Cmb Exam #1

Front 1

Why does water have a high melting and boiling point?

Back 1

Hydrogen bonds must be broken to melt ice and vaporize water.

Front 2

Heat of Vaporization of water?

Shape of water?

Back 2

Heat of Vaporization: 40 kJ/mol

(high heat of vaporization)

Shape: tetrahedryl (104.5)

Front 3

On average, each water molecule is bonded to ____ other water molecules. In ice, water forms ____ hydrogen bonds, increasing it's volume.

Back 3

Average: 3.4 other water molecules

Ice: 4 hydrogen bonds

This is why ice floats-- because its density is decreased by the increase in volume.

Front 4

Hydrogen Bond Characteristics

Back 4

• H bonded with O, N, or F
• 1/20 strength of covalent bond
• Strongest when three atoms lie in straight line

Front 5

When is a hydrogen bond strongest?

Back 5

When three atoms lie in a straight line

Front 6

Water competes with molecules for hydrogen bonds. The strength of hydrogen bonds can be increased by ________.

Back 6

Shielding the bond from the water molecules

e.g., double helix DNA where base-to-base hydrogen bonds are within helix and not exposed to water molecules.

Front 7

DNA is ______ charged. Histones are ______ charged.

They're attracted to one another in water, however the level of attraction is dependent on _______ concentration.

Back 7

DNA = Negative Histones = Positive

Depends on salt concentration. At high concentrations, there are more ions between DNA and histones, decreasing their attraction to one another.

Front 8

Amphipathic molecules

Back 8

Have a polar and nonpolar region

Front 9

Hydrophobic Effect

Back 9

Non-polar molecules forced to cluster.

• Water molecules lose a plane which they can H-bond with one another and therefore it's preferable to push the hydrophobic molecules into one area and decrease the SA as much as possible to maximize the number of H-bonds that can be made.
• This is how proteins fold and cell membranes form.

Front 10

Ionization state of a group is determined by ______ as well as ______.

Back 10

1. Dissociation constant

2. pH of solution

Front 11

Dissociation Constant of Water

Back 11

10^-14

It is small, so only a small fraction of molecule is dissociated

Front 12

Ionization State of Weak Acid and how it relates to pH

Back 12

• If pH is below pKa: protonated
• If pH is above pKa: dissociated
• If pH = pKa: half dissociated; half associated

Front 13

Because pH is a logarithmic scale:

• If pH value is 1 pH unit below pKa:
• If pH value is 1 pH unit above pKa:

Back 13

1. 90% protonated

2. 10% protonated

Front 14

Example of a strong acid and the pH

HCL --> H+ + Cl-

Back 14

The concentration of [H+] = 0.1M.

So the pH = -log(0.1) = 1, and the solution is acidic.

Front 15

Equation to determine the pH

Back 15

pH = -log [H]

Front 16

Common structure of all amino acids?

Back 16

Central carbon (alpha) bound to carboxyl and amino group.

H2N--C--CO2H

Front 17

At a pH of 7, the amino group and the carboxyl group of an amino acid are both ______.

Back 17

Ionized

It's called a Zwitterion

Front 18

All proteins are optically active because it has a central carbon with four different groups.

What is the exception?

Back 18

Glycine is the only AA that is not an optical isomer because it has an H in place of the R group.

Front 19

Are amino acids L- or D- optical isomers?

Back 19

All AAs are L-Amino Acids

Front 20

Which are the non-polar alipathic Amino Acids?

(7)

Back 20

1. Glycine
2. Alanine (CH3)
3. Proline
4. Valine (3 methyl groups)
5. Leucine (secondary butyl group)
6. Isoleucine
7. Methionine

Front 21

What is the unique characteristic of proline?

Back 21

1. Stiff and can change direction of a peptide chain.

2. Only AA with secondary amine on it that folds back on itself (all others have primary amine group).

- So the amino and carboxyl side are on one side and cannot rotate in free space like other AAs.

Front 22

Two Amino Acids with Hydroxyl Groups?

Back 22

1. Serine: essentially an alanine with an OH

2. Threonine

Front 23

Three aromatic amino acids?

Back 23

1. Phenylalanine (alanine w/ phenyl group)

2. Tyrosine (OH on end of the phenyl group)

3. Tryptophan (endol group)

Front 24

What determines the UV absorption of proteins?

Back 24

Aromatic amino acids

Front 25

Intrinsic fluorescence of proteins is mostly due to _______.

Back 25

Tryptophan

Front 26

Positively charged R groups? (3)

(BASIC)

Back 26

1. Lysine

2. Arginine

3. Histidine

• Histidine side chain dissociates at neutral pH
• Lys and Arg are in high amounts in histones

Front 27

Positive charged amino acids are acidic or basic?

Back 27

POSITIVE = BASIC

NEGATIVE = ACIDIC

Front 28

Negative charged amino acids are acidic or basic?

Back 28

POSITIVE = BASIC

Negative = Acidic

Front 29

Negatively Charged Amino Acids (2)?

(Acidic)

Back 29

1. Aspartate (aspartic acid w/ H)

2. Glutamate (glutaminc acid w/ H)

Front 30

What two amino acids have an amide side chain?

Back 30

1. Asparginine (Asn)

2. Glutamine (Gln)

Front 31

What two amino acids contain sulfur?

Back 31

1. Cysteine (Cys)

- very unstrable

- spontaneously oxidizes to disulfide bonds

2. Methionine (Met)

Front 32

Unique characteristics of Cysteine?

Back 32

1. Unstable

2. Spontaneously OXIDIZES to for disulfide bonds

Front 33

3 Modified Amino Acids

Back 33

1. Hydroxyproline

2. ɤ - Carboxyglutamate

3. O-Phosphoserine

Front 34

In a protein, only the AA side chain can ionize. Why?

Back 34

Because the carboxyl and amine are involved in the peptide bond.

Front 35

Peptide bone is formed by ______.

Back 35

Removal of water between a carboxyl and amino group. It's between the carboxyl carbon of one group and the amine of another.

• They are stable and resist hydrolysis.
• Vulnerable thermodynamically (heat)

Front 36

Typical AA Composition of Proteins

Back 36

1. All 20 AAs are not in all proteins
2. Some AAs are less common
3. 30-40% are non-polar
4. Lysine and arginine are prevalent in histones
5. Little info available from AA composition of protein

Front 37

Characteristic of Peptide Bone

Back 37

1. Partial double bond characteristic

2. Planar because carbon-nitrogen bond has partial double bond character.

3. Rigid and does not rotate around the N-C bond.

Front 38

Alpha helix protein structure is formed by?

Back 38

Rigidity of the peptide bond and hydrogen bonding is used to predict alpha helical structure.

• Chemical properties of alpha helix is determined by side chains
• One side of alpha helix is positively charged and the other side is non-polar

Front 39

Beta sheet protein structure

Back 39

Hydrogen bonding and peptide bond rigidity used to predict B-sheet structure

• Parallel or antiparallel

Front 40

Difference between physical properties of alpha helix and beta sheet?

Back 40

Alpha helix: can be stretched and absorb water

Beta sheet: cannot be stretched and does not absorb water

Front 41

Almost all proteins contain some _____ and _____ structure.

Back 41

Alpha helix and beta sheet structure

Front 42

Proline and Glycine residues often occur at bends in a peptide chain. Why?

Back 42

1. Proline forces the turn

2. Glycine is flexible due to small side chain.

Front 43

Conjugated proteins

Back 43

Contain group other than amino acids (e.g., lipoprotein contains lipid group)

Front 44

Four levels of protein structure:

Back 44

Primary: Sequence of AA and disulfide bridges

• Covalent structure of protein

Secondary: Steric relationship of AA which are close to one another (e.g., B-sheet, A-helix)

Tertiary: Steric relationship of entire chain (e.g., folding)

• 3-D conformation of peptide (e.g., alpha helix may lay on top of a B-sheet)

Quaternary: number of subunits and geometry of their packing

Front 45

In a protein (e.g., myoglobin) with many alpha helices, hydrophobic side chains occur where ________.

Back 45

The helices are in contact.

Front 46

Types of structural patterns in proteins? (4)

Back 46

1. Beta-alpha-beta loop
2. Alpha-alpha corner
3. Beta barrel
4. Twisted beta sheet

Front 47

Structures used to denature proteins? (3)

Back 47

1. Beta-mercaptoethanol

2. Urea

3. Guanidine hydrochloride

Front 48

Proteins can refold spontaneously according to their ______.

Back 48

Amino acid sequence.

Some proteins cannot refold (e.g., if there are post-translational modifications)

Front 49

Immunoglobulins on protein function and structure

Back 49

1. Specificity coded in AA sequence

2. IgG has quaternary structure and domains

3. Domains have different functions

4. Domains have same structure, evolution by gene duplication

Front 50

In contrast to most proteins, IgG ________.

Back 50

Does not have a unique amino acid sequence.

Sequence is variable in N-terminal of heavy and light chains.

Front 51

The variable amino acid sequence occurs in the ______ of the heavy and light chains of IgG.

Back 51

First domain of the heavy and light chains

The hyper-variable regions provide binding specificity.

Front 52

Operon

Back 52

Unit of DNA containing a cluster of genes under the control of a single promoter. These genes are co-transcribed into an mRNA strand.

Genes in an operon are expressed all together or not at all.

Front 53

Lac Operon (E. coli) and it's regions

Back 53

An example of an inducible set of genes. It has three genes in the operon. The regions, from left to right are:

1. Repressor Promoter
2. Repressor Protein Gene: gene for transcribing repressor
3. Promoter
4. Operator
5. Lac Z Gene: codes for beta-galactosidase
6. Lac Y Gene: lactose permease
7. Lac A Gene: don't worry about it

Front 54

Difference in DNA location for prokaryote and eukaryote

Back 54

Prokaryote doesn't have nucleus, so DNA is free floating.

Eukaryote has nucleus, so DNA is separate.

Front 55

Gene expression in prokaryotes is coupled with translation

Back 55

Transcription is tightly coupled with translation

Two Regulation Points:
1. Initiation inhibition: prevent RNA polymerase from attaching

2. Attenuation: start transcribing but stop before elongation

Front 56

Prokaryotic gene structure

Back 56

1. Core promoter: -35 to -10 region UP element

2. Promoter: DNA sequence around site of transcriptional initiation involved in binding of RNA Polymerase

3. Operator: DNA element that binds specific repressor tightly and prevents transcription

4. Activator binding sites

5. Structural Genes: usually transcribed together in a POLYCISTRONIC mRNA (encodes several proteins from one RNA molecule)

Front 57

Positive and Negative Regulation of Transcription

Back 57

POSITIVE: Induction--activator facilitates transcription initiation.

• Small molecule causes binding of regulator to DNA site.
• Small molecule causes dissociation of regulator from DNA site.

NEGATIVE: Repression--repressor inhibits transcription initiation.

• Small molecule causes binding of regulator to DNA site
• Small molecule causes dissociation of regulator from DNA site.

Front 58

Three structural genes of the Lac Operon

Back 58

1. LacZ: B-galactosidase breaks down lactose into glucose and galactose

2. LacY: Permase increases uptake of lactose from outside of cell

3. LacA: Transacetylase

Front 59

Negative Regulation of Lac Operon

Back 59

NO Lactose present

1. LacL gene (repressor gene) synthesizes repressor molecule
2. Repressor binds to operator as tetramer and prevents transcription

Note: NO absolute turnoff for the ZYA genes because LacL is not always repressing at the operon.

Front 60

Depression occurs in the presence of lactose (the inducer)

Back 60

If lactose is present, the Lac Operon is working at low levels but needs glucose to have high levels of activity.

1. Lactose converted to allolactose by B-galactosidase
2. Allolactose binds LacL repressor molecule, causing it to fall off operator.
3. RNA Pol binds to promoter and transcription begins

Front 61

Catabolic Repression

If lactose and glucose is present, the cell will ______ before the Lac operon is turned on.

Back 61

Cell will use all glucose before Lac operon is turned on.

Front 62

What molecule signals the presence or absence of glucose in the cell for the Lac Operon?

Back 62

cAMP.

When glucose is high, cAMP is low.

The Lac promoter is weak. Without CRP-cAMP binding the lacZYA promoter is weak.

Front 63

Positive Regulation of Lac Operon

Catabolite Repression

Back 63

If glucose present, catabolite repression prevents transcription even if the inducer (e.g., lactose) is present.

1. Mediated by cAMP and CRP
2. Low glucose increases cAMP levels
3. cAMP is coactivator for CRP
4. cAMP-CRP binds activator site.

Front 64

Summary of Lactose Operon

Back 64

BOTH NEGATIVE AND POSITIVE TOGETHER (with lactose and no glucose)

1. Lactose is converted to -> allolactose by β-galactosidase, repressor dissociates from operator (transcription increases 20-fold)

2. Without glucose [cAMP] increases in the cells, cAMP binds to CRP

3. cAMP-CRP binds to CRP site in lac promoter and further stimulates transcription about 50 fold to achieve 1000 fold in total.

Front 65

Regulatory Schemes of Prokaryote Gene Regulation

Back 65

Anabolic Pathway: turned on by substance it breaks down

Catabolic Pathway: turned off by product

Front 66

Tryptophan Characteristics

Back 66

1. Essential AA
2. Deficiency causes niacin deficiency. Can lead to Pellagra
3. Excess causes mental symptoms

Front 67

Difference between Tryptophan and Lac Operons

Back 67

Lac Operon: inducible

• Normally off, but can be turned on

Tryptophan Operon: repressible

• Normally on, but can be turned off
• So here, the repressor is NORMALLY OFF.

Front 68

Two ways the Tryptophan Operon is regulated

Back 68

1. Negative Regulation: high Trp binds repressor and stops transcription.

2. Attenuation: inhibition of elongation

• Four regions in LEADER sequence that can bind with one another (2&3, or 3&4)

Front 69

Explain Attentuation in Trp Operon

1. High [Trp]

2. Low [Trp]

Back 69

High [Trp] Level:

1. Ribosome moves along mRNA
2. Since [Trp] is high, ribosome moves through region 1 (which has 2 Trp codons) fast because Trp is readily available to be used by tRNA
3. This doesn't give section 2&3 time to bond.
4. Section 3&4 bond and create an ATTENUATOR STRUCTURE that causes polymerase to fall off
5. This is a form of Rho-independent termination

Low [Trp] Levels:

1. Ribosome moves along RNA
2. Since [Trp] is low, ribosome stalls at region 1. This allows polymerase to transcribe region 2, 3, and 4.
3. Region 2&3 bind together while waiting for tRNA to incorporate Trp. This is not an attenuating structure. This prevents region 3&4 from binding together.
4. Ribosome moves through with no problem.

Front 70

Difference between Trp repressor (TrpR) and Lac repressor (LacL)

Back 70

Trp binds to TrpR and makes it active. Now it can bind to operator and repress. It's normally off.

Allolactose binds LacL repressor and makes it inactive. It's normally on.

Front 71

Structure of the Trp Operon

Back 71

TrpR: repressor (unlinked to TrpE-A and has own promoter)

TrpL: leader sequence. Required for attenuation

P: Promoter for RNA Pol

O: Operator: Binding site for trpR

TrpE-A: structural genes

Front 72

Eukaryotic Gene Expression is Regulated Largely at the Transcriptional Level through what mechanisms (5)?

Back 72

1. Rate gene is transcribed: rate of Pol II activity initiated and promoter AND rate of elongation of transcript
2. Rate of RNA splicing
3. Rate of transport of RNA from nucleus to translational machinery
4. Rate of translation of mRNA into protein at ribosome.

REMEMBER: Gene transcription (mostly initiation) is the predominant regulatory mechanism.

Front 73

Effect of euchromatin and heterochromatin on transcription?

Back 73

Heterochromatin: Tightly wound and coiled. Inhibitory because transcription factors (TF) cannot gain access to DNA sequence and trigger transcription initiation.

Euchromatin: open and accessible to TF. This allows transcription to occur.

Front 74

Nucleosomes

Back 74

Composed of octomers of histones (4 dimers)

Front 75

How is heterochromatin turned into euchromatin?

Back 75

1. SWI/SNF factor is an ATPase that drives dissociation of nucleosomes, opening up the heterochromatin.

Front 76

Structure of SWI/SNF Complex

Back 76

Has a CENTRAL ATPase subunit in the complex. This can be one of two proteins (but not both):

1. BRG-1
2. BRM

SWI/SNF complex integrates:

1. Many cell signals from cell and
2. Chromatin structure/gene transcription programs

Front 77

Post-translational modification of histones and two types of epigenetic marks?

1. Repression mark
2. Activation mark

Back 77

A histone PTM can be repressive or activating for chromatin structure--depends. These are called EPIGENETIC MARKS. They occur in the AMINO-TERMINAL TAIL.

Lysines: can be acetylated or methylated

Serines and threonines: phosphorylated or ubiquinated

REPRESSION MARK: tri-methylation of lysine #9 on histone #3 (H3K9Me3)

• promotes heterochromatin

ACTIVATION MARK: acetylation of lysine #8 on histone 3 (H3K8Ac)

• promotes euchromatin

Front 78

DNA is negatively charged and histones are positively charged. This is how they stay together.

How does acetylation effect this?

Back 78

Acetylation of LYSINES block positive charge, disrupting interaction with negative phosphate backbone of DNA and promotes euchromatin.

Histone Acetylations (HACs) add acetyl groups

Histone Deacetylations (HDACs) remove acetyl groups.

Front 79

Difference between transcription factor and coregulator?

Back 79

Transcription Factor: bind directly to DNA at specific sequences.

• Binding is required for initiation of transcription, but not always sufficient. Sometimes the TF must also be activated.

Coregulators: DO NOT bind to DNA

• bridge connection between TF and the chromatin modifying protein (e.g., HAT, HDAC)

Both regulate transcription

Front 80

Examples of Transcription Factors (from the eukaryotic gene regulation lecture)

Tata-Binding Protein (TBP

Back 80

TBP binds to TATAA box and serves as landing pad for RNA Pol II next to transcription initiation site.

Front 81

Examples of Transcription Factors (from the eukaryotic gene regulation lecture)

AP-1 Protein

"TRE Response Element"

Back 81

Binds to consensus sequence TGATCA.

AP-1 is a heterodimer of protooncogene products FOS and JUN. It targets genes by:

• Being quickly synthesized in response to growth signaling pathways
• Phosphorylation on the JUN subunit

Front 82

Examples of Transcription Factors (from the eukaryotic gene regulation lecture)

Vitamin D Receptor (VDR)

Vitamin D Response Element (VDRE)

Back 82

Heterodimer of Vitamin D Receptor (VDR) and Retinoid-X-Receptor (VDR/RXR) is a transcription factor.

It is activated by the ligand Vitamin D (Calcitriol).

Front 83

Eukaryotic RNA Polymerase II Promoter:

Elements can function both independently and cooperatively.

Back 83

Independently

Cooperatively: transcription factors form more highly active complexes when simultaneously bound to adjacent response elements.

Front 84

Four Superclass of Transcription Factors

1. TF Class

2. Examples

Back 84

Based on their DNA-binding motifs

Superclass I: Leucine Zipper Factors (bZIP)

• AP-1: FOS and JUN
• CREB (cAMP response element binding protein)

Superclass II: Zinc-Coordinating DNA Binding Domain

• Nuclear receptors: estrogen receptor, retinoic acid receptor (RAR)

Superclass III: Helix-Turn-Helix

• Homeodomain proteins (HOX genes)

Superclass IV: Beta-Scaffold Factors with Minor Groove Contacts

• STAT Family
• NF-kB/IkB

Front 85

Superclass I

Leucine Zipper (bZIP)

Back 85

Two subunits. Hydrophobic interactions between to Leucines hold then together and form "zipper". The basic region interacts with DNA.

Examples:

1. AP-I: FOS and JUN
2. Cyclic AMP Response Element Binding Protein (CREB): cAMP/PKA pathway

AP-1 is a heterodimer of FOS and JUN proteins. AP-1 binds TREs (TPA response element). TPA activates phosphorylation cascade, which ends by activating JUN.

Front 86

AP-1 is a transcriptional effector of MAPK signaling. How does this work?

Back 86

Mitogens/Growth factors lead to activation of AP-1 and transcription of TRE-containing genes (e.g., c-fos gene) by initiating the mitrogen-activate protein kinase (MAPK) signaling cascade.

1. Induction of c-fos expression via transcription ERK is the MAPK that phosphorylates TCF, which with SRF and EVI-1, activates c-fos transcription.

2. Newly synthesized FOS protein combines with JUN, followed by phosphorylation of dimer by same cascade, binding to AP-1 elements (TREs)--induction of transcription.

Front 87

Superclass II

Zinc Coordinating DNA Binding Domain

Nuclear Receptors

Back 87

Nuclear receptors (NRs) are prototypes of Zn-Coordinating DNA Binding TFs. Can be heterodimer or homodimer.

All heterodimer NRs have common RXR unit, with other subunit providing unique activity.

NRs (and all other TFs) have two functional domains:

1. DNA Binding Domain
2. Transcription Activating Domain

Most NRs also have a Ligand-Binding Domain that induces a conformational change that opens the DNA-binding interface of the molecule.

Activating domains in NRs serve as interaction sites for other transcription activating proteins that form the bridge to RNA pol II at the start.

Front 88

Zinc Finger Motifs

Back 88

Zn confers alpha helices (with connecting beta strands) that insert into the MAJOR groove of DNA for nucleotide-specific binding.

All members of SUPERCLASS 2 use this basic Zn finger motif for DNA binding.

Front 89

Superclass III

Heliz-Turn-Helix Transcription Factors

Back 89

Includes Homeodomain TFs (HOX), Forkhead TFs (FOX), and ETS factors.

Helix-turn-helix motif provides DNA binding interface of 3 alpha helices that bind to major groove of DNA.

Front 90

Superclass IV

Beta-Scaffold Factors w/ Minor Groove Contacts

Back 90

Examples: p53 tumor suppressor, STAT family

Note, they contact the MINOR and major groove of the DNA.

Front 91

What are the two obligatory domains of all transcription factors (TFs)?

Back 91

1. DNA binding domain

2. Transcription-activating domain

Front 92

Nuclear Receptors (NRs) use recruitment of _______ to promote assembly and activation of the RNA Pol II initiation complex.

Back 92

Mediator complex.

Mediator is last part of the bridge. It extends from regulatory DNA element to the RNA Pol II.

• Recruits RNA Pol II and other TFs

Front 93

Binding of an activating transcription factor (TF) to its target DNA sequence elicits three kinds of transcription-activating machinery. They are?

Back 93

1. SWI/SNF

2. HAT complexes

3. Mediator

Front 94

In eukaryotes, _____ domain is the kinase that phosphorylates the C-terminus of Pol II. This is the last step in activating the Pol II to begin transcription.

Back 94

CDK-8 domain

Front 95

Alpha Helix

Back 95

Peptide chain wrapped into a helix and hydrogen bonded to itself.

Front 96

Beta Sheet

Back 96

Peptide chains are in extended line conformations. Adjacent chains are hydrogen bonded to each other.

Front 97

Collagen Helix

Back 97

Linear helix created by three peptide chains wrapped around each other.

Front 98

Beta Turns

Back 98

Conformations adapted when a peptide chain has to sharply change directions in a folded protein.

Front 99

A zymogen is

Back 99

Self-inhibiting

Front 100

Enzyme Kinetics

Pre-Steady State Burst

Back 100

Intermediate lasts on order of milliseconds of seconds. Initial burst is evidence of an ACYL INTERMEDIATE in the reaction

Presteady state burst continues until product to enzyme ratio is 1:1 and then it begins a steady state reaction. When [P]/[E] > 1, RXN is in steady state.

Front 101

Serine Protease contain a CATALYTIC TRIAD comprised of what three amino acids?

Back 101

Serine

Histidine

Aspartate

Front 102

Active site of an enzyme has residues important for what?

Back 102

1. Chemistry of RXN

2. Binding of substrate

Front 103

Binding Residues in R1 and R2 position for:

1. Trypsin

2. Chymotrypsin

3. Thrombin

Back 103

Trypsin:

R1: K, R

R2: --

Chymotrypsin:

R1: F, W, Y

R2: --

Thrombin:

R1: R

R2: P

Front 104

Drug inhibitor design is based on _______.

Meaning of a

• High Ki
• Low Ki

Back 104

Shape complimentarity.

Low Ki: only a little bit of drug needed to get 50% inhibition (GOOD)

High Ki: takes a lot of drug to get to 50% inhibition (BAD)

Front 105

Steps to make an inhibitor protein (3)

Back 105

1. Computer search of shape
2. Proposed ligands made based on structure and mechanism
3. Test the Ki of ligand

Low Ki: keep it

High Ki: discard

Front 106

A common combination of amino acids that should be included in drug design?

Back 106

1. Proline

2. Arginine

"it didn't take a computer to figure that out"

Front 107

Thrombin PPACK

Back 107

PPACK
Arginine, Proline, Phenylalanine

Catalytic Triad:

1. Serine 195
2. Histidine 57
3. Aspartate 102

Front 108

A peptide bond cannot be hydrolyzed without the help of _____.

Back 108

An enzyme (e.g., Serine Protease)

Front 109

Hydrolysis of a peptide bone by serine protease

Back 109

1. Hydrolysis of peptide bond

2. Hydrolysis of bond with serine protease, releasing the enzyme.

Enzymes must always be released--not used up during reaction.

Front 110

Summary of how an enzyme helps get from substrate to product.

Back 110

Enzyme must be complementary to transition state to work. If enzyme is complementary to substrate, transition state cannot form and RXN won't occur.

The order of the reaction is:

1. Enzyme [E] + Substrate [S]
2. Enzyme-Substrate [ES] Complex
3. Transition State [ⱡ]
4. Enzyme-Product [EP] Complex
5. Enzyme [E] + Product [P]

Transition state stabilization:

• Bent stick within the enzyme
  • Negative charge on transition state [ⱡ] of substrate attracted to positive charge on the enzyme.

Front 111

Difference in Delta G (free energy) for catalyze vs. uncatalyzed reaction

Back 111

Delta G for uncatalyzed reaction: much higher

Delta G for catalyzed reaction: low

Front 112

What can proteins do to stabilize the transition state (4)?

Back 112

1. Make H+ bonds

2. Ionic interactions (neutralize charge)

3. Hydrophobic interactions

4. Chemistry

Front 113

Define an acid and base

Back 113

ACID:

1. Proton donor
2. Electron acceptor

BASE:

1. Proton acceptor
2. Electron donor

Front 114

Linear order of catalytic triad

Back 114

Serine-->histidine-->aspartate

Front 115

General base catalysis

Back 115

General Base Catalysis

• The electrons from the histidine pull the hydrogen (proton) away from the oxygen, making the oxygen more reactive (more like hydroxide) as it reacts with the carbonyl carbon
• The aspartate aims the histidine towards the "hydroxide" of the serine

Front 116

What is the pKa of:

• -COOH R-group of Glu and Asp
• -NH R-group of His

Back 116

Glu, Asp: 4.5

His: 6.5

Front 117

Steps of Serine Protease Function (8)

Back 117

1. General base catalysis
2. General acid catalysis
3. Rearrangement of breaking of the peptide bond
4. Product #1 is released
5. Hydrolysis (water enters active site)
6. Start of deacylation
7. General acid catalysis (His-57) and breaking of acyl enzyme intermediate bond to Ser-195
8. Product #2 is released

Front 118

Serine Protease Function: Step 1

General Base Catalysis

Back 118

The shape and charge of the transition state is different than the shape and charge of the ES.

Peptide bond shape: planar

Transition state shape: tetrahedral

The two dashed lines represent two hydrogen bonds that stabilize the extra negative charge in the transition state. This is worth 4 orders of magnitude.

• Enzyme really likes the tetrahedral shape and the hydrogen bonds provides hydrogen bonds from two backbone protons, so it stabilizes the charge and the shape.
• So here we have had general base catalysis and transition state stabilization (8 orders of magnitude of rate acceleration alone).

Front 119

Serine Protease Function: Step 2

General Acid Catalysis

Back 119

Proton is added to the N of the serine, making it (+) charged.

The (-) charge of the oxygen is now removed and the bond between the carbon and the positively charged N is broken

Front 120

Serine Protease Function: Step 3

Rearrangement of Breaking of Peptide Bond

Back 120

Rearrangement of breaking of the peptide bond

Front 121

Serine Protease Function: Step 4

Product #1 is Released

Back 121

Product 1 (P1) is released

• Half of the broken stick is stuck to the enzyme and the other half is free to float away.
  • The part stuck to the enzyme is the ACYL ENZYME INTERMEDIATE (EI) which hangs around for a long time.
• This is the end of the initial burst. After this, we're starting to get into the steady state and we are now starting to find the rate limiting step.

Front 122

Serine Protease Function: Step 5

Water Enters Active Site (Hydrolysis)

Back 122

Water enters the active site (hydrolysis)

• This is the steady state rate limiting step
• Hydrolysis is slower than the initial serine attack because it's not as easily oriented because serine is stuck in there. Water is also not as good of a nucleophile.

Front 123

Serine Protease Function: Step 6

Start of Deacylation

Back 123

Start of Deacylation

• General base catalysis and formatting of 2nd tetrahedral enzyme transition state

Front 124

Serine Protease Function: Step 7

General Acid Catalysis (His-57) and Breaking of the Acyl Enzyme Intermediate Bond to Ser-195

Back 124

General acid catalysis (His-57) and breaking of the acyl enzyme intermediate bond to Ser-195

Front 125

Serine Protease Function: Step 8

Product #2 is Released

Back 125

Product #2 is released.

Front 126

Serine Protease Transition State Analogues (2)

Back 126

Bind 2-3 times orders of magnitude better than substrate because enzyme wants to entice/stabilize transition state.

1. Phosphodiester
2. Boron complex

(Tetrahedral negatively charged shape)

Front 127

Parts of a Free Energy Diagram

Back 127

Note, delta Gcat is much lower than delta G uncat.

Also note location of the transition state.

Front 128

First Rate law for first and second order

Back 128

First Order Rate Law: (s^-1)

v = k[S]

Second Order Rate Law: (M^-1s^-1)

v = k [S1][S2]

Front 129

Units of first and second order rate laws?

Back 129

First order: s^-1

Second order: M^-1s^-1

Front 130

First order equation

Back 130

h: Plancks' constant

K: Boltzmann constant

T: Temperature

R: Gas constant

Units: s^-1

*Lowe the delta G, higher the rate

Front 131

In first order equation, how does delta G and the rate (k) relate?

Back 131

Lower the delta G, higher the rate (k)

Front 132

What does rate (k) in enzyme kinetics represent?

Back 132

How many micromolar of substrate can get transformed per minute.

Front 133

Reasons why you want to get an assay as soon as possible?

Back 133

They level off:

1. Product inhibition
2. Substrate depletion
3. Enzyme inactivation

Front 134

Methods for measuring enzyme rate? (6)

Back 134

1. Spectrophotometric methods
2. Fluorescent methods
3. Manometric methods (gas pressure)
4. Electrode methods
5. Polarimetric methods (optical isomer)
6. Sampling methods

Front 135

You can only use the Michaelis-Menten equation to get Vmax if _________.

Back 135

If the graph is a hyperbolic curve.

Front 136

Units of Km (@ 1/2 Vmax) and units of Vmax?

Back 136

Km (@ 1/2 Vmax): = Millimolar (mM)

(note this is the substrate concentration)

Vmax = Micromolar/minute (uM/min)

(note this is the rate)

Front 137

Be careful, doesn't equal 1/2 Vmax because the units wouldn't be correct.

Back 137

Km is the [substrate] that gives 1/2 Vmax.

Front 138

Dependence of Initial Velocity (Vo) on [Substrate]

Back 138

Note, at low [S] we can eliminate [S] from the equation because it's negligible.

Note, at high [S] we can eliminate Km from the equation because it's negligible.

• Now, Vo = Vmax (the two [S] cancel out)

Front 139

Using a graph to determine Vo and Vmax

Back 139

Note, at Vo the substrate is essentially zero, so [S] is eliminated from the bottom of the equation.

You get Vo from the data.

At high [substrate], note that Vo = Vmax

Front 140

Steady State Kinetics

Back 140

Rate of ES formation is equal to the rate of ES breakdown

• [S]total >> [E]total and [S] = [S]total
• So the change in ES over time is zero

As soon as S leaves the ES complex, a new S immediately comes into the spot.

• Waiting for product to form (k2 or kcat) is really slow.

Front 141

Remember Km is the substrate concentration to give 1/2 Vmax. It is not always the binding constant.

Back 141

Remember Km is the substrate concentration to give 1/2 Vmax. It is not always the binding constant.

Front 142

Association constant and dissociation constant using k values.

Back 142

k1/k-1 = association constant

k-1/k1 = dissociation constant

Front 143

Initial rate (Vo) is proportional to the rate limiting step (k2) and the amount of substrate present (ES)

Back 143

Vo = K2 [ES]

Initiate rate (Vo) is proportional to rate limiting step (k2 or kcat) and the amount of substrate present [ES]

Front 144

Memorize this equation:

Vmax = k2 [Et]

k2 = rate limiting step

[Et] total enzyme concentration

Back 144

Vmax = k2 [Et]

So if you know Vmax and k2, you'll be able to determine the amount of [Et] in the blood.

k2 is found experimentally

Front 145

When you flip over a hyperbolic curve, you get a _________.

This changes the Michaelis-Menten equation used with the hyperbolic curve to the _______, which is used with new graph.

Back 145

Double Reciprocal Plot (straight line)

Lineweaver-Burk Equation

Front 146

Michaelis-Menten Equation

Back 146

V1 should really be Vo.

Front 147

Lineweaver Burk Equation

Back 147

This is in the form of Y= mx + b

Slope: Km/Vmax

y-int: 1/Vmax

x-int: 1/km

Front 148

Comparing high and low km values

Back 148

Low km = HIGH Affinity

• Faster reaction at lower [S]
• Takes less time to reach 1/2 Vmax

High km = LOW affinity

• Slower reaction at lower [S]
• Takes longer to reach 1/2 Vmax

Front 149

General Equation for MM and enzyme kinetics

Back 149

You can get Kcat, Km, and Vmax from steady state kinetics

Front 150

Not all enzymes follow the Michaelis-Menten Enzyne Steady State Kinetics.

Anything that's not a perfect hyperbolic curve cannot use MM equation

Back 150

Allosteric enzymes show a sigmoid curve (or anything other than hyperbolic)

It can bind via either a cooperative process or anticooperative process

Front 151

Allosteric enzymes

Back 151

• Catalytic and regulatory subunit
• Modulator
• Binds at regulatory site which activates catalytic site

Heterotropic modulator

Homotropic modulator:

• Hgb is an example
• 4 subunits: each subsequent molecule binds tighter
• Substrate acts as the modulator

Front 152

Types of inhibition (2)

Back 152

1. Irreversible inhibition (suicide substrates)

• Mimic substrate and/or transition state
• Covalently attack an enzyme functional group

2. Reversible Inhibition

• Competitive
• Non-competitive
• Uncompetitive

Inhibitors can have linear, hyperbolic, or parabolic effects

Front 153

Competitive Inhibitor on Lineweaver Burke Plot

Back 153

• Vmax = same despite amount of inhibitor
• Km = increase

MORE SUBSTRATE has an effect

• Add enough substrate, you can overcome inhibitor and achieve Vmax

Front 154

Non-competitive Inhibitor on Lineweaver Burke Plot

Back 154

• Vmax changes---Vmax is lowered.
• Km stays the same
• Substrate has NO effect, so no Km effect.

Front 155

Non-Competitive Inhibitor Site vs. Allosteric Binding

Back 155

Non-Competitive Inhibitor Site:

• Effects residues involved in catalysis
• Does not effect residues involved in binding

Allosteric Binding:

• Effects binding

Front 156

Uncompetitive Inhibitor on Lineweaver-Burke Plot

Back 156

• Vmax and Km lowered equally
• No change in slope

Inhibitor binds ONLY to the ES complex. It cannot bind in the absence of substrate.

Front 157

On Lineweaver Burke Plot, how do you find Vmax?

Back 157

Do 1/Y-intercept. This is the same as 1/Vmax.

Front 158

Equation to determine concentration from spectroscopy?

Back 158

A = εbC or C = A/εb

where, ε = 1.4 x 10^7 M^-1 cm^-1

b = 1 cm

C is the concentration of substrate

Front 159

Impact of genetic disorders

Back 159

3-7% Dx with genetic disorder

• 3% of newborns have significant disorder
• Account for 10% of pediatric admissions in U.S.

Front 160

Define:

1. Gene
2. Locus
3. Mutation

Back 160

Gene: functional and physical unit of heredity passed from parent to offspring

- hereditary unit: molecularly a sequence of DNA required for the production of a functional product.

Locus: position of a gene on a chromosome

Mutation: change in DNA sequence

- permanent, heritable

Front 161

Types of Mutations (outcomes)

(4)

Back 161

1. Loss of function: reduction or loss of Fx

- often recessive phenotypes

2. Gain of function: increase in NORMAL Fx

- often dominant phenotypes

3. Novel property: confers new property/Fx

- often cancer, rearrangement

4. Dominant-negative allele: disrupts Fx of a wild type allele in same cell

- frequently dominant phenotypes

Front 162

Information about human genome

Back 162

Genome: all DNA in organism or cell

• Around 3.4 billion base pairs
• Around 21,000 genes
• Average gene size: 3,000 bp
• Only about 5% codes for proteins

Front 163

Define:

1. Allele
2. Polymorphism

Back 163

Allele: alternative form of genetic information at a particular locus

Polymorphism: at least two relatively common alleles at a locus (e.g., blood type)

Front 164

Define:

1. Genotype
2. Phenotype

Back 164

Genotype: individuals genetic makeup--his or her DNA sequence at given locus

Phenotype: OBSERVABLE expression of a genotype

Front 165

Pedigree symbols

1. Twins
2. Divorce
3. Miscarriage
4. Unknown Sex

Back 165

Twins

Shown as a carrot

If line connects carrot, they're IDENTICAL

Divorce

Double lines through connection

Miscarriage

Triangle

Unknown Sex

Diamond

Front 166

Single Gene Traits/Disorders

Back 166

Traits that are determined by alleles at a single locus.

Front 167

Define:

1. Dominant
2. Recessive
3. Codominant

Back 167

Dominant: phenotypically expressed in heterozygotes

Recessive: phenotypically expressed only in homozygotes for mutant allele

Codominant: alleles that are both expressed when they occur together (e.g., ABO blood groups)

Front 168

Define:

1. Homozygous
2. Heterozygous
3. Compound Heterozygote

Back 168

Homozygous: identical alleles at a given locus

Heterozygous: different alleles at a given locus

Compound heterozygote: two different mutant alleles at a given locus

Front 169

Degree of relationships:

1st, 2nd, and 3rd degree

Back 169

1st Degree: parents, siblings, offspring

2nd Degree: grandparents, aunts, nieces

3rd Degree: first cousins, etc.

Front 170

Pedigree Pattern of

Autosomal Dominant Inheritance

Back 170

• Males and females equally affected
• Vertical transmission: involves every generation (no skipping)
• Around 50% of offspring affected
• Unaffected individual doesn't transmit trait
• e.g., Achondroplasia (FGFR3 mutation)

Front 171

Factors complicating pedigree analysis:

1. New mutations
2. Germline mosaicism
3. Penetrance
4. Variable expressivity

Back 171

New mutation: frequent cause for affected individual with no family history (AD)

Germline Mosaicism: mutation in a germline in one parent

Penetrance: proportion of individuals with a disease genotype who express the disease phenotype. If it's not 100%, it's reduced.

Variable Expressivity: severity of disease may vary greatly. Extent of expression of phenotype.

Front 172

Factors complicating pedigree analysis:

1. Pleiotropy
2. Genetic Heterogeneity
3. Linkage

Back 172

Pleiotropy: when a gene has multiple, seemingly different unrelated, phenotypic effects

Genetic Heterogeneity: similar phenotype caused by different genotypes

• Locus heterogeneity: mutations at different loci
• Allelic heterogeneity: different mutant alleles at same locus
• Clinical heterogeneity: different clinical phenotypes caused by mutations in same gene

Linkage: co-inheritance of two or more non-allelic genes at nearby loci

Front 173

Homozygous Autosomal Dominant Genotype

Back 173

Typically much more severe than heterozygote

Consider in rare situation where two heterozygotes mate

Front 174

Examples of genetic disorders with delayed age of onset

Back 174

Huntington disease

Myotonic dystrophy

Familial Alzheimer disease

AD breast CA

AD Parkinson disease

Front 175

Examples of Autosomal Dominant Conditions

Back 175

Marfan syndrome

Achondroplasia (dwarfism)

Familial (early) Alzheimer Disease

Huntington Disease

Familiar Hypercholesterolemia

Familial Breast CA (BRCA 1 or 2)

Front 176

Pedigree pattern of Autosomal Recessive Inheritance

Back 176

• Horizontal transmission: disease in siblings, but usually not in earlier generations
• 25% (1/4) recurrence risk
• Males and females equally affected
• Increased consanguinity in pedigree

Front 177

Examples of Autosomal Recessive Conditions

Back 177

Sickle cell disease

Cystic fibrosis

Tay-Sachs disease

Hemochromatosis

Phenylketonuria (PKU)

Thalassemias

Front 178

X-Linked Recessive Inheritance Pattern

Back 178

• Males affected more than females
• 100% of daughters of affected males are carriers
• 50% of sons of carrier females are affected (hemizygous) and 50% of daughters are carriers
• NO FATHER-TO-SON transmission
• May be transmitted through series of carrier female

Front 179

Lyon Hypothesis

Back 179

One X chromosome in each somatic cell is randomly inactivated early in embryonic development.

• Random, fixed, and incomplete

Ensures dosage compensation

Front 180

X-Linked Recessive Example Disorders

Back 180

Duchenne muscular dystrophy

Hemophilia A

Color blindness

Glucose-6-phosphate dehydrogenase (G6PD) deficiency

Front 181

X-Linked Dominant Inheritance Pattern

Back 181

• Affected fathers: 100% of daughters affected and NO sons affected.
• Affected mothers: 50% of sons and 50% of daughters affected
• No male-to-male transmission
• Twice as many females with disorder as males (unless disease is lethal in males)
• Expression in females usually less severe

Front 182

X-Linked Dominant Condition Example

Back 182

Hypophosphatemic (Vit-D resistant) Rickets

Incontinentia Pigmenti Type I

Front 183

Other factors to consider in genetic patterns

1. Sex-limited traits
2. Sex-influenced traits
3. Phenocopy

Back 183

Sex-limited traits: autosomal genes expressed in only one sex

Sex-influenced traits: autosomal genes where the same genotype is expressed in different frequencies, depending on sex

Phenocopy: phenotype produced by environmental factors that mimics a genetically determined trait.

Front 184

Define:

1. Polygenic
2. Multifactorial

Back 184

Polygenic: Inheritance of a trait determined by many genes at different loci--each with a small additive affect.

Multifactorial: inheritance of a trait determined by combination of genetic and environmental factors. Complex inheritance.

Front 185

Qualitative vs. Quantitative Trait

Back 185

Qualitative: individual either has it or they don't (e.g., pyloric stenosis)

Quantitative: trait with measurable quantity that differs among individuals.

• Measurable on a continuous numerical scale
• e.g., BP, height, cholesterol
• Variations due to differences in genotype AND environmental factors
• Follows NORMAL distribution in population
• Abnormal is extreme variant of normal range

Front 186

Threshold Model

Back 186

You either have it or you don't (Yes or No)

• Assumes underlying continuous variation in liability for the disease but no clinical effect until exceeds a threshold.
• If frequency of sexes are different, sexes may have different thresholds.
• Child of the LEAST frequently affected sex (high threshold) have a higher risk

Front 187

Multifactorial (Complex) Inheritance

Back 187

Cause most common adult conditions and congenital malformations

1. Are not single gene disorders and don't demonstrate Mendelian pattern of inheritance
2. Familial aggregation because family members are more likely to have genes in common w/ affected person than unrelated person.
3. More common in close relative. Recurrence risk rapidly decreases in distant relatives
4. Recurrence risk is higher if more than one family member is affected.
5. More severe the disease, higher the recurrence risk
6. Consanguinity increases risk

Front 188

Recurrence risk in Multifactorial (Complex) Inheritance

Back 188

Risk to 1st degree relative: square root of population risk

Recurrence risk is higher if > 1 family member affected.

• One child affected: recurrence risk is 1-5%
• 2nd Child affected: recurrence risk is 6-10%

Front 189

Examples of multifactorial malformations

Back 189

1. Cleft lip with or without cleft palate
2. Neural tube defects
3. Congenital dislocation of hip
4. Congenital heart defect
5. Pyloric stenosis

Front 190

Empiric risk

Back 190

Based on observational data from many studies

• Best estimate of recurrence risk

Front 191

Probability Calculations

"and" vs. "or"

Back 191

And: probability of two independent events (A and B) both occurring. Multiply.

p(A) x p(B) e.g., 1/2 x 1/2 = 0.25 (1/4)

-------------------------------------------------------------

Or: probability of either of two mutually exclusive events occurring. Addition.

p(A) + p(B) e.g., 1/4 + 1/4 = 0.5 (1/2)

Front 192

Hardy-Weinberg Law

(and both equations)

Back 192

In absence of influences, allele and genotype frequency will remain constant from one generation to the next.

p + q = 1 p^2 + 2pq + q^2 = 1

• Determines allele frequency and carrier frequency when incidence of disorder is known
• Relates gene frequency to genotype frequency

Front 193

Hardy-Weinberg Law Summary of Steps (4)

Back 193

1. Determine genotype frequency (count people)

e.g., 100 people: 64 are AA; 32 are Aa; 4 are aa

2. Count genes (remember, two genes per person)

• 100 people, so 200 alleles
• A: (2)64 +32 = 160 a: 32 + 2(4) = 40
• 160 + 40 = 200 total

3. Calculate gene frequency

• Remember, out of 200 total genes/alleles
• p = 160/200 = 0.8 q = 40/200 = 0.2

4. Calculate genotype frequency

• p^2 = 0.8^2 = 0.64
• 2pq = (2)(0.8)(0.2) = 0.32
• q^2 = 0.2^2 = 0.4

Front 194

Hardy-Weinberg Principle

Assumptions (5)

Back 194

1. Infinitely large population so no random fluctuation
2. No migration or emigration
3. No mutation
4. Random mating with respect to locus
5. No selection for one gene over another

Front 195

Genetic Drift

Back 195

Chance fluctuation of allele frequencies over time in a small population

Front 196

Founder effect

Back 196

In small population, distribution of genes can be determined by founders, the people at the top of the pedigrees that started the kindred or population.

Front 197

Gene flow

Back 197

Slow diffusion of genes across a barrier (physical or cultural)

Front 198

Define

1. Assortative mating
2. Fitness

Back 198

Assortative Mating: non-random mating based on phenotype

Fitness: measured by number of an individual's offspring; depends on differences in survival and fertility.

Front 199

Fitness differences in Dominant vs. Recessive

Back 199

• Selection and mutation are more obvious for dominant trait

• Selection occurs in hemizygous males for X-linked recessive trait, but not for females

• Heterozygote Advantage: heterozygotes have increased fitness over both genotypes (e.g., sickle cell disease

Front 200

Autosomal Recessive Disease

Hardy-Weinberg Information

Back 200

RULE: frequency of mutant allele (q) is the square root of the disease incidence (q^2)

• Carrier (2pq) much more common than affected (q^2)

e.g., disease incidence (q^2) is 1/40,000

• q = 1/200 so p ~ 1 (199/200)
• 2pq = 2(1)q = (2)(1/200) = 1/100

RULE: carrier frequency is 2q (really 2pq, but p=1)

Front 201

Autosomal Dominant Disease

Hardy-Weinberg Information

Back 201

RULE: frequency of mutant allele is about 1/2 the incidence of the trait.

• Homozygotes for autosomal dominant conditions are VERY rare (usually fatal) and can be ignored during calculation.
• Since p^2 is very small (rare), p is very small AND q is ~1.

e.g., If disease incidence is 1/500

1/500 = 2pq = 2p and p = 1/1000

Front 202

X-Linked Recessive Disorders:

Hardy-Weinberg Information

Back 202

Males are hemizygotes (XY), so disease frequency equals gene frequency

• disease frequency = q (not q^2)
• Disease frequency (males) = q

Females have two genes (XX), so Hardy-Weinberg can be used to calculate carrier frequency.

• Carriers in females: 2pq Afflicted: q^2

Front 203

X-Linked Dominant Disorders:

Hardy Weinberg Information:

Back 203

Incidence of affected males = q

Incidence of affected females = 2(p)(q)

• Females are twice as likely to be affected

Front 204

Bayes Analysis

Back 204

• Adjusts PRIOR risk by considering additional information

• Provides modified or posterior risk

Front 205

Define genetic terms:

1. Presymptomatic
2. Predisposition

Back 205

Presymptomatic: eventual development of disorder is certain if mutation present

Predisposition: eventual development of symptoms is likely, but not certain, in presence of mutation.

Front 206

Breast Cancer

Back 206

• 5-10% is familial

• These are frequently caused by mutations in BRCA1 and BRCA2 genes

• If carry one copy of mutation (autosomal dominant), increased risk for developing breast cancer).
• Also risk for ovarian and prostate CA

Front 207

BRCA1 and BRCA2

Lifetime Cancer Risks

Back 207

BRCA1 Mutation

• 50-85% Breast CA
• 40-60% Ovarian CA

BRCA2 Mutation

• 50-85% Breast CA
• 15-27% Ovarian CA
• Melanoma risk

Both have risk for prostate, pancreatic, and male breast CA.

Front 208

Most common genetic cause of mental retardation and birth defects?

Back 208

Chromosome abnormality

Front 209

Each DNA molecule has been packaged into a mitotic chromosome that's ______ fold shorter than it's length.

Back 209

10,000

Front 210

Chromosome Abnormalities

1. Constitutional
2. Acquired

Back 210

Constitutional: present at birth, usually at conception of shortly thereafter. Leading cause of pregnancy loss and retardation.

Acquired: develop in somatic cells; associated with cancers. Can be solid or fluid.

1. Solid changes: not very specific
2. Fluid changes: very specific

Front 211

Mammalian Cell Cycle

Interphase

Back 211

During interphase, DNA is decondensed. Cell spends significant amount of time here.

(n = haploid chromosome; c = # of chromatids)

G1 Phase: interval between mitosis and replication

• 12-24 hours
• Cell growth and preparation for DNA replication
• 2n, 2c (46, each chromosome with 1 chromatid)

S Phase: DNA replication

7 hours

• DNA replication and CHROMATID duplication
• 2n, 3c

G2 Phase: interval between S and mitosis

• 4-6 hours
• Growth and DNA repair
• 2n, 4c

Front 212

Steps in Mitosis

PMAT

Back 212

Prophase:

• Chromatin condenses into chromosomes
• Nuclear envelope disapears

Metaphase:

• Chromosomes align at metaphiseal plate

Anaphase:

• Sister chromatid separate
• Centromeres divide

Telophase:

• Chromatin expands
• Cytoplasm divides
• Two daughter cells await cytokinesis

Front 213

Three types of chromosome structures (3)

Back 213

1. Metacentric

2. Submetacentric

3. Acrocentric

Front 214

Metacentric Chromosome

Back 214

Two sister chromatids connected in middle by a centromere.

• Also call the two arms short and long even though they're the same size.

Front 215

Submetacentric Chromosome

Back 215

Centromere that is closer to one side than the other.

This is how most human chromosomes look.

Short arm = p Long arm = q

(think, "petite")

Front 216

Acrocentric Chromosome

Back 216

Have a unique short arm side.

• 5 pairs in humans: 13, 14, 15, 21, 22
• Include satellite and stalk region
• Short arm don't have coding genes.
• Only have repetitive sequence to code ribosomal RNA. They're the DNA sequence to code ribosomal RNA--called NUCLEI ORGANIZING REGIONS (NOR).
• Very polymorphic--difference lengths of short arms are present in different in different normal individuals
• Not associated with any abnormal phenotypes because they don't have coding genes

Front 217

What are the acrocentric chromosome (5)?

Back 217

Chromosome #: 13-15, 21, and 22

Front 218

Short arm of acrocentric chromosomes don't have _________.

They do have _________ that code for ______. These are called _______.

Back 218

DO NOT have coding genes.

They do have repetitive sequences to code ribosomal RNA.

These are called Nuclei Organizing Regions.

Front 219

Mitotic Nondisjunction

Back 219

Mitotic error that results in: trisomy, monosomy, and normal chromosome distribution.

3 Cell Lines:

1. Trisomy (13, 18, 21 can survive)
2. Monosomy (fatal)
3. Normal distribution

Front 220

Mosaicism

Back 220

Coexcistence of two different cell lines

Trisomy cell line (from non-disjunction) coexists with the normal mitotic cell (disomy)

• Mosaic 47/46 chromosomes, meaning some have 47 and some have 46.

Front 221

Meiosis

Back 221

Comprised of ONE ROUND of DNA replication but TWO cell divisions (M1 and M2).

• Divides genetic material in half
• Shuffles genetic material by recombination

Front 222

Meiosis I

Back 222

Two chromosomes (maternal and paternal) become two cells with one chromosome (2n-->1n).

1. DNA replication
2. Recombination
3. Cell division of two homologous chromosomes

Front 223

Meiosis II

Back 223

Two cells with one chromosome each comprised of two sister chromatids split into four cells

1. Two sister chromatids separate and cells divide
2. Results in FOUR cells, each with it's own chromosome.

Note: there is no DNA replication in 2nd meiotic phase.

Front 224

Meiotic Nondisjunction

Meiosis I

Back 224

If it occurs during 1st cell division, one normal and one without any chromosomes.

• During 2nd division, two cells will receive two chromosomes and one will receive none (just like in mitosis)
• Fails to separate two homologous chromosomes.
• After fertilization w/ normal gamete, the cell with 2 chromosomes will be trisomy. The cell with no chromosomes will become monosomy (fatal)

Front 225

Meiotic Nondisjunction

Meiosis II

Back 225

If it occurs during 2nd cell division, two of the four cells will be normal. One will have two chromosomes (receives both chromatids).

• One will have no chromosomes.
• Fails to separate two sister chromatids
• After fertilization with normal gamete, cell with two chromatids become trisomy; cell with no chromatids becomes monosomy (fatal)

Front 226

Autosome

Back 226

A chromosome that isn't sex (X or Y)

e.g., a somatic cell

Front 227

Numerical chromosome abnormalities that can exist

Back 227

Autosomes: 13, 18, and 31

Sex chromosomes

• Only monosomy that can survive (45, X--Turner Syndrome)

Front 228

Aneuploidy

Back 228

State of having a chromosome number that's not a multiple of the haploid number.

Loss or gain of a single chromosome.

• Monosomy: chromosome # is 45
• Trisomy: chromosome # is 47

Front 229

Euploidy

Back 229

Exact multiple of haploid set of chromosomes

Haploid: normal number of gametes (n=23)

Diploid: normal number in zygote and somatic cells (2n = 46)

Polyploidy: complete set(s) of extra chromosomes

• Triploidy: 3n = 23 x 3 = 69 so 3n=69
• Tetraploidy: 4n = 23 x 4 = 92 so 4n=92

Front 230

Structural Chromosome Abnormalities can be one of two types

Back 230

Balanced: no loss or gain of DNA

• Reciprocal translocation
• Inversion

Unbalanced: abnormalities in DNA

• Duplications
• Deletions
• Marker chromosomes

Front 231

Down syndrome incidence

Back 231

As maternal age increases, incidence increases.

Down Syndrome incidence peaks @ 40 years old

Front 232

Chromosome Testing

1. Requirements (2)
2. Sources

Back 232

Requirements

Must be (1) mitotically active and dividing, and (2) have a nucleus.

Sources

1. Blood (WBCs because RBCs are anucleic)
2. Bone marrow
3. Fibroblasts
4. Amniocentesis
5. Chorionic villus sampling

(1st trimester 9-11wks from

extraembryonic tissue from chorion)

**Chemicals are used to make mature WBCs divide

Front 233

Amniocentesis

Back 233

10-15 mL of amniotic fluid is used.

• Performed at 2nd trimester (different than chorionic villus)
• Fetal cells from babies urine and skin
• Centrifuge to remove cells
• Can do BIOCHEMICAL (enzyme activity) and CHROMOSOME analysis.

Front 234

Procedure for chromosome analysis

Back 234

1. Add mitogen (e.g., colchicine) to arrest cells as they enter mitosis
2. Incubate for 2-3 days
3. Mitotic inhibitor added
4. Add hypotonic solution to hydrate cells
5. Add fixative and make membrane brittle/fragile to break open.
6. Make cell membrane burst and put on slide to count chromosome #

Front 235

G-Banding Techniques for Chromosomes

Back 235

G-Banding: trypsin with Giemsa staining

Different staining pattern (dark and light) based on chromosome contents and structures:

G+ are Dark Bands

• AT rich
• Replicate late; condense early
• Few genes; mostly repeats

G- are Light Bands

• GC rich (CpG islands)
• Replicate early; condense late
• Mostly genes, few repeats
• Loss of G- band is usually more severe than loss of G+ band

Front 236

Banding for chromosome location:

e.g., Band 6p23

Back 236

6p23

Chromosome number = 6

Arm = short (p)

Chromosome region = 2

Band in chromosome region = 3

Centromere is considered zero (0) and the number gets higher as you move away from the centromere either up or down the chromosome.

Front 237

Human Karyotyping

Back 237

Chromosomes are numbered (1-22) based on their size. They get smaller as the number increases.

Exception: chromosome 22 is larger than 21

Chromosome 21 is smallest chromosome in the human genome. This is why trisomy 21 is most prevalent--because it is most tolerable because it has less genes than other chromosomes.

Front 238

Karyotype Descriptions

Back 238

Chromosome #, sex designation, abnormalities

Normal male: 46,XY Normal female: 46,XX

• Turner Syndrome: 45,X
• Klinefelter Syndrome: 47, XXY
• Down Syndrome: 47, XY, +21
• 6p;18q translocation: 46,XX,t(6;18)(p22;q21)
• Deletion in 5p: 46,XY,del(5)(p14)

(deletion in short arm of chromosome 5)

• Mosaic (two cell lines): 47,XX,+21/46,XX

- abnormal listed 1st: 47,XX,+21 (trisomy)

- normal listed 2nd: 46,XX

Front 239

Karyotype for the following diseases:

• Turner syndrome
• Klinefelter syndrome
• Down syndrome

Back 239

Turner: 45,X

Klinefelter: 47,XXY

Down: 47,XY,+21 (if male)

Front 240

Triploidy

Back 240

Gain of an extra SET of chromosomes (note this is different than triSOMY)

Types: 69,XXX/XXY/XYY

• 60% double fertilizations (two sperm): 69,XYY
• 40% are diploid egg: 69,XXY

Symptoms

• Severe intrauterine growth retardation (IUGR)
• Rare mosaics may survive w/ moderate mental retardation
• When consulting, you are only getting one type of tissue (e.g., blood). So this doesn't give you whole clinical picture

Front 241

How many genes are in the human genome?

Back 241

22,000

Front 242

Within each cell, genome is packed as ________.

Back 242

Homogenous chromatin (DNA and histones)

Front 243

When cell divides, it's DNA _______ and is visible as chromosome under microscope.

Back 243

Condenses

Front 244

Chromosome abnormalities are seen in ____ % of 1st trimester and ____% of 2nd trimester losses.

Back 244

1st Trimester: 50%

2nd Trimester: 20%

Front 245

In interphase, chromosome is highly _______ and cannot be visualized.

Back 245

Interphase: decondensed (euchromatin)

• G1: single chromatid
• S: single chromatid replicates
• G2: chromosome comprised of two chromatids

Front 246

When nondisjunction occurs during mitosis, one cell gets ____ chromosomes and one cell gets ____ chromosomes.

Back 246

One gets 45 and one gets 47.

• If both abnormal cells are viable, a mixture of 3 cells develop: (1) normal (2) 45 cell line (3) 47 cell line.
• Only one of the two abnormal cell lines may be viable, leading to a mixture of 46/45 or 46/47. This is called mosaicism.

Front 247

Remember, Mosaicism is caused by nondisjunction during _____.

It results in ___________.

Back 247

MITOSIS

Results in coexistence of two cell populations.

Front 248

Full summary of Meiosis

Back 248

Meiosis reduced chromosome number from diploid (46) to hapolid (23). Composed of DNA replication followed by two divisions.

Meiosis I

• reduction division in which homologous chromosomes pair (UNIQUE TO MEIOSIS--doesn’t occur in somatic cells), disjoin, and move to opposite daughter cells.
  • 2n-->n

Meiosis II

• No DNA replication. Two sister chromatids of each chromosome separate and go to opposite daughter cells just like in mitosis.
  • Chromosome number remains 23 but the DNA amount is reduced to half.

Front 249

Two types of Chromosome Deletions?

Back 249

Terminal Deletion: end of telomere is lost.

• Loss of terminal DNA
• No centromere here
• Daughter cell will have a normal chromosome and a short chromosome.

One breakpoint indicates the terminal segment of chromosome has been lost: 46, XY, del(5)(p14)

Interstitial Deletion: middle of the telomere is lost (two breaks).

• Two breakpoints indicates that the segment between them has been lost and the terminal segment of the chromosome is retained.

e.g., 46,XY,del(5)(p14p15.3)

- male with interstitial deletion of 5p segment; only part of short arm of chromosome 5, from p14 o p15.3 is lost.

Front 250

Both types of chromosome deletions can occur at both mitosis and meiosis.

Unequal crossover will occur ____...

Back 250

Unequal crossover occurs at meiosis during pairing and recombination of homologous chromosomes.

• Can result in deletion or duplication.

Front 251

Three common deletion syndromes

Back 251

1. Wolf-Hirshhorn Syndrome

• terminal del4p

2. Williams Syndrome

• interstitial del7q11.2

3. Cardiofacial Syndrome

• del22q11.2

Front 252

Wolf-Hirshhorn Syndrome

1. Mutation

2. Presentation

Back 252

Wolf-Hirshhorn Syndrome

Mutation: terminal del4p

Presentation

• Low birthweight, microcephaly
• Beaked prominent nose
• Profound MR; heart disease (50%); seizures (50%)

Front 253

Williams Syndrome

1. Mutations

2. Presentation

Back 253

Williams Syndrome

Mutation: interstitial del7q11.2

Presentation:

• Overfriendliness (unique personality)
• Cardiac disease, CT abnormalities, intellectual problems, endocrine/growth problems

Front 254

Cardiofacial Syndrome

1. Mutation

2. Presentation

Back 254

Mutation: deletion 22q11.2

Most common deletion syndrome in human

Front 255

Isochromosome

Back 255

When two sister chromatids break apart at the centromere.

The two short arms combine and the two long arms combine.

Front 256

Ring Chromosome

Back 256

Two terminal ends of chromosome break off and the remaining ends join together to form ring.

Unstable during mitosis. Possibilities:

• No crossing over event: separation resulting in two rings
• 1 crossing over event: connected double ring
• 2 crossing over events: broken ring after they're interlocked.

Ring chromosome phenotype: growth retardation and mild developmental delay

Front 257

Types of Chromosome Inversions (2)

Back 257

Both inversions occur without loss or gain of DNA

1. Paracentric: one arm without centromere

2. Pericentric: involves centromere

memory aid:

• "a" is latin for without, so parAcentric is without involved centromere
• "perIcentric Involves the centromere"

Front 258

Paracentric Inversion

Back 258

Happens in one arm without involving centromere.

Inversion loop occurs to allow for pairing of homologous chromosomes.

• Recombination outside the loop: normal or inverted
• Recombination inside loop: both lost in cell division.

- Dicentric gametes: two centromeres

- Acentric gametes: one centromere

Front 259

Pericentric Inversion

Back 259

Involves the centromere

Recombination outside inversion loop: normal and inverted

Recombination insider loop leads to four possible gametes:

1. Normal
2. Inverted
3. Dup p / delq and dup q / del p

Front 260

Difference between peri- and paracentric inversion?

Back 260

Products of PERIcentric inversion have ONE centromere and are therefore viable.

Front 261

Reciprocal Translocation

Back 261

Breaks at different chromosomes that switch chromosomes.

No loss or gain in genetic material, so they're balanced. Carriers are phenotypically normal but are at risk for having abnormal kids.

One parent contributes two normal chromosomes. Parent with translocation has two options:

1. Give two abnormal chromosomes: results in phenotypically normal baby that can potentially have bad kids when they reproduce. This is because all genetic material is present.
2. Give one abnormal and one normal chromosome: baby will be unbalanced and abnormal--results in partial monosomy and partial trisomy.

Front 262

Difference between inversions and translocations

Back 262

Inversions are at the same chromosome.

Translocations are at different chromosomes.

Front 263

Robertsonian Translocation

Back 263

Involves ACROCENTRIC chromosomes

(13-15, 21, 22)

Chromosomes join together at the centromere after ends break off of them. This fusion reduces chromosome number from 46 to 45.

This is a balanced translocation and they're phenotypically normal--this is only situation where you can reduce chromosome # from 46 to 45 and be phenotypically normal.

• At risk for abnormal kids
• Between 13 and 14 is most common type
• 14 and 21 is Down Syndrome

Front 264

What is the only situation that you can reduce chromosome number from 46 to 45 and still be phenotypically normal?

Back 264

Robertsonian translocation involving acrocentric chromosome (13-15, 21, and 22).

Front 265

Translocation-type Down Syndrome

(Robertsonian Translocation)

Back 265

Carriers of Robertsonian Translocation involving chromosome 21 are at risk for producing child with translocation type down syndrome.

• Mother carrier: recurrence risk is 10-15%
• Father carrier: recurrence risk is 0-2%

It's important to note that having the Robertsonian Translocation doesn't cause Down Syndrome. It means you're a carrier. It's how this is passed down to the kid that determines whether they'll get it.

Front 266

Female vs. Male Chromosome

Back 266

Females down regulate genes on their X-chromosome to equal male levels of expression.

Only one X-chromosome with be active. The other is a Barr body.

Barr Body: inactive X-chromosome in female

Front 267

Lyon hypothesis

Back 267

In females, one X chromosome is active.

• X-inactivation occurs early in embryonic life, around 2 weeks after fertilization.
• Note: inactive X must be reactivated in females germ line so each egg can receive an active X
• X-inactivation is random for the maternal or paternal X
• X-inactivation is incomplete
• X-inactivation is clonal--descendants of a cell with an inactive X will have the same inactive X

Front 268

X-Inactivation is incomplete

Back 268

Several genes on inactive X are know to ESCAPE INACTIVATION

• Many of these genes are homologs on the Y chromosome (dosage compensation)

• Terminal regions of either end of the X and Y are identical (pseudo-autosomal region) and recombine in meiosis.

Front 269

Dosage Compensation

Back 269

XXY vs XXXY

Front 270

Mosaicism in females

Back 270

Each tissue is mosaic in the female.

This is not true mosaicism

Modulational change to control what chromosome is activated or inactivated. It's an EPIGENETIC CHANGE.

Front 271

Variable expression in heterozygote X chromosome

Back 271

Diseased X is "turned off" to prevent expression of disease.

Unfortunate Lyonization: mutant allele on active X chromosome (skewed X inactivation)

If there has been a translocation, normal chromosome is inactivated in order to retain the genetic material that has been translocated between the other two chromosomes.

Front 272

Incidence of different Chromosome Abnormalities

Back 272

35% = sex chromosome aneuploidies

30% = balanced structural abnormalities

25% = autosomal aneuploidies

10% = unbalanced structural abnormalities

Front 273

What are the names of the three trisomy conditions?

Back 273

Trisomy 13: Patau syndrome

Trisomy 18: Edwards syndrome

Trisomy 21: Down syndrome

Front 274

Trisomy 21: Down Syndrome

Back 274

• Most common chromosome abnormality
• Short neck, flat face, low set ears, protruding tongue, heart defects (50%); 15-fold increase in leukemia.
• Most cases (95%) due to trisomy 21 with MEIOTIC nondisjunction; 3-4% from Robertsonian translocation; 2% from mosaicism (mitotic disjunction)

Translocation recurrence risk:

• Mother carrier: 10-15%
• Father carrier: 0-2%
• With 21;21 translocation, recurrence risk is 100%

Front 275

Trisomy 18: Edwards Syndrome

Back 275

Second most common chromosome abnormality

• IUGR and low birth weight
• Heart defects (80-90%), severe MR, small head, mouth, and jaw
• CLENCHED fists are diagnostic
• Many die in first month or before birth

Front 276

Trisomy 13: Patau Syndrome

Back 276

Third most common chromosome abnormality

• Severe intellectual problem, physical problems
• Heart defect (80%); holorosencephaly; poorly developed eyes; cleft lip/palate
• Many die early

Front 277

Sex Chromosome Anueploidies

Back 277

Do better than autosomal aneuploidies because:

• Y chromosome is small and has very few genes (largely repetitive sequences), so extra copies don't have profound effects.

• Only one active X chromosome in a cell, so extra Xs are inactivated.

Note, some genes on inactive Xs can escape inactivation, so they may not be completely harmless.

Front 278

Two types of sex aneuploidies?

Back 278

1. Turner Syndrome: 45,X

2. Klinefelter Syndrome: 47,XXY

Front 279

Turner Syndrome

Back 279

FEMALE ONLY: 45, X

• Gonadal dysgenesis and sexual immaturity
• Short stature, webbed neck, amenorrhea, cardiovascular problems
• Most due to paternal nondisjunction
• 60% are 45,X karyotype
• Some form isochromosomes of Xq (Xp deleted and Xq duplicated)
• Mosaic: 45, X/46XX or 45,X/46 XY

Front 280

Klinefelter Syndrome

Back 280

MALE ONLY: 47,XXY

• Postpubertal testicular failure
• Tall, small testes, gynecomastia, infertile
• More complex karyotypes (XXXY, XXXXY) are more severe.

Front 281

Genomic Imprinting and two examples

Back 281

Genomic Imprinting: certain genes expressed in a parent-of-origin specific manner

1. Prader-Willi syndrome

2. Angleman Syndrome

Front 282

Prader-Willi Syndrome

Back 282

Example of genomic imprinting.

Loss of active genes in 15q11-q13 (paternal)

Normally, inherit one copy of chromosome 15 from each parent. Some genes on this chromosome are only active on paternal copy.

• 70%: 15q11-q13 region of paternal deleted
• 25%: inherit two maternal copies of chromosome 15-- called MATERNAL UNIPARENTAL DISOMY (UPD).
• Small number from abnormal methylation

Front 283

Fluorescence in Situ Hybridization (FISH)

Back 283

Analyzes chromosome at molecular level

• Supplemental tool to conventional banding
• 100-fold improvement in detecting chromosome rearrangements
• Can visualize during interphase and metaphase (other techniques can only see during metaphase)

Does not give info about location on chromosome.

Front 284

Conventional Banding Method

Back 284

Allows a glance at complete genome in single test.

Limited resolution for detection of small deletions and other subtle rearrangements.

Front 285

DiGeorge Velocardiofacial Syndrom (VCFS)

Back 285

Autosomal dominant with VARIABLE EXPRESSIVITY.

Deletion within chromosome 22q11

• Craniofacial anomalies and cardiac problems
• Most common microdeletion in humans

Front 286

Preimplantation Genetic Diagnosis with FISH

Back 286

Alternative to prenatal diagnosis. Involves selecting preimplantation embryos from cohort generated by assisted reproduction technology.

• Prescreen couples with previous affected kids or IVF
• Probes for chromosomes X,Y, 13-18, 21, and 22
• Can detect 70% of aneuploidies

Front 287

Array-Based Comparative Genomic Hybridization (aCGH)

Back 287

Allows high-resolution scanning of genome.

• Multiple FISH tests performed simultaneously
• Reveal genomic imbalance (deletions, duplications, marker chromosomes)
• Can detect small deletions that are difficult to pick up by G-banding.

e.g., label patient DNA green and the control another color. Mix the two and observe whether color mixes appropriately. If preference for one color, microdeletion.

Front 288

FISH vs. Array-Based Comparative Genomic Hybridization (aCGH)

Back 288

FISH: one probe

• Reveals BALANCED abnormalities (translocation and inversions)

CGH: hundreds of probes

• Reveals IMBALANCED abnormalities

Front 289

Structural rearrangement should always be analyzed during what cycle of cell?

Back 289

Metaphase cells

Front 290

What is gold standard for chromosome analysis and provides view of entire genome?

Back 290

G-banding

Front 291

_____ can use a unique-sequence probe to detect microdeletion.

Back 291

FISH

Front 292

Multicolor paining allows detection of _____, but not for _______.

Back 292

Detection of structural changes.

Not good for small deletions

Front 293

_______-fold improvement between chromosome study and microarray

Back 293

1000 fold improvement