What are the three classes of small-scale mutations?

A

Base substitutions/modifications, deletions, insertions

208
Q

What class of mutation can result in a frameshfit?

A

Deletions or insertions can lead to a shift in the reading frame, affecting the codons downstream from the mutation

209
Q

What imaging/visualization technique is useful for detecting large deletions/insertions?

A

SKY chromosome painting. Different chromosomes appear in different colors making it easy to see when chromosomes have exchanged material

210
Q

What are the classes of large scale mutations?

A

Deletions, duplications, inversions, insertions, and translocations

211
Q

What are SNPs?

A

Single nucleotide polymorphisms that are responsible for normal allelic sequence variation throughout humans. 1:250 to 1:1000 base pairs differ between allelic sequences

212
Q

What is a de novo mutation?

A

A new mutation in an individual with no family history of that genetic defect. The mutation arises in the germ line of the parent of the affected individual.

213
Q

What are the sources of de novo mutations?

A

1) Arise from copying errors 2) Spontaneous chemical attack (depurination, deamination) 3) Environmental exposure

214
Q

What happens in a depurination?

A

Purines are spontaneously destroyed via hydrolytic attack, so the DNA loses an A or a G. Happens in 5000 bases per day per cell

215
Q

What happens in a deamination?

A

Amino group is lost via hydrolytic attack, so a C is converted to a U (OR A to hypoxanthine; G to xanthine). Happens in 100 base pairs per cell per day

216
Q

Which nucleotide base cannot be deaminated?

A

Thymine

217
Q

Describe the mutation that can occur to methylated cystosine bases (5-methy cytosine)

A

Cytosines can be methylated at carbon 5 as a means to control gene expression. The 5mC can be deaminated, converting the base into thymine. This T will be opposite a G in the DNA strand creating a mismatched base pair.

218
Q

A C is deaminated. Describe the resulting base pairs at that location following DNA replication.

A

Deamination of C produces a U. The two strands will be used as templates to produce new DNA strands, so the U will pair with an A (mutation). The other strand, with the G will pair with a C (correct pairing)

219
Q

An A is depurinated. Describe the resulting base pairing at that location following DNA replication.

A

A gap in the base pairing will exist with an unpaired T across from the location of the removed A. One new strand will skip the T-A pair completely, while the other will replicate successfully.

220
Q

How do cells deal with mutated DNA that is not repaired?

A

Pathology occurs when the rate of DNA damage > rate of repair. (ex: cancer, senescence, apoptosis)

221
Q

How are deleterious mutations avoided?

A

1) Natural selection selects against undesireable mutations 2) Repair/recombination ensure that mutations are repaired or eliminated before they can cause harm

222
Q

What are the three general classes of DNA repair mechanisms?

A

1) Direct repair 2)Excision repair 3)Double-strand break repair

223
Q

Describe the direct repair mechanism

A

Damage is reversed in an energetically costly manner. Example: A methylated form of G can bind to C or T and blocks replicative polymerases. An enzyme named MGMT reverses the methylation to quickly repair

224
Q

What are the three main types of excision repair.

A

Base excision repair (BER), nucleotide excision repair (NER) and mismatch repair (MMR)

225
Q

Describe the common mechanism of DNA excision repair.

A

Recognition of damage, cleavage by endonuclease, removal by nuclease or helicase activity, replacement by polymerase and ligase activity.

226
Q

How does base excision repair differ from nucleotide excision repair?

A

Base excision repair uses specific glycosylases to remove individual bases. Nucleotide excision repair is less specific and removes a chunk of DNA that includes the mutated base(s).

227
Q

What is transcription coupled repair?

A

As RNA polymerase is transcribing DNA, it can stall and back up at sites of damage. Elongation factors are able to repair the damage and then proceed with normal transcription.

228
Q

Describe mismatch repair.

A

Proteins (MutS and MutL) can recognize errors in newly synthesized DNA strands. This protein complex removes the DNA from the site of the mutation to the next nick. DNA is then re-synthesized properly following strand removal.

229
Q

What property of newly synthesized DNA directs the MMR system to attach to the appropriate strand.

A

The lagging strand transiently contains nicks that have not yet been sealed by DNA ligase

230
Q

In which stages of the cell cycle does homologous end joining (HEJ) occur? What about nonhomologous end joining (NHEJ)?

A

HEJ occurs in S and G2 phases. NHEJ occurs in Go and G1 phases

231
Q

How does nonhomologous end joining work?

A

An accidental double-strand break occurs in a DNA strand. The ends are recognized by Ku proteins. Nucleotides are degraded from both ends in order to produce a ‘clean break’. The ends are then joined, resulting in the deletion of DNA sequence.

232
Q

Describe the homologous end joining process.

A

This occurs when sister chromatids exist. An accidental double-strand break occurs in one crhomosome. Nucleotides are degraded to make a clean break, and then the sister chromatid is used to guide the successful repair of the other chromatid.

233
Q

How is strand invasion catalyzed during recombination?

A

Double strand breaks initiate general recombination. RecA/Rad51 proteins catalyze the strand invasion.

234
Q

What is crossing over?

A

The general recombination process that occurs during meiosis. Homologous DNA double helices exchange portions of their DNA by breaking and rejoining.

235
Q

In what stage of meiosis do homologs separate?

A

Anaphase I

236
Q

In what stage of meiosis do sister chromatids separate?

A

Anaphase II

237
Q

If there are 3 pairs of homologous chromosomes in a cell, how many possible gametes exist if no recombination occurs? How does this change with recombination?

A

Without recombination: 2^3 = 8 possible gametes; with recombination: infinite

238
Q

What is a chiasma?

A

The point of crossing over that is observed at the end of prophase I in meiosis

239
Q

What proteins are involved with the general recombination process in meiosis?

A

Spo11 breaks the dsDNA to begin the process. MRE11 uses nuclease activity to remove Spo11 and process the ends. A RecA-like protein begins strand exchange.

240
Q

What is a Holliday junction?

A

A DNA intermediate that forms during recombination that consists of 4 strands of DNA shared between two DNA helices. This structure can rotate (and is stabilized by proteins) into a symmetrical conformation. This can be cleaved in 2 different ways resulting in either the parental DNA configuration or an exchanged DNA configuration

241
Q

Describe the two different types of heteroduplexes and how they form.

A

Heteroduplexes are formed during recombination whether or not therecombination results in a cross-over. Heteroduplexes are regions in which bases from one parent are matched with bases from the other parent. They will either be small ‘islands’ of exchanged DNA from non-crossover events, or the beginning of the junction between DNA from each parent. (See figure 5-65)

242
Q

What is gene conversion?

A

When the distribution of alleles following meiosis differs from the expected ratio. 3:1 instead of 2:2. This is caused by mismatch repair excising a portion of one parents DNA and filling in the gap with an extra copy of the other parent’s allele.

243
Q

How does loss of heterozygosity occur in somatic cells?

A

Recombination can occur during S/G2. If the gene of interest is exchanged during recombination, then the segregation during mitosis can result in two dominant alleles in one cell and two recessive alleles in the other cell.

244
Q

How do RNA and DNA differ chemically?

A

RNA contains ribose sugars while DNA contains deoxyribose

RNA strands contain uracil bases while DNA strands contain thymine bases

245
Q

How does the DNA containing T and RNA containing U protect from mutations?

A

Cytosine can spontaneously deaminate into a U. If U were normally present in DNA, then this deamination product would be difficult to detect in DNA.

246
Q

What are the 3 major types of RNA and how prevalent is each type in terms of present produced?

A

Messenger RNA (mRNA): 20-40%

Ribosomal RNA (rRNA): 50-70%

Transfer RNA (tRNA): 10%

247
Q

What are the major steps in transcription?

A
  1. Initiation
  2. Elongation
  3. Termination
248
Q

How does transcription differ from DNA replication?

A

Transcription produces RNA from DNA instead of copying DNA

Transcription is caltalyzed by RNA polymerase

No primers are needed

Entire strand does not have to be transcribed, the cell can choose which genes to transcribe

Error rate in transcription is much higher

249
Q

Which RNA polymerase is responsible for the production of Pre-mRNA?

A

RNA polymerase II

250
Q

Which RNA polymerase is responsible for the production of Pre-rRNA 5.8S, 18S and 28S?

A

RNA polymerase I

251
Q

Which RNA polymerase is responsible for producing Pre-tRNA and 5S rRNA?

A

RNA polymerase III

252
Q

Are promoters typically located upstream or downstream from the transcription start site?

A

Promoters are typically located upstream from the transcription start site

Example: TATA box typically found at -25 bp and CAAT box found at -75 bp

253
Q

If the template strand of DNA has the following base sequence:

3’-ATCGA-5’

What will the synthesized mRNA base sequence look like?

A

If Template: 3’-ATCGA-5’, then

(Non-template: 5’-TAGCT-3’)

mRNA: 5’-UAGCU-3’

254
Q

True or false: noncoding material is not transcribed into pre-mRNA

A

False. Coding and noncoding material is transcribed into pre-mRNA

Pre-mRNA is then processed and non-coding regions are removed by splicing.

255
Q

What DNA elements are often found in promoters for RNA polymerase II and where are they located?

A

BRE (upstream)

TATA (upstream)

INR (at transcription start point)

DPE (downstraem)

256
Q

How does RNA polymerase know which strand to attach to and which direction to transcribe?

A

RNA polymerase always moves 3’—-> 5’ on the template strand

Promoters mark the location and orientation of the gene to transcribe

The direction is determined by the asymmetry of the promoters

257
Q

Describe how TFIID initiates transcription in eukaryotes.

A

The TBP region of TFIID bnds to the TATA box which distorts the DNA backbone slightly unwinding the double helix and marks the location of an active promoter

TFIID binds to the minor groove in DNA and widens the groove which allows RNA polymerase and transcription factors to bind more closely than if the DNA was linear

258
Q

What role does TFIIH play in initiation of transcription?

A

TFIIH uses ATP to pry apart the DNA helix to expose the template strand

TFIIH also phosphorylates the C-terminal tail of RNA polymerase which causes a conformational change that allows RNA polymerase to detach from the general factors and begin the elongation phase of transcription

259
Q

How does termination of transcription differ between different RNA polymerases?

A

RNA polymerase I: needs specific termination factor downstream of transcription unit

RNA polymerase II: can terminate at multiple sites between 0.5-2 kb beyond the polyA addition site

RNA polymerase III: terminates after a series of U residues

260
Q

What enzyme facilitates pre-mRNA processing?

A

RNA polymerase II facilitates pre-mRNA processing in addition to catalyzing the synthesis of pre-mRNA

261
Q

Describe the modifications that occur on the ends of mRNA

A

5’ end: 7-methylguanosine cap; connects to 5’ terminal via a 5’ to 5’ triphosphate bridge

3’ end: poly adenine tail

262
Q

Describe the process of poly-A addition

A

A cleavage and polyadenylation specificity factor (CPSF) attaches to a cleavage sequence (example: AAUAAA) that is 10-30 bases upstream from the cleavage site

The cleavage site occurs after a CA residue

30 nucleotides downstream from the cleavage site, a cleavage stimulation factor binds to a GU-rich or G-rich sequence.

Poly-A can be added to the 3’ end of the mRNA CA-OH left following cleavage

263
Q

What are the 3 critical parts that define the boundaries of an intron?

A

1) 5’ splice site (splice donor)
2) Branch point adenine
3) 3’ splice site (splice acceptor)

264
Q

Describe the mRNA splicing reaction.

A

Consists of 2 transesterification reactions forming an excised lariat intron

The hydroxyl group from the branch point adenine attacks the 3’PO4 of Exon 1 which leaves the 3’ end of exon 1 with a hydroxyl group

The 3’ hydroxyl group of exon 1 attacks the 5’ phosphate of exon 2, thus connecting the exons and leaving a looped lariat intron detached

265
Q

What enzyme catalyzes the mRNA splicing reaction?

A

The spliceosome

A ribonucleoprotein complex made up of U1, U2, U4, U5 and U6 snRNPs

266
Q

Describe the individual roles of the U snRNPs of the spliceosome.

A

These snRNPs base pair with the mRNA transcript

U1: Binds the 5’ splice site

U2: Binds the branch point and forms part of the catalytic center

U5: Binds the 5’ splice site; then the 3’ splice site

U4: Masks catalytic activity of U6

U6: catalyzes splicing

267
Q

How can splicing mutations cause disease?

A

If the mutation inactivates a normal splice site within an exon, then cryptic splice sites will be activated leading to abnormal mRNA transcripts

268
Q

Describe how the ribosomal RNA subunits are transcribed and assembled.

A

The precursor 45S rRNA (13,000 bp long) is chemically modified and then cleaved into 3 separate rRNA strands.

The 18S rRNA is incorporated into the small ribosomal subunit

The 5.8S rRNA and the 28S rRNA are incorporated into the large ribosomal subunit (along with the 5S rRNA which is made elsewhere)

269
Q

Where are ribosomes synthesized and assembled?

A

The subunits are synthesized and assembled in the nucleolus

The small and large subunits are transported into the cytoplasm where they are assembled into complete ribosomes