Biochemistry Exam 2 Bchm561

Front 1

DNA Polymerase Catalyzes

Back 1

Catalyzes phosphodiester bond formation

- Catalyzes the step-by-step addition of deoxyribonucleotide units to a DNA chain

Front 2

(DNA)n + dNTP <--->

Back 2

(DNA)n+1 + PPi

(dNTP=deoxyribosenucleotide, PPi= pyrophospate ion)

Front 3

Characteristics of DNA synthesis
1) The reaction requires all _ activated precursors: and _

Back 3

The reaction requires all four activated precursors:

deoxyribonucleoside 5'-triphosphates:
dATP, dGTP, dCTP, TTP.
and Mg2+

Front 4

Characteristics of DNA synthesis:
2) DNA polymerase requires a ___ to synthesize a product:
DNA polymerase is a ____ directed enzyme.

Back 4

2) DNA polymerase requires a template to synthesize a product that is complementary to the template.

DNA polymerase is a template-directed enzyme.

Front 5

DNA replication proceeds from _' to _'

Back 5

DNA replication proceeds from 5' to 3'

Front 6

DNA Replication:

DNA must be bound by a ____ having a free _'- __ group

This group nucleophiliclly attacks ____________ and forms a __________ and releases a ______.

This reaction is driven by ______.

Back 6

DNA must be bound by a primer having a free 3'-OH group.

The free 3'-OH group on the primer nucleophiliclly attacks the innermost phosphorous on the nucleoside aand forms a phosphodiester bond and releases pyrophosphate.

the reaction is driven by hydrolysis of pyrophosphate

Front 7

RNA polymerase that copies viral RNA from an RNA template

Back 7

RNA-directed RNA polymerase copies viral RNA from an RNA template.

Front 8

Genetic information in retroviruses flows from ____ to ____.

Back 8

Genetic information in retroviruses flows from RNA to DNA.

Front 9

Retroviruses contain _ strands of ___ which upon entering a cell it will infect is copied into ___ through the action of a viral enzyme called ________.

Back 9

Retroviruses contain 2 copies of a single stranded RNA molecule. On entering the cell the RNA is copied into DNA through the action of a viral enzyme called reverse transcrpitase.

Front 10

mRNA:

Back 10

1) photocopy of genetic information
2)this genetic information is translated to synthesize functional proteins using different RNAs

Front 11

Restriction Enzymes are:

Back 11

Restriction Enzymes are precise, molecular scalpels that allow an investigator to manipulate DNA segments.

Front 12

Restriction Enzymes recognize:

Back 12

Restriction Enzymes recognize specific base sequences in a double helical DNA and cleave, at specific locations, both strands of that duplex.

Front 13

Restriction Enzymes are ___ about their cleaving locations.

Back 13

Restriction Enzymes are symmetrical about their cleaving locations

Front 14

RFLP

Back 14

RFLP (Restriction fragment length polymorphism)- can detect diseases

Front 15

___ Blotting

Back 15

can be used to follow the inheritance of certain DNA genes.

Front 16

DNA sequencing: the precise ____ sequence can be determined and from that 3 things can be determined:

Back 16

DNA sequencing: the precise nucleotide sequence can be determined and from that
-gene architecture
-control of gene expression
and
-protein structure can be estimated

Front 17

PCR (______): amplifies

Back 17

PCR(Polymerase chain reaction): amplifies specific DNA sequences

Front 18

PCR is carried out by adding the following to a solution containing___

Back 18

PCR is carried out by adding

-the target sequence

- a pair of primers which will hybridize with the flanking sequences of the target

-all four dNTP's

-a heat stable DNA polymerase

Front 19

A PCR cycle contains the following steps:

Back 19

1) Strand Separation: The two strands of the parent DNA are separated by heating to 95*C for 15 seconds

2) Hybridization of the primers: the solution is then abruptly cooled to 54*C to allow each primer to hybridize to a DNA strand. One primer hybridizes tothe 3' end of the target on one strand and the other primer hybridizes to the 3' end on the complementary target strand. (parent DNA duplexes do not occur because of the large excess of primesr)

3)DNA synthesis: The solution is then heated to 72*C, the optional tempurature for Taq DNA polymerase. This heat stable polymerase elongates both primers in the direction of the target sequence (5' to 3'). This takes place on both strands.

This is repeated again and again. Increases exponentially.

Front 20

For PCR:
The sequence of the target needs/does not need to be known:

the target can/cannot be much larger than the primers

primers do/do not need to be perfectly matched to flanking sequences in order to amplify targets

Back 20

For PCR the sequence of the target does not need to be known; only the sequence of the flankers needs to be known.

primers do NOT have to be perfectly matched to flanking sequences in order to amplify targets.
For PCR the target can be much larger than the primers

Front 21

Key tools for Recombinant DNA technology:

Back 21

Restriction Enzymes and DNA Ligase are key tools for Recombinant DNA technology.

Front 22

Vector(biology):

Back 22

A vector (in biology) is a DNA molecule that is used to transfer foreign genetic material into another cell)

Front 23

Good vectors for cloning in E.Coli:

Back 23

Plasmids(naturally occuring circular DNA in bacteria) and bacteriophage lambda( a virus)

Front 24

The vector is prepared for accepting the new DNA by:

Back 24

The vector is prepared for accepting the new DNA by cleaving it at a single specific site with a RESTRICTION ENZYME.
This creates staggard cuts and these cuts are cohesive(sticky-will bond with itself). so.. ANY DNA fragment with the same sticky ends can be inserted in.

Front 25

Fragments which bond with the plasmid can be made by:

Back 25

using the same restriction enzyme that was used to open the plasmid DNA

Front 26

The DNA fragment and the cut plasmid can be joined by using ______, which catalyzes the formation of a phosphodiester bond at a break in a DNA chain.

Back 26

The DNA fragment and the cut plasmid can be joined together using DNA Ligase, which ____

Front 27

DNA Ligase requires:

Back 27

DNA Ligase requires a free 3'-hydroxyl group and a 5'-phosphoryl group

-the chains joined by a ligase must be in a double helix

-ATP and NAD+ are required

Front 28

Insertional Inactivation:

Back 28

Insertional Inactivation occurs when a section of DNA is inserted into a Plasmid at a restriction site which interferes with the gene.

Front 29

pBR322:

Back 29

this plasmid has an origin of replicatoin and a selectable marker based on ampicillian resistance

aka.. this plasmid can be effected by DNA insertion only.

Front 30

PCR requires:

Back 30

PCR requires the primers sequences

Front 31

(Recombinant DNA)

___________: allow specific DNA to be identified (restriction maps) and manipulated.

Back 31

Restriction enzymes allow specific DNA to be identified and manipulated

Front 32

restriction enzymes recognize;

Back 32

restriction enzymes recognize specific base sequences in a double helical DNA and cleave, at a specific location, both strands of the duplex

Front 33

Restriction fragments can be separated by

Back 33

gel elecrophoresis

Front 34

Southern Blotting:

Back 34

DNA

Front 35

Northern Blotting

Back 35

RNA

Front 36

DNA sequencing: the precise nucleotide sequences in DNA can determine:

Back 36

-gene architecture

- controll of gene expression

- protein structure (function)

Front 37

MUST know for PCR

Back 37

flanking sequence

Front 38

Joins DNA fragments and cut plasmids

Back 38

DNA Ligase

Front 39

DNA ligase requires

Back 39

a 3'-OH and 5'phosphoryl group

energy (ATP)

Front 40

Are descended from a common ancestor:

Back 40

homologs

Front 41

paralogs:

Back 41

same species-different function

Front 42

orthologs:

Back 42

different species but similar function

Front 43

understanding the homology reaveals:

Back 43

evolutionary history and information about their function

Front 44

If a substrate has a high kM that means:

Back 44

the enzyme probably has a low affinity for the substrate

LOWER THE KM THE HIGHER THE AFFINITY FOR THE SUSTRATE

Front 45

2 evolutionary events are represented by branch points:

Back 45

specization

genetic duplication

Front 46

Reverse transcriptase requires:

Back 46

- all 4 dNTPs

- a primer

-an RNA template

Front 47

How are sequence allignments made:

Back 47

Two sequences are alligned to maximize the number of "matches" (base matches). Sometimes gaps are inserted to maximize the number of matches. statistical analysis reveals the significance of the matches by determining the probability that the matches were random or not.

Front 48

Molecule by which most of the CO2 in the blood is transported through

Back 48

Bicarbonate

Front 49

An example of paralogs include:

Back 49

hemoglobin, myoglobin

Front 50

The biological role of restriction enzymes in bacteria is to:

Back 50

cleave foreign DNA

Front 51

Can be used to add nucleotides to the 3'end of DNA during cDNA synthesis

Back 51

Terminal Transferase

Front 52

site of O2 binding:

Back 52

Heme group

Front 53

2,3 BPG ___ O2 affinity because

Back 53

2,3-BPG decreases O2 affinity because it stabilizes th T-state

Front 54

HbF has a ___ affinity because:

Back 54

HbF has a higher affinity than Hb because it has a substitution which makes it more likely to bond with oxygen (hist=ser)

Front 55

The Bohr effect results from formation and breaking of:

Back 55

ionic bonds

Front 56

The Bohr Effect:

Back 56

higher CO2 levels decrease the affinity of Hb for O2 by lowering the pH (increasing H+ atoms from carbonic acid).

Front 57

Catalyst in Bohr effect:

Back 57

carbonic anhydrase

Front 58

-ly charged carbamates form ___ which ___

Back 58

-ly charged carbamantes form salt bridges which stablize the T-state

Front 59

A steric inhibitor binds to:

Back 59

a steric inhibitor binds to the active stite

Front 60

an allosteric inhibitor binds:

Back 60

not to the active site

Front 61

Competitive inhibitor:

Back 61

binds to substrate at the active site

Front 62

Competative inhibitor results:

can be overcome by:

Back 62

Km increases, Vmax not effected

can be overcome by adding more solute

Front 63

uncompetitive inhibitor:

can be overcome by:

Back 63

binds to ES complex,

can be overcome by nothing

Front 64

uncompetitive inhibitor results:

Back 64

Vmax= lowered

Km= lower

Front 65

Noncompetitive inhibitor:

can be overcome by:

Back 65

binds to a site other than the active site

can be over come by nothing

Front 66

noncompetative inhibitor results:

Back 66

Vmax=lower
Km= same

Front 67

myglobin binds;

Back 67

o2 in muscle, storage

Front 68

hemeglobin binds:

Back 68

main oxygn transport

Front 69

distant evolutionary relationships can be detected through:

Back 69

substitution matricies

Front 70

BLAST

Back 70

basic local allignment search tool

Front 71

Which structure (3* or 1*) is more conserved? Why?

Back 71

3 is more conserved than 1 because a proteins 3D shape determins its function. Sometimes substitutions and changes in the 1* chain of amino acids can result in similiar 3Dstructure which whill retain the same functio

Front 72

divergent evolution:

Back 72

same ancestor

Front 73

convergent evolution

Back 73

different pathways which lead to the same solution (function)

Front 74

Hglob and Mglob are ___ related to each other

Back 74

evolutionarly

Front 75

definition of allosteric enzyme;

Back 75

when a nonsubstrate molecule inducesa conformational change

when effectsthe affinity of the enzyme for a substrate

Front 76

Proteolytic enzymes

Back 76

catalyze the hydrolysis of peptide bonds

Front 77

the interaction of the enzyme and substrate at the active site promotes the formation of the:

Back 77

transition state

Front 78

the active site of an enzyme is the part:

Back 78

which most greatly lowers the activation energy

Front 79

mRNA

Back 79

messenger RNA, template

Front 80

tRNA

Back 80

transfer RNA, transfer active amino acids to the ribosome

Front 81

rRNA

Back 81

ribosomal RNA, catalytsts for protein synthessi

Front 82

snRNA

Back 82

Small Nuclear RNA, splice exons

Front 83

miRNA

Back 83

MicroRNA; non coding DNA which bind to mRNA and inhibit their translation

Front 84

siRNA

Back 84

Small Interfeering RNA
bind to mRNA and facilitate its degredation; inhibits the expression of genes

Front 85

RNA polymerase requires:

Back 85

a template
activated precursors
Mg2+