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BIO 214 > Midterm #2 > Flashcards

Midterm #2 Flashcards

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1
Q

what is DNAs function

A
  • it carries the genetic instructions required for the expression of traits in all known living organisms
  • it serves as the hereditary unit of life
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2
Q

what is a double helix

A

two spiral backbones

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3
Q

describe what it means for DNA to be in a double helix

A

DNA is a twisted ladder with the rungs being the nitrogenous bases (A, C, T, G) aka the sugar phosphate backbone

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4
Q

what are the individual units of DNA (nucleotide)

A

individual units of DNA are made up of nucleotides
1 phosphate
1 sugar
1 nitrogenous base

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5
Q

what holds the double helix together

A

hydrogen bonds (IMFs) hold the two strands together

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6
Q

what are the complimentary base pairs

A

nucleotides follow complimentary base pairing rules:
- adenine pairs with thymine
- guanine pairs with cytosine

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7
Q

nucleic acid

A

large biomolecule composed of many nucleotides (DNA and RNA)

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8
Q

what is a nucleotide

A

can refer either to deoxyribonucleotide –> monomer of DNA
ribonucleotide –> monomer of RNA

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9
Q

draw a nucleotide

A

see first page of notes

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10
Q

draw cytosine

A

see first page of notes

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11
Q

draw thymine

A

see first page of notes

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12
Q

draw uracil

A

see first page of notes

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13
Q

draw adenine

A

see first page of notes

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14
Q

draw guanine

A

see first page of notes

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15
Q

which nitrogenous bases are purines

A

adenine and guanine

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16
Q

what nitrogenous bases are pyrimidines

A

cytosine, thymine, and uracil

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17
Q

what are the bonds in a nucleotide

A
  • phosphodiester bonds
  • glycosidic bond
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18
Q

how do the nucleosides differ from ribonucleoside and deoxyribonucleoside

A

ribonucleoside
- found in RNA
- name (ex. cytosine)
- possess two OHs on the bottom of the ring (pg. 2)

deoxyribonucleoside
- found in DNA
- name (ex. deoxycytosine)
- possess one OH on the bottom of the ring

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19
Q

when a new strand of DNA or RNA is being synthesized, which side is it added to?

A

synthesized from 5’ to 3’
5’ has the phosphate 3’ has the sugar
(see pg. 2)

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20
Q

function of DNA polymerase

A

hydrolyzes dNTPs between alpha and beta phosphate

DNA polymerases are enzymes that create DNA molecules by assembling nucleotides

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21
Q

how do two nucleotides link

A

the hydrolysis of a high energy bond provides the energy necessary to link the two nucleotides together

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22
Q

what does ATP break down to

A

ATP often breaks down to ADP and an inorganic phosphate

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23
Q

during DNA synthesis, what is released as a result of hydrolysis

A

pyrophosphate is released

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24
Q

when one phosphodiester bond is created, what happens

A

another phosphodiester bond is created (pg. 3)

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25
Q

describe the base pairing of DNA

A

DNA has complimentary and antiparallel base pairing

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26
Q

what are watson crick bonds

A

adenine (A) forms a base pair with thymine (T) using two hydrogen bonds, and guanine (G) forms a base pair with cytosine (C) using three hydrogen bonds

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27
Q

draw the hydrogen bond acceptors and donors on the watson crick bonds

A

see pg. 3

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28
Q

what handedness does DNA have

A

right handed (bottom of the staircase, left hand on the wall, going up) –> imagine a screw

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29
Q

describe the major and minor grooves

A
  • grooves occur from the way the two phosphate sugar backbones curve
  • angle of the gycosidic bonds are what create major vs. minor grooves
  • smaller angles make the minor grooves
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30
Q

length of major and minor grooves

A

~10 nucleotides
the length of one turn is 3.4 nm (34 angstroms)

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31
Q

where does the DNA binding proton bind to

A

the major groove of DNA via hydrogens
- this is because the major groove is chemically rich

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32
Q

what is the significance of the major groove

A

allows unique DNA sequences that allow for specificity for DNA binding proteins (ex. turning on/off genes)

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33
Q

what is tautomerization

A

unstable isomer caused by the shift of a hydrogen
- results in alternative base pairing may cause a change in the sequence (mutation)

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34
Q

what tautomer does thymine bond to

A

enol tautomer of guanine

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35
Q

what tautomer does guanine bond to

A

enol tautomer of thymine

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36
Q

what tautomer does adenine bond to

A

imino tautomer of cytosine

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37
Q

what tautomer does cystosine bond to

A

imino tautomer of adenine

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38
Q

draw where the hydrogen goes from the amino to the imino form

A

pg. 7

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39
Q

draw where the hydrogen goes from the keto to the enol form

A

pg. 7

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40
Q

what is the tautomerization in the TEMPLATE

A

pg. 7

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41
Q

what is the tautomerization in the SUBSTRATE

A

pg. 7

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42
Q

define gene

A

the sequence of DNA that codes for a trait

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43
Q

nucleotide

A

the individual molecules that make up the DNA (A, T, C, G)

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44
Q

double heliz

A

the three dimensional structure of the DNA

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45
Q

genome

A

the complete set of genetic material

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46
Q

chromosome

A

the compacted DNA structure wound around proteins (histones)

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47
Q

allele

A

the specific version of the gene

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48
Q

DNA

A

the series of nucleic acids that contains our genes

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49
Q

how many chromosomes do humans have

A

2 pairs of 23 chromosomes
- one set of chromosomes is from the mother, the other from the father
- 22 pairs of non-sex chromosomes (autosomes) and one pair of sex chromosomes

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50
Q

XY

A

phenotypic male

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51
Q

XX

A

phenotypic female

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52
Q

klinefelter’s syndrome

A

XXY chromosomes

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53
Q

turner’s syndrome

A

X chromosome

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54
Q

down syndrome

A

have a 3rd chromosome at chromosome 21 (trisomy 21)

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55
Q

aneuploidy

A

atypical number of chromosomes

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56
Q

draw the anatomy of homologous chromosomes

A

pg. 9

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57
Q

where are alleles located

A

the gene locus

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58
Q

what is the haploid genome

A

1 set of chromosomes –> ~3 billion base pairs

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59
Q

where is our genetic information stored

A

the nuclei of most of our cells in our body (except for red blood cells)
- DNA is the same in every cell (even though DNA is the same in every cell, different genes are expressed (turned on) in different cells)

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60
Q

what do genes make

A

proteins

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61
Q

proteins

A

proteins are the workhorse of the cell and they come in all varieties:
- enzymes catalyze rxns (ex. lactase breaks down lactose)
- structural proteins provide support and shape to the cell
- cell surface proteins receptors (acid binding)

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62
Q

why do genes typically have two different alleles (versions) of a gene

A

a parent contains two alleles of a gene but randomly contributes just one of those alleles to each of their offspring - offspring inherits a set of genes from mom and dad

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63
Q

genotype

A

the specific allele combination for a gene

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64
Q

phenotype

A

the outward expression of the genotype

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65
Q

punnet square PP

A

homozygous dominant

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66
Q

punnet square Pp

A

heterozygous

67
Q

punnet square pp

A

homozygous recessive

68
Q

what is the function of the dominant allele

A

makes working protein

69
Q

what is the function of the recessive allele

A

makes broken protein

70
Q

what do metabolic pathways do

A

help cells break down macromolecules, such as, proteins, nucleic acids, polysaccharides, and lipids

71
Q

what do catabolic pathways achieve

A

important for releasing energy and heat

72
Q

describe the structure of metabolic pathways

A

pg. 15
a precursor molecule (the substrate for the enzyme) is catalyzed by an enzyme that was encoded for by an organism. then, intermediates are formed until the product is achieved

73
Q

alkaptonuria

A

a recessive disease hallmarked by black pigment accumulation in the tissues due to the build up of homogentisic acid. urine turns black upon exposure to air (oxidation)

74
Q

how can you determine which enzyme in the metabolic pathway was broken or missing

A

by examining which metabolite was bulding up

75
Q

how do defects in different steps of a pathway effect phenotypes

A

defects in different steps lead to different phenotypes

76
Q

phenylketonuria

A

build up of phenylalanine can damage the nervous system, which leads to many symptoms

77
Q

PKU and AKU are caused by blocks in steps of a pathway, what is the likely outcome of a person who is heterozygous for both genes

A

normal phenotype – heterozygous for each means they have a working enzyme for each, so pathway can be completed as normal.

78
Q

PKU and AKU are caused by blocks in steps of a pathway, what is the likely outcome of a person who is heterozygous for both genes

A

some combination of both genes

79
Q

how do we discover what genes are involved in a process

A

genetic screen
- random mutagenesis in a model organism
- search for mutations that exhibit a loss-of-function phenotype

determine gene identity of mutant
- complementation test –> only useful to determine gene identity when you have another mutant whose identity you know
- gene mapping

80
Q

phototroph

A

wild type
- can synthesize what it needs (AAs, nitrogenous bases, or vitamins)

81
Q

auxotroph

A

mutant
- cannot synthesize what it needs due to mutated gene in an essential metabolic enzyme; these require addition of a particular nutrient

82
Q

why do we do a genetic screen for auxotroph mutants

A

to uncover all of the genes / enzymes for all of the steps in the pathway

83
Q

conditional mutants

A

under one condition (the permissive condition) they exhibit wild type phenotype (grow / survive) and under another condition (restrictive condition) the show their mutant phenotype (die in this case)
- these allow you to have control on whether the mutant or wild type is shown

84
Q

describe the restrictive condition for the genetic screening of conditional mutants

A

contains minimal media
- requires the organism to synthesize metabolites
- wild type phototrophs show growth
- mutant auxotrophs show no growth

85
Q

what is minimal media composed of

A

Media that contains the minimum nutrients possible for colony growth, generally without the presence of amino acids, and are often used by microbiologists and geneticists to grow “wild type” microorganisms.

86
Q

describe the permissive condition for the genetic screening of conditional mutants

A
  • contains all the metabolites
  • does not require the organism to synthesize metabolites
  • wild type phototrophs show growth
  • mutant auxotrophs show growth
87
Q

a mutant of yeast unable to make guanine should grow on what media

A

pg 18

minimal + nitrogenous base

88
Q

what does no growth indicate

A

that the mutant carries mutation in the synthetic pathways for more than one amino acids

89
Q

complementation test

A

method to determine whether two mutants with the same phenotype are alleles of the same or different genes

90
Q

draw the difference between alleles with mutants in different genes vs. alleles with mutants in the same genes

A

pg. 21

91
Q

what is it called when two mutants are in different genes

A

we say they compliment one another because they each can provide the wild type

92
Q

fail to complement

A

groups that fail to complement one another because they have mutations in the same gene

93
Q

draw complementing strands

A

pg 21

94
Q

why and when must you test for dominance

A

test for dominance before complementation test because complementation tests are only useful for mutants that are recessive
- a test for dominance is performed by crossing the mutant with the wild type and checking first that the wild type will rescue the mutant
- if something is recessive, you will see growth on minimal media

95
Q

draw recessive auxotroph in the test for dominance

A

pg. 22

96
Q

do practice problems on pg. 22

A
97
Q

do practice problem on pg. 23

A
98
Q

do practice problems on pg. 24

A
99
Q

draw the anatomy of a gene

A

pg. 25

100
Q

what are the components of a gene

A

pg. 25
- control elements
- promoter
- +1 site
- 5’ UTR
- coding region
3’ UTR

101
Q

describe transcription on a coding strand

A

pg. 25
transcription creates a replica
- transcription haults where there is a transcription termination signal (typically a long row of As)

102
Q

how does space effect the coding region

A

pg. 25
- the extra space helps preserve the coding region

103
Q

coding region

A

pg. 25
stretch of DNA that contains the coding sequence that will encode the protein (in eukaryotes, this includes both exons and introns, though introns will be removed)

104
Q

coding strand

A

pg. 25
strand of DNA that contains the coding sequence of the gene of interest

105
Q

UTR

A

pg. 25
untranslated region
- at the beginning of the 5’ UTR, we have the +1 site
- 5’ UTAR is important in facilitating contact with the ribosome

106
Q

+1 site

A

pg. 25
represents the first nucleotide that is translated

107
Q

promoter

A

pg. 25
region of DNA upstream of transcription start site that is recognized by RNA polymerase

108
Q

draw the promoter on a region of DNA

A

pg. 25

109
Q

control elements

A

pg. 25
regions of DNA that increase or decrease the rate of transcription

110
Q

describe the mechanism of trancription

A

NN pg. 1

when transcirption is hindered, RNA polymerase can associate with the promoter by binding to another factor called the sigma factor

once RNA makes contact, transcription is initiated

the two strands of the DNA separate to make a transcription bubble

RNA pol transrcibes the DNA by reading the template from 3’ to 5’ which allows RNA pol to make an exact replica of the coding strand (replacing Ts with Us)

RNA transcribes the new transcript from 5’ to 3’

RNA pol runs into the termination site and then released from the DNA

111
Q

sigma factor

A

kind of transcription initiation factor that helps RNA polymerase recognize the promoter
- releases once RNA polymerase makes contact and initiates transcription

112
Q

draw the interaction between the prokaryotic promoter and its RNA polymerase

A

NN pg. 1

113
Q

holoenzyme

A

made of an apoenzyme (RNA pol) and its cofactor (sigma factor)

114
Q

apoenzyme

A

the inactive form of the enzyme and requires a cofactor to be active

115
Q

what is the role of the sigma factor

A

allows RNA pol to make contact with the promoter. it then releases after RNA pol initiates transcription

116
Q

what sites does RNA pol make contact with

A

-35 and -10 sites

117
Q

what is the purpose of the space between the -35 and -10 sites

A

the distance of +/- 17 nucleotides of any variety provides optimal space needed for functioning

118
Q

concensus sequence

A

sequence of DNA recognized by prokaryotic RNA polymerases
- these are the optimal sequences
- the closer the -10 and -35 sequence match the consensus sequence, the tighter the RNA polymerase binds and the more transcript that is made for that gene.

119
Q

what effect do changes to DNA binding site have

A

any changes to DNA binding site (-10 and -35 sites) weakens the sigma factors association which weakens the RNA pol holoenzyme association. This means the the RNA pol cannot transcribe at the same rate, which affects the number of transcripts that can be made from it

120
Q

how can you tell the direction of transcription

A

NN pg. 2

find the consensus sequence - that is the coding strand which is transcribed from 5’ to 3’

121
Q

answer practice questions on NN pg. 2

A
122
Q

what is the primary energy source of E. coli

A

glucose

123
Q

what happens when glucose is absent from the E. coli

A

E. coli expresses the genes of the Lac Operon to use lactose

124
Q

B-galactosidase

A

an enzyme in E. coli that can break down lactose into monosacharides galactose and glucose

125
Q

what gene expresses B-galactosidase

A

Lac-Z

126
Q

describe Lac-Z

A

expresses B-galactosidase
- Lac-Z expression is inducible (lactose serves as the indcer for B-galactosidase expression)

127
Q

draw the lac operon

A

NN pg. 3

128
Q

operon

A

set of linked genes under the control of the same promoter

129
Q

repressor protein

A

aka Lac I

binds to a segment of DNA called the operator

130
Q

operator

A

a segment of DNA found between promoter and structural gene that acts as a control element

131
Q

Lac Y

A

makes lactose permease –> allos the absorption of lactose

132
Q

Lac A

A

encodes the enzyme galactosidase transacetylase –> breaks down a toxic compound that gets imported into the cell because of permease

133
Q

linked genes

A

overall they are called structural genes
- Lac Z
- Lac Y
- Lac A

134
Q

what happens when the repressor is bound to the operator and how does the repressor do this

A

OCCURS WHEN LACTOSE IS ABSENT
prevents transcription of structural genes
- Lac repressor works by looping DNA at the opperator (physically impedes RNA from progressing) –> prevents transcription

135
Q

what happens when the repressor is not bound to the operator and how does the repressor do this

A

OCCURS WHEN LACTOSE IS PRESENT
lactose is converted to allolactose which binds to the repressor protein, which prevents the repressor from binding to the operator. This allows RNA pol to properly bind to its propmoter

136
Q

negative regulation

A

when a repressor protein binds to the operator to prevent transcription
- repressed by the lac I repressor, which represses transcription when lactose is absent. this is regulated because it would be wasteful to transcribe when lactose is absent

137
Q

polycistronic

A

lac operon mRNA is polycistronic, where the mRNA contains the coding sequence for the three different structural genes

138
Q

what is the effect of a recessive-loss-of-function mutation on Lac I on the expression of the structural genes

A

NN pg. 4
they are constitutively expressed

139
Q

what is the effect of a mutant whose operator is missing the key binding sites for the Lac I repressor protein

A

NN pg. 4
they are consitutively expressed

140
Q

Lac I-

A

NN pg.5
Lac I never bound, repressor cannot bind to protein, Lac operon never expresed whether or not lactose is present

CONSTITUTIVE PHENOTYPE

141
Q

Lac I+

A

NN pg.5
wild type

presence of lactose: lac operon expressed - inhibition relieved
absence of lactose: inhibits lac operon

can turn expression on / off, which means it has an INDUCIBLE PHENOTYPE

142
Q

Lac I^s

A

NN pg.5
Lac I always bound whether or not lactose is present

UNINDUCIBLE PHENOTYPE

143
Q

Lac O^c

A

NN pg.5
continual expression of the lac operon whether or not lactose is present

CONSTITUTIVE PHENOTYPE

144
Q

Lac O+

A

NN pg.5
wild type

presence of lactose: lac operon expressed - inhibition relieved
absence of lactose: inhibits lac operon

can turn expression on / off, which means it has an INDUCIBLE PHENOTYPE

145
Q

draw the answer: you found a mutant in E.coli that has a constitutive phenotype. how would you know whether it was Lac I- or Lac O^c mutation

A

NN pg.6

146
Q

explain the answer: you found a mutant in E.coli that has a constitutive phenotype. how would you know whether it was Lac I- or Lac O^c mutation

A

NN pg.6

  • each cell could be given a wild type copy of the Lac I+ gene and of the operator region (Lac O+) to make partial diploids (A partial diploid is a bacterial cell that contains two copies of some genes.)

IF…
Lac I- mutation, the wild type copy can make a repressor protein which can bind to the operator and work as normal
(INDUCIBLE PHENOTYPE)

IF…
Lac O^c mutation, the wild type copy can also make a repressor protein, but it doesn’t matter because the protein cannot bind to the O site, which causes it to continuously be made
(CONSTITUTIVE PHENOTYPE)

147
Q

is a repressor cis or trans acting and explain what this means

A

NN pg. 6
trans
- wild type copy on the strand can recuse loss of function on other strand because its making a protein (proteins are soluble –> they can move around and repress their own operator or another operator)

148
Q

is an operator cis or trans acting and explain what this means

A

NN pg. 6
cis
- wild type copy on the strand cannot recuse loss of function on other strand because they do not move - they control expression locally (same DNA)

149
Q

haploid

A

Haploid refers to the presence of a single set of chromosomes in an organism’s cells.

150
Q

diploid

A

(of a cell or nucleus) containing two complete sets of chromosomes, one from each parent.

151
Q

consider a haploid E.coli LacI mutant with a dominant gain of function mutation in which it loses its affinity for lactose. no other portion of the protein has been effected. how does that affect expression of the structural genes

A

NN pg. 6

they become permanently inhibited (uninducible)

152
Q

do practice problems on NN from pg 7 through pg. 9

A
153
Q

positive regulation

A

NN pg. 10
- deals with activating transcription Lac operon activates transcription when glucose is absent. this occurs because its wasteful to transcribe the lac operon (lactose when glucose is present, because glucose is the preferred energy source for E. coli.

154
Q

how does glucose regulate the lac operon

A

through cyclic AMP –> a cellular messenger that is a derivative of ATP

155
Q

if you have high levels of glucose and low levels of lactose, how much cAMP do you have

A

low levels of cAMP (cAMP is inversely related to glucose)

156
Q

if you have low levels of glucose and high levels of lactose, how much cAMP do you have

A

high levels of cAMP (cAMP is inversely related to glucose)

157
Q

if you have low levels of glucose and low levels of lactose, how much cAMP do you have

A

low levels of cAMP (cAMP is inversely related to glucose)

158
Q

if you have low cAMP how does that effect how much the lac operon is expressed

A

low levels of cAMP is correlated with low levels of lac operon expression

159
Q

if you have high cAMP how does that effect how much the lac operon is expressed

A

high levels of cAMP is correlated with high levels of lac operon expression

160
Q

adenylate cyclase

A

NN pg 11

resides near plasma membrane and functions to alert the rest of the cell when glucose is present / absent. it does this by converting ATP to cAMP. it only does this when glucose is absent

161
Q

describe a working catabolite pathway when lactose is present and glucose is absent

A

NN pg 11

glucose absent –> ATP –> cAMP –> activate lac operon

162
Q

how does cAMP / CAP complex activate the lac operon

A

NN pg. 11

cAMP creases, which allows for it to bind to CAP (catabolyte activation protein)

when cAMP is high, it can bind to CAP protein. when it binds, cAMP activates CAP and allows it to now bind to the region of the DNA called the CAP site.

the binding of the RNA pol at its promoter initiates transcription and activates the lac operon

163
Q

answer practice problems on NN from pg. 12-13

A

BIO 214 (2 decks)

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