For exam 2 Flashcards

1
Q

Learning Goals of cell cycle

A

Cell division is the basis of growth, development, tissues repair, and reproduction of living organisms
Mitosis coordinates nuclear division in eukaryotic cells to produce genetically identical daughter cells
The eukaryotic cell cycle consists of several phases and is regulated by a molecular control system

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Types of cell division

A

Prokaryotic cell: binary fission as a mechanism of reproduction
Eukaryotic cells: mitosis as a mechanism of reproduction (single-celled eukaryotes) or growth/repair (multicellular eukaryotes)
Meiosis: as a mechanism of specialized reproductive cells(gametes)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

The 4 events that must occur for cell division

A

Reproductive signal: to initate cell division
Replication: of the DNA
Segregation: distribution of the DNA into the two new cells
Cytokinesis: separation of the two new cells

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Interphase and M phase (mitosis/cytokinesis)

A

Interphase: being in’s after cytokinesis, ends when mitosis starts, cell nucleus is visible and cell functions occur, indicating DNA replication, divided into sub phases: G1, S, G2 (defined by DNA replication status)
M(mitosis) phase: Nuclear membrane dissolves fully

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

G1, S, and G2

A

G1: getting ready to make DNA
S: duplicating DNA
G2: double DNA in cell

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

Interphase

A

DNA exists as long, threadlike “chromatin”
G1: each chromosome consists of one double strand DNA
S; DNA replication produces 2 identical double stands of DNA (sister chromatids) for each chromosome
G2: each chromosome consists of two associated dsDNA molecules(sister chromatids)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

M-phase

A

Chromosomes befoul e visible as dense, compact rods, each consisting of 2 chromatids held together at the certeromere(until separation)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Mitosis phases

A

-prophase/pre metaphase: compaction of replicated DNA into visible chromosomes; breakdown of nuclear envelope
-metaphase: duplicated chromosomes line up in middle of cell
Anaphase: sister chromatids separate and move to opposite sides of cell (now are daughter chromosomes)
-telophase: deco patio n and formation fo new nuclear envelope around the two separated sets of daughter chromosomes
- cytokinesis: division of the cytoplasm (forms two cells)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Spindle fibers

A

Micro tumbles fui cation as spindle fibers which orient and more chromosomes in the dividing cell
Positions of the centro Somme’s define the poles adn plane of division
Polar micrtubles overlap in center
Kinetochore micro tubules attach to kinetochores on the chromatids, sister chromatids attach to opposite halves of the spindle
Micro tubules form and attach to chromosomes during pro metaphase

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

G1-S Cdk phosporylates RB protein

A

Unphosphorylated (active) RB inhibits the cell cycle at Restricion Point, cell does no tenter S phase, when RB is inactivated and no longer blockers the cell cycle, the cell can go to DNA replication

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

MTOC and Centrosome

A

MTOC= microtubule organizing center
-surrounded by high conversation of tubulin dimmers
-forms/orients mito tic spindle that will attach to and more the duplicated chromosomes during M phase
Centrosome= MTOC of animal cells
-consist of 2 centrioles- hollow tubes formed by micro tubules at R angles
-doubles during S phase, each will move to opposite ends of nuclear envelope during G2-M transition
-positions determine the spindle orientation and plane of cellular division

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Cytokinesis

A

In animal cells, a contractile ring of actin and myosin micro filaments pinches in the cell membrane
In plan cells vesicles form the Golgi

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

Transitions depend on activity of enzymes calles

A

cdks= cyclin-dependent kinases
This is only active when bound to its partner protein called cyclin

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Unregulated cell division: Cancer

A

Normal positive regular OTs such as growth factors or their receptors stimulant the cell cycle
Normal negative regulators that inhibit the cell cycle

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

DNA involvement with cell division

A

Binary Fission and Mitosis: DNA copied and complete copy segregated to each ‘daughter cell’
Products identical to the ‘mother cell’
Meiosis: DNA copied, followed by 2 rounds of division and nuclear segregation, DNA content reduced by 1/2, each product is unique

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Sexual reproduction

A

Systematic joining of gametes to produce a diploid phase of life cycle, coupled with meiosis that reduces chromosome number in the haploid phase.
Meiosis is a specialized cell division where a single round of DNA synthesis is followed by two stages of chromosome segregation( diploid mother cell(pairs of chromosomes)) to haploid daughter cheeks (each with one of each kind of chromosome)
Shuffles genetic variation- offspring are not identical to parents or each other

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

Homologous chromosomes

A

Appear the same and contain the same genes except for sex chromosomes

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

Summary of meiosis

A

Functions
-reduce chromosome number from dipoloid to haploid, ensure that each haploid cell has a complete set of chromosomes, generate diversity amount daughter cells (hamate’s or spores)
Key Features
-2 nuclear divisions but DNA is replicated only once- begins in a diploid cell (Meiocyte) with all chromosomes in pairs, ends with haploid produces (4 possible)
-homologous chromosomes pair and exchange genetic information, then segregate from each other in Meiosis 1, sister chromatids separate from each other in meiosis 2

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

Uniques events of meiosis 1

A

Duplicated homologous Pairs of chromosomes come together and pair along their entire lengths
-paring occurs during prophase 1, it is called synapsids, the 4 chromatids of each homologous pair form a tetra d or bivalvet, can lead to crossing over between non-sister chromatids
After metaphase 1 the homologous pairs separate, maternal and paternal centromeres of each pair segregate to opposite poles, cells at the end of meiosis 1 are haploid but each chromosome still contains 2 chromatids

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

Sex and Meiosis learning goals

A

Meiosis has 2 consecutive nuclear divisions, resulting in daughter cells with half the number of chromosomes as the parent cell

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

Crossing over

A

Exchange of genetic material during prophase 1

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

Events of meiosis

A

Meiosis 2:
-duplicated cells at end of Meiosis 1 are haploid, but each chromosome still consists of 2 chromatids
-critical event of meiosis 2 is separation of the sister chromatids, similar to mitosis, sister chromatids segregate to opposite poles

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

Timing of events of meiosis

A

Prophase 1 may last a long time: males 1 wk-1 month, females: in utero, pause, resume at puberty

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

Nondisjunction

A

Homologous pairs fail to separate at Anaphase 1 or sister chromatids fail to separate at anaphase 2, either results in and upload y- chromosomes missing or presents in excess
Potential causes:
-aneuploidy is sometimes caused by lack of cohesion’s that hold the homologous pairs together. Without cohesions, both homologous segerate at random
-failure to undergo crossing over
-frequency of nondisjunction goes up as female ages

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
Q

Trisomic

A

If both homologs go to the same pole and the resulting egg is fertilized

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
26
Q

Monosomic

A

A fertilized egg that does not receive a copy of a particular chromosome

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
27
Q

Crossing over

A

Exchange between non sister chromatids produces recombination between DNA molecules

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
28
Q

Independent assortment

A

Haploid sets of chromosomes inherited from parents mixed by segregation of homologs during meiosis 1

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
29
Q

Meiosis, Mendel, and linkage learning goals

A

Segregation of chromosomes in meiosis accounts for mendel’s laws of segregation and independent assortment
Genes in physical proximity on the same chromosome exhibit linkage

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
30
Q

Mendel’s first law

A

The law of segregation: the two alleles of a gene separate and are transmitted individually and equally to gametes
A gene is a sequence within a DNA molecule and resides at a particular site on a chromosome(locus). Funcation of the gene influences characteristics of the organism. Because genes are shared by homologous chromosomes, different alleles segerate equally to gametes during meiosis
The transmisión of chromosome pairs (homologs) through meiosis is the mechanism

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
31
Q

Mendel’s second law

A

The law of independent assortment: alleles of different genes assort independently during gamete formation

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
32
Q

Linkage

A

Alleles of separate loci were transmitted together to offspring

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
33
Q

Recombinant Types

A

Nem combinations of alleles/phenotypes

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
34
Q

Meiosis has 2 consecutive nuclear divisions, resulting in daughter cells with half the number of chromosomes as the parent cell

A
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
35
Q

Linkage learning goals

A

Genes in physical proximity on the same chromosome exhibit linkage
The frequency of crossing over between linked genes is a measure of their relative distance
Sex chromosomes contain the gene that determine sex and exhibit unique patterns of inheritance

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
36
Q

Frequency of crossing over between two linked genes is proportional to the distance between them

A

Frequencies of recombinant gametes and resulting (non=parental) offspring are greater for loci that are further apart
Recombinant frequency = # of recombinant offspring/ total # of offspring
Maximal recombinant frequency is .5 also the expectation under independent assortment

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
37
Q

Absolute linkage

A

Rare
Even alleles of different loci that are very close on the same chrome se are sometimes recombined by crossing over

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
38
Q

Linkage group

A

All of the loci on a chromosome

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
39
Q

Genetic maps

A

Recombinant frequencies can be used to make this, showing the arrangement of genes along a chromosome

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
40
Q

Map unit

A

Distance between genes = 100x recombinant frequency
Also called a centimorgan (cM)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
41
Q

Sex chromosomes

A

Sex determination varies among species
In most dioecious organisms (2 sexes), sex is determined by a gene or genes
The gene with primary control of sexual development are present on the sex chromosomes
Other chromosomes are called autosomes

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
42
Q

SRY

A

Sex determing region on the Y- is the part of the Y chromosome that encodes a protein that initiates male development

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
43
Q

Nondisjunction of Sex Chromosomes

A

Sex chromosome abnormalities can result from nondisjunction in meiosis: pair of homologous chromosmes fail to separate in meiosis 1 or pair of sister chromatids fail to separate in meiosis 2
Result is aneuploidy- abnormal number of chromosomes
XO- the individual has only one sex chromose (Turner syndrome)
XXY- Klinefelter syndrome, affects males and results in sterility and overlong limbs

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
44
Q

Human sex chromosome

A

Genes on sex chromosmes exhibit sex-linked inheritance
The Y chromosome carries few genes; the X chromosome carries many genes involved in a variety of functions
Thus males have only one copy of the genes on the X (hemizygous) and express the phenotype of that allele

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
45
Q

X-linked recessive phenotypes

A

Appear much more often in males than females; heterozygous females are often CARRIERS
Phenotype can skip a generation if it passes from a male to his daughter and then grandson

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
46
Q

Genetics learning goals

A

Sex chromosomes contain the genes that determine sex and exhibit unique patterns of inheritance
Dominance is not always complete and it depends on the interaction between alleles
Alleles of different genes can interact to affect the phenotype
Genotype and phenotype of Mendelian traits are predictable

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
47
Q

Complete dominance

A

Heterozygotes appear similar to one of the homozygotes; used to define a dominate allele and a recessive allele

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
48
Q

Incomplete dominance

A

Sometimes heterozygoetes have an intermediate phenotype
You mix a red flower with a white flower and get a pink flower
One allele is insuffienct to produce the same phenotype as the two alleles of either homozygote, so the phenotype lies between

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
49
Q

Co dominance

A

Phenotypes of both alleles appear in the heterozygote
Each allele is expresses in the heterozygotes, think about the blood type example

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
50
Q

Many traits are influenced by the genotype of more than a single gene

A

Physical characteristics reflect underlying cellular functions, such as en y antic activity within biochemical pathways

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
51
Q

Epistasis

A

Phenotypic expression of one gene is influenced by genotype of another gene
Typical results in modification of the usual 9:3:3:1 ratio of di hybrid cross- 9:3:4
Think about dog coat color

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
52
Q

Probability Rules

A

Probablility of an event that can occur in two different (mutually exclusive) ways is the sum of the individual properties

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
53
Q

Single gene disorders

A

Most are rare in the general population
Caused ny a mutant allele of a single gene
The genetic change can result ina change in phenotype

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
54
Q

Single gene disorders

A

Most are rare in the general population
Caused ny a mutant allele of a single gene
The genetic change can result ina change in phenotype

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
55
Q

Recessive Disorders

A

Both alleles have to be mutant: albinism, CF, PKU,

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
56
Q

Dominate disorders

A

One mutant allele is enough: huntington disease, achondroplasia

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
57
Q

Frederick Griffith

A

Trying to find vaccine for pneumonia by isolating 2 types of bacteria smooth and rough(not dangerous) S strain had polysaccharide capsule around cells
Identified transforming principle

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
58
Q

Hershey and Chase Experiment

A

Used virus to determine where the genetic material was located by injecting radioactive phosphate and suffer
Conclusion: DNA contained the information needed to make the next generation of the phage

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
59
Q

DNA Structure/ Chargaff’s Rules

A

Determined that in DNA molecule the amounts of purines were present in equal amount to Pyrimidines, A=T and G=-C(triple H bond, more stable)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
60
Q

Rosalind Franklin

A

Used x-ray cystrallography to be able to display DNA and determine that DNA is a spiral or helical molecule and nitrogenous bases are interior

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
61
Q

Watson and Crick

A

Combined all knowledge about DNA to determine structure

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
62
Q

Antiparallel strands

A

Polarity of stand is determined by the sugar-phosphate bonds
Phosphate groups connect to the 3’C of on e sugar adn the 5’C of the next sugar
3’ end has hydroxial group

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
63
Q

Minor vs major groove

A

The minor groove does not have the nitrogenous bases exposed
Major groove is wider/more exposed as the nitrogenous bases are exposed

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
64
Q

Protein-DNA

A

Interactions plays a crucial role in many biological processes such as regulation of gene expression, DNA replication, repair, transportation, recombination, and packaging of chromosomal DNA

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
65
Q

4 key structures of DNA

A
  1. It is double stranded helix of uniform diameter
  2. It is right handed
  3. Strands in antiparallel orientation based on 5’ and 3’ carbons of deoxyribose sugar
  4. Outer edges of nitrogenous bases are exposed in the major and minor grooves
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
66
Q

DNA’s 4 important funcation

A

(Double- helical structure is essential)
-genetic material stores genetic information- millions of nucleotides; base sequences encodes huge amounts of information
-genetic material is susceptible to Mutation- change in information/message- possible through a simple alteration to a sequence
Genetic material is preciesely replicated in cell division- by complementary base pairing
- genetic material is expressed as the phenotype- nucleotide sequence determines sequence of amino acids in proteins
- the strucutre of DNA suggested a way in which the information in DNA might be copied so that it could be passed down to cells produced in mitosis and meiosis
- because of complementary base pairing, the information is contained in both strands; each strand can act as a template to make a new strand

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
67
Q

New nucleotides are added to the new strand at the 3’ end

A

Sequence is determined by it complementary base pairing with template strand

68
Q

Phosphodiester bond

A

Created by enzyme DNA, it is between internal phosphate at 5’ carbon and hydroxyl at 3’ carbon

69
Q

Semi conservative

A

Each parental strand is a template for a new strand

70
Q

Conservative

A

The two parental strands remain together in one daughter molecule while serving as a template for another daughter molecule

71
Q

Dispersive

A

Parent molecule is dispersed among both strands in the two daughter molecules

72
Q

The meselson-stahl experiment

A

Showed that DNA replication is semi conservative
If conservative the first generation of DNA molecules would have been both high and low density but no intermediate density
If dispersive: the density of all DNA molecules in the first generation would be intermediate but this density would not be present in subsequent generations; would shift closer to light

73
Q

3 steps of DNA replication

A

Initiation
Elongation
Termina nation

74
Q

Initiation

A

Unwinding (denaturing) the DNA double helix and synthesizing to RNA primers

75
Q

Elongation

A

Synthesizing new strands of DNA using each of the parental strands as templates
-synthesis occurs discontinuously in a series of fragments called Okazaki fragments

76
Q

Termination

A

Synthesis ends

77
Q

Replication Fork

A

The site where DNA unwinds to expose bases
One new strand, the leading strand, can grow continuously at its 3’ end as the fork opens
The other strand cannot be made that way
It is made in small pieces (Okazaki fragments) in the 5’ to 3’ direction
Cannot being until fork has advanced a little ways- is called the lagging strands
Is oriented so that its exposed 3’ end gets farther from the fork

78
Q

DNA Replication

A

DNA replication begins with a short primer- a stater stand of RNA complementrary to the DNA template
Required for leading strand and each Okazaki fragment
-primase (an enzyme) synthesizes the primer RNA; one nucleotide at a time
-DNA polymerase then add nucleotides to the 3’ end of teh RNA primer
-DNA polymerase is processive; catalyzes many polymerizations each time they bind to DNA; VERY RAPID

79
Q

DNA Replication: Initiation

A

The first DNA stands must be separated by an enzyme called DNA Helicase- it separates the two strands so the nucleotides are no longer base paired
As the strands separate a protein called topoisomerase protects the rest of the DNA molecule from being wound together
The stands will want to rejoin after separation- to prevent that there are single stranded binding proteins I between each base that prevents them from coming back together

80
Q

DNA Primase

A

Primase is the starter- it connects a few complementary RNA bases to the template strand- this is called an RNA Primer to the template stands

81
Q

DNA Replication: Elongation

A

DNA polymerase only works in one direction- 5’ to 3’

82
Q

Problems with DNA replication

A

Copying also proceeds in opposite direction due to anti-parallel strands

83
Q

DNA ligand is what bond

A

Covalent

84
Q

Lagging stand

A

Synthesis is opposite direction to fork movement and requires constant repriming

85
Q

DNA Replication: Termination

A

In eukaryotic cells occurs when the two replication forks meet
Last primer is removed from lagging strand, no DNA synthesis occurs because there is not 3’ end to extend a single strand bit of DNA is left at each end

86
Q

Telomeres

A

Are repetitive sequences (TTAGGG) at the ends of eukaryotic chromosomes. These are DNA sequences that do not encode proteins
Functions:
-telomeres protect the important protein coding DNA in the chromosome from being lost
-enxtend the chromosome and prevent coding regions of DNA from being cut off
Prevent DNA repair mechanism from mistakenly joining chromosome ends

87
Q

Telomerase

A

Adds telomeres back onto the end of the chromosome
It has RNA template that base pairs with the single stand overhang. Telomerase works as a DNA polymerase to synthesis new dna
If too many are lost cell undergoes apoptosis

88
Q

DNA Replication process

A

Proceeds at a replication fork requiring many proteins for distinct patterns of synthesis of leading and lagging strands
The polymerase chain reaction (PCR) generates a large number of copies of a targeted genome region through cycles of DNA replication

89
Q

Semi conservative DNA replication

A

A large muli-protein complex interacts with the parent (template) DNA stands
All chromosomes have at least one region called the ORIGIN of REPLICATION
Proteins in the replication complex bind to a DNA sequence in ori. Replication proceeds in both directions from there

90
Q

Replication fork

A

Bidirectional from origin
Leading and lagging strand in both directions

91
Q

Mutations sub objectives

A

Explain why DNA mutations are ultimately the source of phenotypic variation
Explain why many DNA mutations do not have a phenotypic effect

92
Q

Mutations

A

The changes of the nucleotide sequence of the genome (genetic materials) of an organism
Caused by any agents that damage DNA (UV, chemicals, viruses)
Mutations can alter phenotypes by: generating a non functional protein, changing how a protein works, altering when and where a gene is expressed

93
Q

Mutations

A

Mutations occur spontaneously at low rate, but can also be induced by mutagens
Mutations change the sequence of DNA and may alter or eliminate encoded proteins with or without phenotypic consequences

94
Q

Repair mechanisms

A
  1. Proofreading- by DNA polymerase
  2. Mismatch repair- corrects new strand
  3. Excision repair- removes damaged bases
95
Q

DNA proofreading

A

As DNA polymerase adds a nucleotide to a growing strand, it has a proofreading function
If bases are paired incorrectly the nucleotide is removed by DNA polymerase
DNA polymerase recognizes a mismatch, backs up, removes mismatched nucleotide, then recommences synthesis.

96
Q

Mismatch repair

A

The newly replicated DNA is scanned for mismatched bases, this mechanism recognizes old and new strands by modifications present on the template strand
If the repair fails to distinguish old and new strands, the DNA sequence may change
After replication, a protein complex scans for mismatched bases, recognized because of abnormal hydrogen bonding. The mismatched fragment is removed and replaced
A few mismatches will lead to base-pair substitutions, a type of point mutation: a single base is changed, inserted, or deleted
In some cases, mismatch may not be repaired before the next round of replication

97
Q

Excision repair

A

DNA can be damaged by radiation, chemicals in the environment, and random spontaneous chemical reactions
-Enzymes constantly scan dna for damaged bases- they are excises and DNA polymerase 1 adds the correct ones.

98
Q

Repair of UV Damage

A

Uv radiation (from sun or tanning beds) is absorbed by thymine, causing it to form covalent bonds with adjacent nucleotides- disrupts DNA replication
A mechanism stimulated by light recognizes and repairs Pyrimidines dimers

99
Q

Excision repair

A

Removes damaged nucleotides and replaces them with normal ones
Repair of thymine dimers (covalent linkages between adjacent thymines formed on exposure to UV radiation)
Dimers are converted back by photolyase which uses light

100
Q

Spontaneous mutations

A

Caused by polymerase errors or spontaneous chemical changes in bases
2 types that alter base-pairing properties:
Tautomeric shirft: bases have two isomers (tautomers) when a base temporarily forms its rare tautomer it can pair with a different base, leading to a mismatch
Deaminiation: loss of a NH2 group in cytosine, forming uracil
A will be inserted into the new DNA strand (A pairs with U) instead of G

101
Q

Mutagens

A

Induced mutations are caused by this,

102
Q

Somatic mutations

A

Occur in somatic(body) cells. They may have consequences for the phenotype of an individual but are not passed to offspring

103
Q

Germ line mutations

A

Occur in germ line cells (gametes) are passes to offspring and can have consequences for future generations

104
Q

Silent mutations

A

Do not add etc protein function

105
Q

Loss of function mutations

A

Prevent gene transcription or produce nonfunctional proteins: nearly always recessive

106
Q

Gain of funcation mutation

A

Lead to a protein with altered function
Usually dominate common in cancer cells

107
Q

Missense mutation

A

It is a point mutation in which a single nucleotide change results in a codon that codes for a different amino acid

108
Q

Conditional mutations:

A

Affect the phenotype only under certain environmental conditions
The wild type phenotype is expressed under other conditions

109
Q

Chromosomal mutations

A

Chromosomal mutations are extensive changes in genetic material involving long DNA sequences
They can provide genetic diversity important to evolution by natural selection, but they are often deleterious
Chromosomal rearrangements involve double-strand breaks
An aberrant crossover between homologous or nonhologlous chromosomes can lead to chromosomal rearrangements
Radiation can cause double-strand breaks; repair mechanisms may join non homologous ends

110
Q

Deletions- chromosomal mutations

A

Loss of chromosome segment can have sever or fatal consequences

111
Q

Duplications

A

A portion of a chromosome is replicated resulting in multiple copies

112
Q

Inversions

A

Result from breaking and rejoining, but the segment is “flipped”
Can result in loss of function mutation

113
Q

Translocations

A

Segment of DNA breaks off and is inserted into another chromosome; many involve reciprocal exchanges of chromosome segments

114
Q

Genes code for proteins

A

The early one gene to one enzyme relationship is most commonly expressed as the one gene one polypeptide relationship
Some genes involve controlling other genes
Some genes produce components of cellular structures; some genes code for functional rna and are not translated to polypeptides

115
Q

beanle and Tatum

A

Tested specific gene expression to specific enzyme activity
Discoveries:
-for each mutant strain, the addition of just one compound supported growth
-each mutation caused a defect in only one enzyme in a metabolic pathway
Three different arg mutant strains
-could have mutations in the same gene or in different genes that governed steps of a bio synthetic pathway

116
Q

Transcription

A

Copies information from a DNA sequence (a gene) to a complementary rna sequence

117
Q

Translation

A

Converts RNA sequence to amino acid sequence of a polypeptide

118
Q

Messenger RNA

A

Carries a copy of a DNA sequence to site of protein synthesis at the ribosome; has information for the order of amino acids in a protein

119
Q

Transfer RNA (tRNA)

A

Carries amino acids for polypeptide assembly; decodes the information in mRNA
The adapter molecule associates information in mRNA codons with specific amino acids
3 functions of tRNA:
-it binds to an amino acid and is then “charged”
-it associates with mRNA molecules
-it interacts with ribosomes
3’ end is the amino acid attachment site- binds covalently

120
Q

Ribosomal RNA (rRNA)

A

Catalyzes peptide bonds and provides structure
Doesn’t hold genetic information for making the protein

121
Q

RNA polymerase

A

Catalyze synthesis of RNA from a DNA template
CAN ONLY ADD NEW NUCLEOTIDES TO THE 3’ END OF A GROWING STRNAD- SYNTHESIS IS 5’ TO 3’
Like DNA polymerase- rna polymerase are processive- a single enzyme template binding results in polymerization of hundred of rna bases
Does not need primers but needs promoters
Lacks a proofreading function

122
Q

Transcription DNA to RNA

A

DNA template for base paintings one of the two strands of dna
Nucleoside triphosphates (ATP, GTP, CTP, UTP) as substrates
And RNA polymerase enzyme
Transcription factors (in eukaryotes)
3 phases: initiation, elongation, termination

123
Q

Initation of transcription

A

Requires a promoter- a special sequence of DNA
RNA polymerase binds to promoter
Promoter directs rna polymerase where to start and which direction to transcribe
Part of each promoter is the initation site of transcription

124
Q

Elongation

A

RNA polymerase unwinds dna about 10 base pairs at a time
Erase 3’ to 5’ direction
RNA TRANSCIPT IS ANTIPARALLEL TO THE TEMPLATE DNA STRAND; NUCLEOTIDES ADDED AT 3’ END
RNA polymerase do not proofread and correct mistakes

125
Q

Termination

A

Is specified by a base sequence in DNA that destabilizes the transcription complex
For some genes the transcript falls away from the rna polymerase and dna template for others a helper protein pulls it away

126
Q

Introns and exons

A

Introns- eukaryotic genes may have noncoding sequences
Appear in the primary mRNA transcript-pre-mRNA
Introns are removed from the pre-mRNA in the nucleus
The coding sequences are contained in teh exons that remain
The processed mRNA is exported from the nucleus to the cytoplasm

127
Q

RNA Splicing

A

Removes introns and splices exons others
Newly transcribed pre-mRNA is bound a ends by snRNPs- small nuclear ribonucleoprotein particles
Consensus sequences are short sequences at exon/intron junctions. snRNP binds here and also near the 3’ end of the intron

128
Q

G Cap

A

Modified guanosine Tri phosphate is added to the 5’ end
Facilitates the mRNA binding to ribosome
Protects mRNA from being degreased by ribosomes

129
Q

UTR

A

Untranslated region
Is part of exon- but not being translated

130
Q

Genetic coding

A

The genetic code is nearly universal: the codons that specify amino acids are the same in all organisms
This common genetic code is a common language for evaluation
The code is ancient and has remained intact throughout evolution
The common code also facilitates genetic engineering

131
Q

Degenerate codons

A

From most amino acids there is more than one codon; the genetic code is redundant or degenerate
The genetic code is NOT ambiguous- each codon specifies only one amino acid

132
Q

Conditional mutations

A

Cause phenotypes under restrictive conditions but are not detectable under permissive conditions

133
Q

Point mutation

A

Results from the gain, loss, or substitution of a single base pair of DNA
It may be silent or may alter the sequence of the resulting polypeptide
May change an amino acid or cause the loss of amino acid in the carboxyl terminus in the sequence of the resulting polypeptide

134
Q

Missense point mutation

A

At a non synonymous site changes a single amino acid
Gives different amino acid
Radical change

135
Q

Nonsense point mutation

A

Shortens polypeptide by causing premature termination of translation

136
Q

Loss-of-stop mutation

A

Causes read-through translation to new stop codon

137
Q

Frame shirt point mutation

A

Changes reading frame

138
Q

Anticodon

A

At the midpoint of the tRNA sequence- site of base pairing with mRNA
Unique for each species of tRNA

139
Q

Aminoacyl-tRNA synthetases

A

Activating enzymes- aminoacyl-tRNA synthetase- charge tRNA with correct amino acid
Each enzyme is highly specific for one amino acid and its corresponding tRNA; the process of tRNA charging is called teh second genetic code
The enzymes have three part active sites: they bind a specific amino acid, a specific tRNA and ATP

140
Q

Codon-anticodon wobble

A

Wobble: specificity for the base at teh 3’ end of the codon is not always observed
Wobble allows cells to produce fewer tRNA species, but does NOT create any ambiguity in the genetic code

141
Q

Ribsomes

A

“The workbench” holds mRNA and charged tRNA in the correct positions to allow assembly of polypeptide chain
-ribosomes have 2 subunits: large and small
- in eukaryotes- the large subunits has 3 molecules of ribsomal RNA (rRNA) and more proteins in a precise pattern. The small subunit has one rRNA and less proteins than the large subunit

142
Q

Stages of translation

A

Initiation: formation of initiation complex- a charged tRNA and small ribsomal subunit, both bound to mRNA
Elongation: charged tRNAs enter A site, large subunit acts as peptidyl transferase
Termination: stop codon enters the A site

143
Q

Elongation of translation

A

When the first tRNA hase released its methionine it moves to the E site and dissociates from the ribosome- can then become charged again
Occurs as the steps are repeated assisted by proteins called elongation factors

144
Q

Termination

A

Translation ends when a stop codon enters the A site
Spot codon binds a protein release factor- allows hydrolysis of bond between polypeptide chain and tRNA on the P site
Polypeptide chain separates from the ribosome- C terminus is the last amino acid added

145
Q

General purpose genes

A

Housekeeping
Needed by all cells
Genes for RNA and proteins involved in DNA replication, transcription, translation machinery and other basic functions but not expresses at all time of cell cycle

146
Q

Specialty function genes

A

Needed for response to specific environmental changes or for specialized cell (tissue) functions

147
Q

Control points of gene regulation

A

Transcriptional control: DNA accessibility (1) and transcription initiation (2)
Processing control: RNA processing (3)
Transport control: nuclear export (4) and mRNA stability(5)
Translation Control: translation (6) initation, elongation, and termination
Post-translational control: protein modification (7) and degradation (8)

148
Q

Genotype (DNA) to phenotype (funcational protein)

A

1- transcription control
2- RNA processing control
3- RNA transport control
4- translation control
5- protein activity control
(Prokaryotic cells are 1,4,5)

149
Q

Regulation of gene expression

A

Gene expression begins at the promoter where transcription is initationated (the promoter is where RNA polymerase binds and does its job- makes mRNA)
in selective gene transcription a “decision” is made about which genes to activate
Otherwise, constant gene transcription is know as constitutive expression

150
Q

Regulatory protein

A

Control expression of other genes; most gene are under the control of multiple regulatory proteins

Think about the turning on and off a sink analogy

151
Q

Regulatory protein activity types

A

Repressors and activators:

Negative regulation- binding of a repressor protein to DNA prevents transcription; transcription initation can occur in the absence of the repressor protein
Positive regulation- activator protein binds to DNA and stimulates transcription; transcription initation low in the absence of the activator protein

152
Q

Prokaryotic gene regulation

A

Prokaryotes generally stop synthesis of a protein when it is not needed. The cell can:
Repress mRNA transcription
Hydrolyze mRNA, preventing translation
Prevent mRNA translation at the ribosome
Hydrolyze (degrade) the protein after it is made
Inhibit the protein’s function (something blocking it)

153
Q

Uptake and metabolism of lactose involves 3 proteins:

A

Beta-galacotoside permease- a carrier protien that moves lactose into the cell
Beta-galactosidase- an enzyme that hydrolyses lactose
Beta- galactoside transacetylase- transfers acetyl groups to certain beta-galactosides

154
Q

The lac repressor and lac operan

A

A gene coding for the lac repressor protein; a negative regulator of the lac operon
The lac operon codes for 3 structural proteins needed for utilization of lactose

155
Q

The lac operon

A

The structural genes(encoded proteins) needed to utilize lactose are adjacent on the. E.coli chromosome, share a single promoter and are encoded on a single transcription (different sites of translation initation)
The operon is under coordinate control from the
Promoter- the region of DNA where RNA polymerase binders and initiates transcription
Operator- the region of DNA between the promoter and the structural genes that is bound by the lac repressor
A repressor protein (traffic cop) made by a different regulatory gene binds to the operator to block transcription of the operon when lactose is absent

156
Q

Orginal paradigm for negative control in prokaryotes

A

Repressor protein exerts negative control by blocking transcription when bound at the operator- repressor able to bind in the absence of the inducer allolactose and operon is turned off (repressed)
Inducer changes repressor protien so that it is unable to bind at the operator, operon available for transcription- when lactose is present operon is turned on (expressed)

157
Q

Transcriptional regulation in E.coli

A

Genes that encode proteins that are involved in the same metabolic pathway are organized in Operons
Two kinds of bacterial operons
Indecible- catabolic operons- substrate (the inducer) FOR A CATABOLIC ENZYME BINDS to repressor and changes it so it cannot bind the operator to transcription on
Repressable- anabolic operons- end product of the anabolic pathway acts as a co-repressor to allow repressor to bind operator and repress transcription= transcription off

158
Q

Transcription factors

A

Sequences at and near the promoter control transcription initiation through interactions with this protein
Binds to transcription complex
TF2D binds to the TATA box
Must assemble on the chromosome before RNA polymerase is recruited to the promoter

159
Q

Recognition sequence

A

Recognized by RNA polymerase

160
Q

TATA Box

A

Where DNA begins to unwind and expose the template stand- prokaryotes do not need this as they already do it on its own

161
Q

Enhancers and Silencers

A

Enhancers- positive regulators
Silencers- negative regulators (turns it off)
The combination of these determines the rate of transcription

162
Q

Epigenetic

A

Can be passes onto daughter cells but are reversible

163
Q

alternative splicing

A

When different mRNAs are made from the same gene
Introns and exons spliced differently, distinct proteins can be made
Can be a deliberate mechanism for generating proteins with different functions from a single gene

164
Q

Posttranslational

A

Phosphorylation: added phosphate groups alter the shape of the protein
Glycosylation: adding sugars is important for targeting and recognition
Proteolysis: cleaving the polypeptide allows the fragments to fold into different shapes (insulin)

165
Q

Points of eukaryotic gene regulation

A

Transcriptional control: regulates access (chromatin) and recruitment of RNA polymerase to the promoter in nucleus
Processing control: regulates splicing, capping, and tailing of pre-mRNA in nucleus
MRNA transport and stability: regulates nuclear export and localization in cytoplasm and 1/2 life of mRNA
Translation Control: regulates mRNA assembly with ribosome and polypeptide synthesis
Posttranslational processing: regulates protein activity through processing, folding, joining, and making chemical modifications