Exam 4 Flashcards

Question

backbone of DNA helix is __

Answer 1

sugar-phosphate backbone

Answer 2

- Nucleosomes wrapped into higher order coils called solenoids a) Leads to a fiber 30 nm in diameter b) Usual state of non-dividing (interphase) chromatin - During mitosis, chromatin in solenoid arranged around scaffold of protein to achieve maximum compaction a) Radial looping aided by condensin proteins

Answer 3

Nucleosomes wrapped into higher order coils

Answer 4

- Particular array of chromosomes in an individual organism a) Arranged according to size, staining properties, location of centromere, etc. - diploid (2n) - Haploid (n) - Pair of chromosomes are homologous a) Each one is a homologue

Answer 5

- 2n - Humans are diploid - 2 complete sets of chromosomes - 46 total chromosomes

Answer 6

- n - 1 set of chromosomes - 23 in humans

Answer 7

homologous

Answer 8

- Prior to replication, each chromosome composed of a single DNA molecule - After replication, each chromosome composed of 2 identical DNA molecules a) Held together by cohesin proteins - Visible as 2 strands held together as chromosome becomes more condensed a) One chromosome composed of 2 sister chromatids

Answer 9

cohesin proteins

Answer 10

INTERPHASE: 1) G1 (gap phase 1) 2) S (synthesis) 3) G2 (gap phase 2) - ------------------- 4) M (mitosis) 5) C (cytokinesis)

Answer 11

gap phase 1 - primary growth phase - longest phase

Answer 12

synthesis | - DNA replication

Answer 13

gap phase 2 - organelles replicate - microtubules organize - chromosomes coil more tightly using motor proteins - centrioles replicate to produce spindle fibers - tubulin synthesis occurs

Answer 14

mitosis | - subdivided into 5 phases

Answer 15

cytokinesis | - separation of 2 new cells

Answer 16

Time it takes to complete a cell cycle varies greatly - fruit fly embryos = replicate every 8 min - Mature cells take longer to grow a) typical mammalian cell takes 24 hrs b) liver cell takes over 1 year - Growth occurs during G1, G2, and S phases a) M phase takes abt 1 hr - Most variation in length of G1 a) resting phase (G0)

Answer 17

resting phase | - cells spend more or less time here

Answer 18

point of constriction - kinetochore - each sister chromatid has a centromere - chromatids stay attached at centromere by conhesin a) replaced by codensin in metazoans

Answer 19

attachment site for microtubules

Answer 20

PMAT 1. prophase 2. prometaphase 3. metaphase 4. anaphase 5. telophase

Answer 21

* packaging - Individual condensed chromosomes first become visible with the light microscope a) Condensation continues throughout prophase - Spindle apparatus assembles a) 2 centrioles move to opposite poles forming spindle apparatus (no centrioles in plants) b) Asters – radial array of microtubules in animals (not plants) - Nuclear envelope breaks down

Answer 22

- Transition occurs after disassembly of nuclear envelope - Microtubule attachment a) 2nd group grows from poles and attaches to kinetochores b) Each sister chromatid connected to opposite poles - Chromosomes begin to move to center of cell – congression a) Assembly and disassembly of microtubules b) Motor proteins at kinetochores

Answer 23

- alignment of chromosomes along metaphase plate a) not an actual structure b) future axis of cell division

Answer 24

- begins when centromeres split - key event is removal of cohesion proteins from all chromosomes - Sister chromatids pulled to opposite poles - 2 forms of movements a) Anaphase A b) Anaphase B

Answer 25

kinetochores pulled toward poles

Answer 26

poles move apart

Answer 27

- Spindle apparatus disassembles - Nuclear envelope forms around each set of sister chromatids a) Now called chromosomes - Chromosomes begin to uncoil - Nucleolus reappears in each new nucleus

Answer 28

- Cleavage of the cell into equal halves - Animal cells - Plant cells - Fungi and some protists

Answer 29

constriction of actin filaments produces a cleavage furrow

Answer 30

cell plate forms between the nuclei

Answer 31

nuclear membrane does not dissolve; mitosis occurs within the nucleus; division of the nucleus occurs with cytokinesis

Answer 32

- current view integrates 2 concepts 1) cell cycle has 2 irreversible points a) Replication of genetic material b) Separation of the sister chromatids 2) Cell cycle can be put on hold at specific points called checkpoints a) Process is checked for accuracy and can be halted if there are errors b) Allows cell to respond to internal and external signals

Answer 33

- maturation-promoting factor (MPF) discovered in frog oocytes - Cytoplasm containing MPF can induce cell division - MPF activity varies throughout the cell cycle - MPF enzymatic activity involved the phosphorylation of proteins

Answer 34

- Research in sea urchin embryos discovers proteins that are produced in synchrony with the cell cycle - Named cyclins - 2 forms 1) one that peaks at G1/S 2) another that peaks at G2/M

Answer 35

- Yeast mutants that halted during cell division were used to identify genes necessary for cell cycle progression - In yeast there were two critical control points: commitment to DNA synthesis, and commitment to mitosis - cdc2 gene is critical for passing both boundaries

Answer 36

boundaries

Answer 37

- Protein encoded by cdc2 was shown to be a kinase - MPF was shown to be composed of a cyclin and cdc2 kinase - cdc2 was first identified cyclin-dependent kinase (cdk) - Cdks are the key positive drivers of the cell cycle

Answer 38

1) G1/S checkpoint 2) G2/M checkpoint 3) late metaphase (spindle) checkpoint

Answer 39

- Cell “decides” to divide - cell is healthy - cell has all components needed - Primary point for external signal influence

Answer 40

- Cell makes a commitment to mitosis - has enzymes necessary to undergo mitosis - spindle fibers have started to form before mitosis - Assesses success of DNA replication

Answer 41

Cell ensures that all chromosomes are attached to the spindle

Answer 42

- Enzymes that phosphorylate proteins - Primary mechanism of cell cycle control - Cdks partner with different cyclins at different points in the cell cycle - For many years, a common view was that cyclins drove the cell cycle – that is, the periodic synthesis and destruction of cyclins acted as a clock - Now clear that Cdk itself is also controlled by phosphorylation

Answer 43

- Also called mitosis-promoting factor (MPF) - Activity of Cdk is also controlled by the pattern of phosphorylation a) Phosphorylation at one site (red) inactivates Cdk b) Phosphorylation at another site (green) activates Cdk

Answer 44

- Once thought that MPF was controlled solely by the level of the M phase-specific cyclins - Although M phase cyclin is necessary for MPF function, activity is controlled by inhibitory phosphorylation of the kinase component, Cdc2 - Damage to DNA acts through a complex pathway to tip the balance toward the inhibitory phosphorylation of MPF

Answer 45

- Also called cyclosome (APC/C) - At the spindle checkpoint, presence of all chromosomes at the metaphase plate and the tension on the microtubules between opposite poles are both important - Function of the APC/C is to trigger anaphase itself - Marks securin for destruction; no inhibition of separase; separase destroys cohesin

Answer 46

- Multiple Cdks control the cycle as opposed to the single Cdk in yeasts - Animal cells respond to a greater variety of external signals than do yeasts, which primarily respond to signals necessary for mating - More complex controls allow the integration of more input into control of the cycle

Answer 47

- Act by triggering intracellular signaling systems - Platelet-derived growth factor (PDGF) one of the first growth factors to be identified - PDGF receptor is a receptor tyrosine kinase (RTK) that initiates a MAP kinase cascade to stimulate cell division - Growth factors can override cellular controls that otherwise inhibit cell division

Answer 48

Unrestrained, uncontrolled growth of cells - Failure of cell cycle control - Two kinds of genes can disturb the cell cycle when they are mutated 1) Tumor-suppressor genes 2) Proto-oncogenes

Answer 49

- p53 plays a key role in G1 checkpoint - p53 protein monitors integrity of DNA a) If DNA damaged, cell division halted and repair enzymes stimulated b) If DNA damage is irreparable, p53 directs cell to kill itself - Prevent the development of many mutated cells - p53 is absent or damaged in many cancerous cells

Answer 50

- Normal cellular genes that become oncogenes when mutated a) Oncogenes can cause cancer - Some encode receptors for growth factors If receptor is mutated in “on,” cell no longer depends on growth factors - Some encode signal transduction proteins - Only one copy of a proto-oncogene needs to undergo this mutation for uncontrolled division to take place

Answer 51

- p53 gene and many others - Both copies of a tumor-suppressor gene must lose function for the cancerous phenotype to develop - First tumor-suppressor identified was the retinoblastoma susceptibility gene (Rb) a) Predisposes individuals for a rare form of cancer that affects the retina of the eye

Answer 52

- Inheriting a single mutant copy of Rb means the individual has only one “good” copy left a) During the hundreds of thousands of divisions that occur to produce the retina, any error that damages the remaining good copy leads to a cancerous cell b) Single cancerous cell in the retina then leads to the formation of a retinoblastoma tumor - Rb protein integrates signals from growth factors Role to bind important regulatory proteins and prevent stimulation of cyclin or Cdk production

Answer 53

1 from mom and 1 from dad

Answer 54

all chromosomes except the sex chromosomes (XX or XY)

Answer 55

person has 3 copies of chromosome 21 instead of 2, resulting in Down Sydrome

Answer 56

2 strands on 1 chromosome | ex: 2 sister chromatids for dad chromosome and 2 sister chromatids for mom chromosome make 1 joined chromosome

Answer 57

quickly bc skin on surface is dead and is constantly doing mitosis

Answer 58

interphase

Answer 59

IPMAT - interphase - prophase - prometaphase - metaphase - anaphase - telophase

Answer 60

we go from 46 chromosomes to 92 chromosomes

Answer 61

suppress formation of tumors - maintain cell health - look at replicated DNA and will repair damage to DNA and will allow cells to work in cell cycle

Answer 62

cell death

Answer 63

turn on the cell cycle

Answer 64

- Made up of meiosis and fertilization - Diploid cells a) Somatic (nonreproductive) cells of adults have 2 sets of chromosomes - Haploid cells a) Gametes (eggs and sperm) have only 1 set of chromosomes - Offspring inherit genetic material from 2 parents

Answer 65

non-sexual

Answer 66

- Reproduction that involves an alternation of meiosis and fertilization is called sexual reproduction - Some life cycles include longer diploid phases, some include longer haploid phases - In most animals, diploid state dominates a) Zygote first undergoes mitosis to produce diploid cells b) Later in the life cycle, some of these diploid cells (called germ-line cells) undergo meiosis to produce haploid gametes

Answer 67

a fertilized egg

Answer 68

sperm + egg -> fertilization -> zygote -> mitosis

Answer 69

fertilization

Answer 70

- Meiosis includes two rounds of division a) Meiosis I and meiosis II b) Each has prophase, metaphase, anaphase, and telophase stages - Synapsis

Answer 71

- During early prophase I - Homologous chromosomes become closely associated - Includes formation of synaptonemal complexes a) Complexes also called tetrad or bivalents

Answer 72

- While homologues are paired during prophase I genetic recombination, or crossing over, occurs Sites of crossing over are called chiasmata - First meiotic division is termed the “reduction division” a) Results in daughter cells that contain one homologue from each chromosome pair - No DNA replication between meiotic divisions - Second meiotic division does not further reduce the number of chromosomes a) Separates the sister chromatids for each homologue

Answer 73

- 2 phases: meiosis I and meiosis II - Meiotic cells have an interphase period that is similar to mitosis with G1, S, and G2 phases - After interphase, germ-line cells enter meiosis I

Answer 74

- Prophase I - Metaphase I - Anaphase I - Telophase I

Answer 75

- Prophase II - Metaphase II - Anaphase II - Telophase II

Answer 76

- Chromosomes coil tighter and become visible, nuclear envelope disappears, spindle forms - Each chromosome composed of 2 sister chromatids - Synapsis a) Homologues become closely associated b) Sister chromatids of each homologue are also joined by the cohesin complex in a process called sister chromatid cohesion

Answer 77

- Genetic recombination between non-sister chromatids - Allows the homologues to exchange chromosomal material - Alleles of genes that were formerly on separate homologues can now be found on the same homologue - Chiasmata – site of crossing over a) Contact maintained until anaphase

Answer 78

site of crossing over | a) Contact maintained until anaphase

Answer 79

non-sister chromatids

Answer 80

* tetrads lined up in middle - Paired homologues locked together following crossing over - Microtubules from opposite poles attach to each homologue a) Not each sister chromatid as in mitosis - Homologues are aligned at the metaphase plate side-by-side - Orientation of each pair of homologues on the spindle is random

Answer 81

* pairs of sister chromatids separated to opp poles - Microtubules of the spindle shorten a) Chiasmata break - Homologues are separated from each other and move to opposite poles a) Sister chromatids remain attached to each other at their centromeres - Each pole has a complete haploid set of chromosomes consisting of one member of each homologous pair - Independent assortment of maternal and paternal chromosomes

Answer 82

- Nuclear envelope re-forms around each daughter nucleus - Sister chromatids are no longer identical because of crossing over (prophase I) - Cytokinesis may or may not occur after telophase I - Meiosis II occurs after an interval of variable length

Answer 83

- Resembles a mitotic division - Prophase II - Metaphase II - Anaphase II - Telophase II - cytokinesis follows

Answer 84

- Four cells containing haploid sets of chromosomes - In animals, develop directly into gametes - In plants, fungi, and many protists, divide mitotically a) Produce greater number of gametes b) Adults with varying numbers of chromosome sets

Answer 85

- Nondisjunction - Aneuploid gametes a) Most common cause of spontaneous miscarriage in humans

Answer 86

- Meiosis is characterized by 4 distinct features: 1) Homologous pairing (synapsis) and crossing over 2) Sister chromatids remain joined at their centromeres and segregate together during anaphase I 3) Kinetochores of sister chromatids attach to the same pole in meiosis I 4) DNA replication is suppressed between meiosis I and meiosis II

Answer 87

failure of chromosomes to move to opposite poles during either meiotic division

Answer 88

gametes with missing or extra chromosomes | * most common cause of spontaneous abortion in humans

Answer 89

nuclear envelopes dissolve and new spindle apparatus forms

Answer 90

chromosomes align on metaphase plate

Answer 91

sister chromatids are separated from each other

Answer 92

nuclear envelope re-forms around 4 sets of daughter chromosomes; cytokinesis follows

Answer 93

- How homologues find each other and become aligned is not understood - Sister chromatid cohesion is similar to that in mitosis, but involves meiosis-specific cohesin proteins - Synaptonemal complex proteins have been identified in diverse species, but proteins show little sequence conservation

Answer 94

- Key distinction between meiosis and mitosis is the maintenance of sister chromatid cohesion at the centromere during all of meiosis I, but the loss of cohesion from the chromosome arms during anaphase I - Protein called Shugoshin protects cohesin from separase-mediated cleavage during meiosis I

Answer 95

- Kinetochores of sister chromatids must be attached to the same pole in meiosis I, in contrast to mitosis and meiosis II - Basis of mono-polar attachment seems to be based on structural differences between centromere-kinetochore complexes in meiosis I and in mitosis

Answer 96

- Detailed mechanisms of suppression of replication between meiotic divisions is not known - Cyclin B is lost completely between mitotic divisions, but not between meiotic divisions - Seems to prevent replication initiation complexes from forming

Answer 97

- Because of random orientation of chromosomes at first meiotic division, and because of crossing over, meiosis rarely produces cells that are identical - Resulting variation is essential for evolution

Answer 98

sperm and egg

Answer 99

2 diploid daughter cells; 4 haploid cells

Answer 100

haploid; diploid

Answer 101

aneuploid gametes

Answer 102

short chromosome; an extra chromosome at chromosome 21 | - as women age, the chances of bearing a child with DS increases

Answer 103

aneuploidy in the gametes that results in defects in the zygote - Klinefelter syndrome in Males - Turner syndrome in Females

Answer 104

- webbed neck - short stature - in women

Answer 105

One or two alternative forms of a gene

Answer 106

2 of the same allele

Answer 107

2 diff alleles

Answer 108

Expression of a characteristic or heritable feature passed from parent to offspring

Answer 109

Transmission of characteristics from parent to offspring

Answer 110

Observable expression of a trait | *physical

Answer 111

Total of all genes present in the cells of an organism, also used to describe a set of alleles at a single gene locus

Answer 112

- Before the 20th century, 2 concepts were the basis for ideas about heredity a) Heredity occurs within species b) Traits are transmitted directly from parent to offspring - Thought traits were borne through fluid and blended in offspring - Paradox – if blending occurs why don’t all individuals look alike?

Answer 113

Chose to study pea plants because: - Other research showed that pea hybrids could be produced - Many pea varieties were available - Peas are small plants and easy to grow - Peas can self-fertilize or be cross-fertilized

Answer 114

- Usually 3 stages 1) Produce true-breeding strains for each trait he was studying 2) Cross-fertilize true-breeding strains having alternate forms of a trait a) Also perform reciprocal crosses 3) Allow the hybrid offspring to self-fertilize for several generations and count the number of offspring showing each form of the trait

Answer 115

- Cross to study only 2 variations of a single trait a) Dominant alleles determines the appearance in a heterozygote b) Recessive alleles are masked in a heterozygote by the presence of the dominant allele - Mendel produced true-breeding pea strains for 7 different traits a) Each trait had 2 variants

Answer 116

``` Flower color - *Purple or White Pea color - *Yellow or Green Pea shape – *Round or Wrinkled Pod color – *Green or Yellow Pod shape – *Inflated or Constricted Flower position - *Axial (middle) or Terminal (end) Plant height – *Tall or Short ```

Answer 117

- First filial generation - Offspring produced by crossing 2 true-breeding strains - For every trait Mendel studied, all F1 plants resembled only 1 parent a) Referred to this trait as dominant b) Alternative trait was recessive - No plants with characteristics intermediate between the 2 parents were produced

Answer 118

- Second filial generation - Offspring resulting from the self-fertilization of F1 plants - Although hidden in the F1 generation, the recessive trait had reappeared among some F2 individuals - Counted proportions of traits a) Always found about 3:1 ratio

Answer 119

determines the appearance in a heterozygote

Answer 120

masked in a heterozygote by the presence of the dominant allele

Answer 121

- F2 plants a) ¾ plants with the dominant form b) ¼ plants with the recessive form c) The dominant to recessive ratio was 3:1 - Mendel discovered the ratio is actually: a) 1 true-breeding dominant plant b) 2 not-true-breeding dominant plants c) 1 true-breeding recessive plant

Answer 122

phenotype: 3:1 genotype: 1:2:1

Answer 123

- His plants did not show intermediate traits a) Each trait is intact, discrete - For each pair, one trait was dominant, the other recessive - Pairs of alternative traits examined were segregated among the progeny of a particular cross - Alternative traits were expressed in the F2 generation in the ratio of ¾ dominant to ¼ recessive

Answer 124

- Parents transmit discrete factors (genes) - Each individual receives one copy of a gene from each parent - Not all copies of a gene are identical a) Allele – alternative form of a gene b) Homozygous – 2 of the same allele c) Heterozygous – different alleles - Alleles remain discrete – no blending - Presence of allele does not guarantee expression a) Dominant allele – expressed b) Recessive allele – hidden by dominant allele - Genotype – total set of alleles an individual contains - Phenotype – physical appearance

Answer 125

- Two alleles for a gene segregate during gamete formation and are rejoined at random, one from each parent, during fertilization - physical basis for allele segregation is the behavior of chromosomes during meiosis - Mendel had no knowledge of chromosomes or meiosis – had not yet been described

Answer 126

- Cross purple-flowered plant with white-flowered plant - P (uppercase) is dominant allele – purple flowers - p (lowercase) is recessive allele – white flowers - True-breeding white-flowered plant is pp a) Homozygous recessive - True-breeding purple-flowered plant is PP a) Homozygous dominant - Pp is heterozygote purple-flowered plant

Answer 127

homozygous

Answer 128

homozygous recessive

Answer 129

heterozygous

Answer 130

1) Human genome 2) Chromosome #11 3) Hemoglobin gene 4) Primary Hb RNA transcript 5) Hb Messenger RNA 6) Hb Polypeptide 7) Folded Hemoglobin protein 8) Breathing

Answer 131

- Some human traits are controlled by a single gene a) Some of these exhibit dominant and recessive inheritance - Pedigree analysis is used to track inheritance patterns in families Dominant pedigree a) Dominant trait appears in every generation

Answer 132

- affected: colored - carrier: half-colored - unaffected: blank - female: circle - male: square

Answer 133

huntington disease

Answer 134

- Condition in which the pigment melanin is not produced - Pedigree for form of albinism due to a nonfunctional allele of the enzyme tyrosinase - Males and females affected equally - Most affected individuals have unaffected parents

Answer 135

- A change (mutation) in a single nucleotide can cause a huge effect on the function of a protein that can cause disease to occur - In the case of Sickle Cell Disease a single nucleotide substitution that leads to amino acid changes of glutamic acid to valine, this negatively affects that structure and function of the hemoglobin molecule

Answer 136

- Examination of 2 separate traits in a single cross - Produced true-breeding lines for 2 traits - RRYY x rryy - The F1 generation of a dihybrid cross (RrYy) shows only the dominant phenotypes for each trait - Allow F1 to self-fertilize to produce F2 - F1 self-fertilizes - RrYy x RrYy - The F2 generation shows all four possible phenotypes in a set ratio a) 9:3:3:1 b) R_Y_:R_yy:rrY_:rryy c) Round yellow:round green:wrinkled yellow:wrinkled green

Answer 137

Gametes located on different chromosomes are inherited independently of one another Thus 9:3:3:1 ratio

Answer 138

- In a dihybrid cross, the alleles of each gene assort independently - The segregation of different allele pairs is independent - Independent alignment of different homologous chromosome pairs during metaphase I leads to the independent segregation of the different allele pairs

Answer 139

- Rule of addition | - rule of multiplication

Answer 140

a) Probability of 2 mutually exclusive events occurring simultaneously is the sum of their individual probabilities - When crossing Pp x Pp, the probability of producing Pp offspring is a) probability of obtaining Pp (1/4), PLUS probability of obtaining pP (1/4) b) ¼ + ¼ = ½

Answer 141

- Probability of 2 independent events occurring simultaneously is the product of their individual probabilities - When crossing Pp x Pp, the probability of obtaining pp offspring is a) Probability of obtaining p from father = ½ b) Probability of obtaining p from mother = ½ c) Probability of pp= ½ x ½ = ¼

Answer 142

- Cross used to determine the genotype of an individual with dominant phenotype - Cross the individual with unknown genotype (e.g. P_) with a homozygous recessive (pp) - Phenotypic ratios among offspring are different, depending on the genotype of the unknown parent

Answer 143

- Mendel’s model of inheritance assumes that a) Each trait is controlled by a single gene b) Each gene has only 2 alleles c) There is a clear dominant-recessive relationship between the alleles * *Most genes do not meet these criteria

Answer 144

- Occurs when multiple genes are involved in controlling the phenotype of a trait - The phenotype is an accumulation of contributions by multiple genes - These traits show continuous variation and are referred to as quantitative traits - For example – human height - Histogram shows normal distribution

Answer 145

- Refers to an allele which has more than one effect on the phenotype - Pleiotropic effects are difficult to predict, because a gene that affects one trait often performs other, unknown functions - This can be seen in human diseases such as cystic fibrosis or sickle cell anemia a) Multiple symptoms can be traced back to one defective allele

Answer 146

- May be more than 2 alleles for a gene in a population - ABO blood types in humans a) 3 alleles - Each individual can only have 2 alleles - Number of alleles possible for any gene is constrained, but usually more than two alleles exist for any gene in an outbreeding population

Answer 147

Heterozygote is intermediate in phenotype between the 2 homozygotes A) Red flowers x white flowers = pink flowers

Answer 148

- The system demonstrates both a) Multiple alleles 3 alleles of the I gene (I^A, I^B, and i) b) Codominance I^A and I^B are dominant to i but codominant to each other c) Codominance -Heterozygote shows some aspect of the phenotypes of both homozygotes -Type AB blood

Answer 149

alleles: I^A I^A, I^Ai sugars exhibited: galactosamine receives A and O donates A and AB

Answer 150

alleles: I^B I^B, I^Bi sugars exhibited: galactose receives B and O donates B and AB

Answer 151

alleles: I^A I^B sugars exhibited: galactose and galactosamine universal receiver donates AB

Answer 152

alleles: ii sugars exhibited: none receives O universal donor

Answer 153

- Coat color in Himalayan rabbits and Siamese cats | a) Allele produces an enzyme that allows pigment production only at temperatures below 30ºC

Answer 154

- Behavior of gene products can change the ratio expected by independent assortment, even if the genes are on different chromosomes that do exhibit independent assortment - R.A. Emerson crossed 2 white varieties of corn a) F1 was all purple b) F2 was 9 purple:7 white – not expected

Answer 155

- First suggests central role for chromosomes | - One of papers announcing rediscovery of Mendel’s work

Answer 156

- Chromosomal theory of inheritance | - Based on observations that similar chromosomes paired with one another during meiosis

Answer 157

- Working with fruit fly, Drosophila melanogaster - Discovered a mutant male fly with white eyes instead of red - Crossed the mutant male to a normal red-eyed female a) All F1 progeny red eyed = dominant trait

Answer 158

- TH Morgan - Morgan crossed F1 females x F1 males - F2 generation contained red and white- eyed flies a) But all white-eyed flies were male - Testcross of a F1 female with a white-eyed male showed the viability of white-eyed females - Morgan concluded that the eye color gene resides on the X chromosome

Answer 159

- Sex determination in Drosophila is based on the number of X chromosomes a) 2 X chromosomes = female b) 1 X and 1 Y chromosome = male - Sex determination in humans is based on the presence of a Y chromosome a) 2 X chromosomes = female b) Having a Y chromosome (XY) = male * *male determines sex

Answer 160

- Humans have 46 total chromosomes a) 22 pairs are autosomes b) 1 pair of sex chromosomes c) Y chromosome highly condensed Recessive alleles on male’s X have no active counterpart on Y d) “Default” for humans is female Requires SRY gene on Y for “maleness”

Answer 161

- Certain genetic diseases affect males to a greater degree than females - X-linked recessive alleles a) Red-green color blindness b) Hemophilia

Answer 162

- Disease that affects a single protein in a cascade of proteins involved in the formation of blood clots - Form of hemophilia is caused by an X-linked recessive allele a) Heterozygous females are asymptomatic carriers - Allele for hemophilia was introduced into a number of different European royal families by Queen Victoria of England

Answer 163

- Ensures an equal expression of genes from the sex chromosomes even though females have 2 X chromosomes and males have only 1 - In each female cell, 1 X chromosome is inactivated and is highly condensed into a Barr body - Females heterozygous for genes on the X chromosome are genetic mosaics - Calico cat

Answer 164

- Mitochondria and chloroplasts contain genes - Traits controlled by these genes do not follow the chromosomal theory of inheritance - Genes from mitochondria and chloroplasts are often passed to the offspring by only one parent (mother) a) Maternal inheritance - In plants, the chloroplasts are often inherited from the mother, although this is species dependent

Answer 165

- Early geneticists realized that they could obtain information about the distance between genes on a chromosome - Based on genetic recombination (crossing over) between genes - If crossover occurs, parental alleles are recombined producing recombinant gametes

Answer 166

- Undergraduate in T.H. Morgan’s lab - Put Morgan’s observation that recombinant progeny reflected relevant location of genes in quantitative terms - As physical distance on a chromosome increases, so does the probability of recombination (crossover) occurring between the gene loci

Answer 167

- The distance between genes is proportional to the frequency of recombination events recombination frequency = recombinant progeny / total progeny - 1% recombination = 1 map unit (m.u.) - 1 map unit = 1 centimorgan (cM) - Perform testcross with doubly heterozygous individuals; count progeny

Answer 168

- If homologues undergo two crossovers between loci, then the parental combination is restored - Leads to an underestimate of the true genetic distance - Relationship between true distance on a chromosome and the recombination frequency is not linear

Answer 169

- Uses 3 loci instead of 2 to construct maps - Gene in the middle allows us to see recombination events on either side - In any three-point cross, the class of offspring with two crossovers is the least frequent class - In practice, geneticists use three-point crosses to determine the order of genes, then use data from the closest two-point crosses to determine distances

Answer 170

- Data derived from historical pedigrees - Difficult analysis a) Number of markers was not dense enough for mapping up to 1980s b) Disease-causing alleles rare - Situation changed with the development of anonymous markers a) Detected using molecular techniques b) No detectable phenotype

Answer 171

- Single-nucleotide polymorphisms - Affect a single base of a gene locus Used to increase resolution of mapping - Used in forensic analysis a) Help eliminate or confirm crime suspects or for paternity testing

Answer 172

- First human disease shown to be the result of a mutation in a protein - Caused by a defect in the oxygen carrier molecule, hemoglobin a) Leads to impaired oxygen delivery to tissues - Homozygotes for sickle cell allele exhibit intermittent illness and reduced life span - Heterozygotes appear normal a) Do have hemoglobin with reduced ability - Sickle cell allele is particularly prevalent in people of African descent a) Proportion of heterozygotes higher than expected b) Confers resistance to blood-borne parasite that causes malaria

Answer 173

- Failure of homologues or sister chromatids to separate properly during meiosis - Aneuploidy – gain or loss of a chromosome a) Monosomy – loss b) Trisomy – gain c) In all but a few cases, do not survive

Answer 174

gain or loss of a chromosome - monosomy - trisomy

Answer 175

loss of chromosome

Answer 176

gain of chromosome

Answer 177

- Smallest autosomes can present as 3 copies and allow individual to survive a) 13, 15, 18, 21 and 22 b) 13, 15, 18 – severe defects, die within a few months c) 21 and 22 – can survive to adulthood - Down Syndrome – trisomy 21 a) May be a full, third 21st chromosome b) May be a translocation of a part of chromosome 21 c) Mother’s age influences risk

Answer 178

- Do not generally experience severe developmental abnormalities - Individuals have somewhat abnormal features, but often reach maturity and in some cases may be fertile - XXX – triple-X females - XXY – males (Klinefelter syndrome) - XO – females (Turner syndrome) - OY – nonviable zygotes - XYY – males (Jacob syndrome)

Answer 179

- Phenotype exhibited by a particular allele depends on which parent contributed the allele to the offspring - Specific partial deletion of chromosome 15 results in a) Prader-Willi syndrome if the chromosome is from the father b) Angelman syndrome if it’s from the mother

Answer 180

- Imprinting is an example of epigenetics a) epigenetic inheritance b) no alteration in the DNA sequence c) DNA methylation d) alterations to proteins involved in chromosome structure

Answer 181

- Pedigree analysis - Amniocentesis - Chorionic villi sampling

Answer 182

collects fetal cells from the amniotic fluid for examination

Answer 183

collects cells from the placenta for examination

Answer 184

used to determine the probability of genetic disorders in the offspring

Answer 185

- Studied Streptococcus pneumoniae, a pathogenic bacterium causing pneumonia - 2 strains of Streptococcus 1) S strain is virulent 2) R strain is nonvirulent - Griffith infected mice with these strains hoping to understand the difference between the strains

Answer 186

- Live S strain cells killed the mice - Live R strain cells did not kill the mice - Heat-killed S strain cells did not kill the mice - Heat-killed S strain + live R strain cells killed the mice

Answer 187

- Information specifying virulence passed from the dead S strain cells into the live R strain cells - Our modern interpretation is that genetic material was actually transferred between the cells

Answer 188

- Repeated Griffith’s experiment using purified cell extracts - Removal of all protein from the transforming material did not destroy its ability to transform R strain cells - DNA-digesting enzymes destroyed all transforming ability - Supported DNA as the genetic material

Answer 189

- Investigated bacteriophages a) Viruses that infect bacteria - Bacteriophage was composed of only DNA and protein - Wanted to determine which of these molecules is the genetic material that is injected into the bacteria

Answer 190

- Bacteriophage DNA was labeled with radioactive phosphorus (32P) - Bacteriophage protein was labeled with radioactive sulfur (35S) - Radioactive molecules were tracked - Only the bacteriophage DNA (as indicated by the 32P) entered the bacteria and was used to produce more bacteriophage - Conclusion: DNA is the genetic material

Answer 191

- DNA is a nucleic acid - Composed of nucleotides a) 5-carbon sugar called deoxyribose b) Phosphate group (PO4) 1. Attached to 5′ carbon of sugar 2. Nitrogenous base Adenine, thymine, cytosine, guanine 3. Free hydroxyl group (—OH) - Attached at the 3′ carbon of sugar

Answer 192

- Bond between adjacent nucleotides - Formed between the phosphate group of one nucleotide and the 3′ —OH of the next nucleotide - The chain of nucleotides has a 5′-to-3′ orientation

Answer 193

- Erwin Chargaff determined that a) Amount of adenine = amount of thymine b) Amount of cytosine = amount of guanine c) Always an equal proportion of purines (A and G) and pyrimidines (C and T)

Answer 194

- Performed X-ray diffraction studies to identify the 3-D structure a) Discovered that DNA is helical b) Using Maurice Wilkins ’DNA fibers, discovered that the molecule has a diameter of 2 nm and makes a complete turn of the helix every 3.4 nm

Answer 195

- Deduced the structure of DNA using evidence from Chargaff, Franklin, and others - Did not perform a single experiment themselves related to DNA - Proposed a double helix structure

Answer 196

- Phosphodiester backbone: repeating sugar and phosphate units joined by phosphodiester bonds - A single strand extends in a 5′ to 3′ direction

Answer 197

repeating sugar and phosphate units joined by phosphodiester bonds

Answer 198

- 2 strands are polymers of nucleotides - Arranged as a double helix - Wrap around 1 axis - Antiparallel strands

Answer 199

- Complementarity of bases - A forms 2 hydrogen bonds with T - G forms 3 hydrogen bonds with C - Gives consistent diameter

Answer 200

- Conservative model - Semiconservative model - Dispersive model

Answer 201

both strands of parental duplex remain intact; new DNA copies consist of all new molecules

Answer 202

daughter strands each consist of one parental strand and one new strand

Answer 203

new DNA is dispersed throughout each strand of both daughter molecules after replication

Answer 204

- Bacterial cells were grown in a heavy isotope of nitrogen, 15N - After several generations, the DNA of these bacteria was denser than normal DNA - Cells were switched to media containing lighter 14N - DNA was extracted from the cells at various time intervals

Answer 205

- Conservative model = rejected a) 2 densities were not observed after round 1 - Semiconservative model = supported a) Consistent with all observations b) 1 band after round 1 c) 2 bands after round 2 - Dispersive model = rejected a) 1st round results consistent b) 2nd round – did not observe 1 band

Answer 206

- Requires 3 things 1. Something to copy - Parental DNA molecule 2. Something to do the copying - Enzymes 3. Building blocks to make copy - Nucleotide triphosphates

Answer 207

1) Initiation 2) Elongation 3) Termination

Answer 208

replication begins

Answer 209

new strands of DNA are synthesized by DNA polymerase

Answer 210

replication is terminated

Answer 211

- Matches existing DNA bases with complementary nucleotides and links them - All have several common features a) Add new bases to 3′ end of existing strands b) Synthesize in 5′-to-3′ direction c) Require a primer of RNA

Answer 212

- E. coli model - Single circular molecule of DNA - Replication begins at the origin of replication - Proceeds in both directions around the chromosome - Replicon: DNA controlled by an origin

Answer 213

DNA controlled by an origin

Answer 214

- DNA polymerase I (pol I) - DNA polymerase II (pol II) - DNA polymerase III (pol III) - All 3 have 3′-to-5′ exonuclease activity – proofreading - DNA pol I has 5′-to-3′ exonuclase activity – removing RNA primers

Answer 215

Acts on lagging strand to remove primers and replace them with DNA - has 5′-to-3′ exonuclase activity- removing RNA primers

Answer 216

Involved in DNA repair processes

Answer 217

Main replication enzyme

Answer 218

removing RNA primers

Answer 219

torsional strain

Answer 220

use energy from ATP to unwind DNA

Answer 221

coat strands to keep them apart

Answer 222

enzymes that prevent supercoiling; DNA gyrase is the topoisomerase involved in DNA replication

Answer 223

5 prime to 3 prime

Answer 224

synthesized continuously from an initial primer

Answer 225

synthesized discontinuously with multiple priming events | - DNA fragments on the lagging strand are called Okazaki fragments

Answer 226

DNA fragments on the lagging strand

Answer 227

- Partial opening of helix forms replication fork - DNA primase – RNA polymerase that makes RNA primer a) Later RNA will be removed and replaced with DNA

Answer 228

RNA polymerase that makes RNA primer

Answer 229

- Single priming event - Strand extended by DNA pol III a) Processivity – ß subunit forms “sliding clamp” to keep it attached to DNA

Answer 230

ß subunit forms “sliding clamp” to keep it attached to DNA

Answer 231

- Discontinuous synthesis a) DNA pol III - RNA primer made by primase for each Okazaki fragment - All RNA primers removed and replaced by DNA a) DNA pol I - Backbone sealed a) DNA ligase - Termination occurs at specific site a) DNA gyrase unlinks 2 copies

Answer 232

- Enzymes involved in DNA replication form a macromolecular assembly - 2 main components 1) Primosome - Primase, helicase, accessory proteins 2) Complex of 2 DNA pol III - One for each strand

Answer 233

- Complicated by a) Larger amount of DNA in multiple chromosomes b) Linear structure - Basic enzymology is similar a) Requires new enzymatic activity for dealing with ends only

Answer 234

1 circular DNA in chromosome

Answer 235

Prophase 1

Answer 236

- Multiple replicons – multiple origins of replications for each chromosome a) Not sequence specific; can be adjusted - Initiation phase of replication requires more factors to assemble both helicase and primase complexes onto the template, then load the polymerase with its sliding clamp unit a) Primase includes both DNA and RNA polymerase b) Main replication polymerase is a complex of DNA polymerase epsilon (pol ε) and DNA polymerase delta (pol δ)

Answer 237

multiple origins of replications for each chromosome

Answer 238

DNA and RNA polymerase

Answer 239

a complex of DNA polymerase epsilon (pol ε) and DNA polymerase delta (pol δ)

Answer 240

- Specialized structures found on the ends of eukaryotic chromosomes - Protect ends of chromosomes from nucleases and maintain the integrity of linear chromosomes - Gradual shortening of chromosomes with each round of cell division a) Unable to replicate last section of lagging strand

Answer 241

- Telomeres composed of short repeated sequences of DNA - Telomerase – enzyme makes telomere section of lagging strand using an internal RNA template (not the DNA itself) a) Leading strand can be replicated to the end - Telomerase developmentally regulated a) Relationship between senescence and telomere length - Cancer cells generally show activation of telomerase

Answer 242

enzyme makes telomere section of lagging strand using an internal RNA template (not the DNA itself)

Answer 243

- Errors due to replication a) DNA polymerases have proofreading ability - Mutagens – any agent that increases the number of mutations above background level a) Radiation and chemicals - Importance of DNA repair is indicated by the multiplicity of repair systems that have been discovered

Answer 244

any agent that increases the number of mutations above background level ex: radiation and chemicals

Answer 245

- Falls into 2 general categories 1) Specific repair 2) Nonspecific repair

Answer 246

Targets a single kind of lesion in DNA and repairs only that damage

Answer 247

Use a single mechanism to repair multiple kinds of lesions in DNA

Answer 248

- Specific repair mechanism - For one particular form of damage caused by UV light - Thymine dimers a) Covalent link of adjacent thymine bases in DNA - Photolyase a) Absorbs light in visible range b) Uses this energy to cleave thymine dimer

Answer 249

Covalent link of adjacent thymine bases in DNA

Answer 250

enzyme that absorbs light in visible range and uses this energy to cleave thymine dimer

Answer 251

- Nonspecific repair - Damaged region is removed and replaced by DNA synthesis - 3 steps 1) Recognition of damage 2) Removal of the damaged region 3) Resynthesis using the information on the undamaged strand as a template

Answer 252

1) Recognition of damage 2) Removal of the damaged region 3) Resynthesis using the information on the undamaged strand as a template

Answer 253

- Early ideas to explain how genes work came from studying human diseases - Archibald Garrod (1902) a) Recognized that alkaptonuria is inherited via a recessive allele b) Proposed that patients with the disease lacked a particular enzyme - These ideas connected genes to enzymes

Answer 254

- Recognized that alkaptonuria is inherited via a recessive allele - Proposed that patients with the disease lacked a particular enzyme

Answer 255

- Deliberately set out to create mutations in chromosomes and verify that they behaved in a Mendelian fashion in crosses - Studied Neurospora crassa a) Used X-rays to damage DNA b) Looked for nutritional mutations - Had to have minimal media supplemented to grow

Answer 256

- Beadle and Tatum looked for fungal cells lacking specific enzymes a) The enzymes were required for the biochemical pathway producing the amino acid arginine b) They identified mutants deficient in each enzyme of the pathway - One-gene/one-enzyme hypothesis has been modified to one-gene/one-polypeptide hypothesis

Answer 257

- First described by Francis Crick - Information only flows from DNA → RNA → protein - Transcription = DNA → RNA - Translation = RNA → protein - Retroviruses violate this order using reverse transcriptase to convert their RNA genome into DNA

Answer 258

RNA; protein

Answer 259

DNA → RNA → protein

Answer 260

reverse the central dogma | ex: HIV

Answer 261

- DNA-directed synthesis of RNA - Only template strand of DNA used - Strand of DNA not used as template is the coding strand - T (thymine) in DNA replaced by U (uracil) in RNA - mRNA used to direct synthesis of polypeptides

Answer 262

- Synthesis of polypeptides - Takes place at ribosome - Requires several kinds of RNA and many proteins

Answer 263

- All synthesized from DNA template by transcription - Messenger RNA (mRNA) - Ribosomal RNA (rRNA) - Transfer RNA (tRNA) - Small nuclear RNA (snRNA) - Signal recognition particle RNA (SRP RNA) - Small RNAs (miRNA and siRNA)

Answer 264

- Francis Crick and Sydney Brenner determined how the order of nucleotides in DNA encoded amino acid order - Codon: block of 3 DNA nucleotides corresponding to an amino acid - Are codons spaced (codon sequence punctuated) or unspaced (codons adjacent to each other)?

Answer 265

block of 3 DNA nucleotides corresponding to an amino acid

Answer 266

- Introduced single nucleotide (nt) insertions or deletions and looked for mutations a) Frameshift mutations - Additions or deletions of one or two nts shifted the genetic message, while addition or deletion of 3 nts resulted in a protein that was normal aside from the addition/deletion - Concluded that the genetic code is read in increments of 3 nts, read continuously

Answer 267

- Marshall Nirenberg identified the codons that specify each amino acid - Stop codons a) 3 codons (UAA, UGA, UAG) used to terminate translation - Start codon a) Codon (AUG) used to signify the start of translation - Code is degenerate, meaning that some amino acids are specified by more than one codon

Answer 268

UAA, UGA, UAG

Answer 269

- Strongest evidence that all living things share common ancestry - Advances in genetic engineering - Mitochondria and chloroplasts have some differences in how codons are read, especially stop

Answer 270

- Single RNA polymerase – 2 forms a) Core polymerase b) Holoenzyme - Holoenzyme needed to accurately initiate synthesis - Initiation of mRNA synthesis does not require a primer

Answer 271

- Requires: 1) Promoter 2) Start site 3) termination site - Promoter a) Forms a recognition and binding site for the RNA polymerase b) Found upstream of the start site c) Not transcribed d) Asymmetrical – indicates site of initiation and direction of transcription

Answer 272

a) Forms a recognition and binding site for the RNA polymerase b) Found upstream of the start site c) Not transcribed d) Asymmetrical – indicates site of initiation and direction of transcription

Answer 273

- Grows in the 5′-to-3′ direction as ribonucleotides are added - Transcription bubble: contains RNA polymerase, DNA template, and growing RNA transcript - After the transcription bubble passes, the now-transcribed DNA is rewound as it leaves the bubble

Answer 274

contains RNA polymerase, DNA template, and growing RNA transcript

Answer 275

- Marked by sequence that signals “stop” to polymerase a) Causes the formation of phosphodiester bonds to cease b) RNA–DNA hybrid within the transcription bubble dissociates c) RNA polymerase releases the DNA d) DNA rewinds - Series of G-C base-pairs followed by a series of A-T base-pairs can form a hairpin, which causes RNA polymerase to pause

Answer 276

- Prokaryotic transcription is coupled to translation a) mRNA begins to be translated before transcription is finished b) Operon - Grouping of functionally related genes - Encodes multiple enzymes for a pathway - Can be regulated together

Answer 277

- Grouping of functionally related genes - Encodes multiple enzymes for a pathway - Can be regulated together

Answer 278

- 3 different RNA polymerases 1) RNA polymerase I transcribes rRNA 2) RNA polymerase II transcribes mRNA and some snRNA 3) RNA polymerase III transcribes tRNA and some other small RNAs - Each RNA polymerase recognizes its own promoter

Answer 279

- Initiation of transcription at Pol II promoters a) Requires a series of transcription factors - Necessary to get the RNA polymerase II enzyme to a promoter and to initiate gene expression - Interact with RNA polymerase to form initiation complex at promoter - Elongation complex factors coordinated by structural feature called the CTD - In eukaryotes the end of the transcript is not the end of the mRNA

Answer 280

- In eukaryotes, the primary transcript must be modified to become mature mRNA a) Addition of a 5′ cap - Protects from degradation; involved in translation initiation b) Addition of a 3′ poly-A tail - Created by poly-A polymerase; protection from degradation c) Removal of non-coding sequences (introns) - Pre-mRNA splicing done by spliceosome

Answer 281

- Introns – non-coding sequences - Exons – sequences that will be translated - Small ribonucleoprotein particles (snRNPs) recognize the intron–exon boundaries - snRNPs cluster with other proteins to form spliceosome a) Responsible for removing introns

Answer 282

non-coding sequences

Answer 283

sequences that will be translated

Answer 284

recognize the intron–exon boundaries - cluster with other proteins to form spliceosome a) Responsible for removing introns

Answer 285

- Single primary transcript can be spliced into different mRNAs by the inclusion of different sets of exons - Up to half of known human genetic disorders may be due to altered splicing - One estimate suggests 95% of mammalian genes exhibit some form of alternative splicing - Explains how 20,000 genes of the human genome can encode the more than 100,000 different proteins

Answer 286

tRNA molecules carry amino acids to the ribosome for incorporation into a polypeptide a) Aminoacyl-tRNA synthetases add amino acids to the acceptor stem of tRNA b) Anticodon loop contains 3 nucleotides complementary to mRNA codons

Answer 287

- Each aminoacyl-tRNA synthetase recognizes only 1 amino acid but several tRNAs - Charged tRNA – has an amino acid added using the energy from ATP a) Can undergo peptide bond formation without additional energy - Ribosomes do not verify amino acid attached to tRNA

Answer 288

recognizes only 1 amino acid but several tRNAs

Answer 289

has an amino acid added using the energy from ATP | a) Can undergo peptide bond formation without additional energy

Answer 290

The ribosome has multiple tRNA-binding sites - P site - A site - E site

Answer 291

binds the tRNA attached to the growing peptide chain | * peptide site

Answer 292

binds the tRNA carrying the next amino acid | * acceptor site for next tRNA

Answer 293

binds the tRNA that carried the previous amino acid added | * exit site

Answer 294

- The ribosome has two primary functions a) Decode the mRNA b) Form peptide bonds - Peptidyl transferase a) Enzymatic component of the ribosome b) Forms peptide bonds between amino acids - Activity of ribosomes mostly thought to be carried out by rRNAs

Answer 295

a) Decode the mRNA | b) Form peptide bonds

Answer 296

- Enzymatic component of the ribosome | - Forms peptide bonds between amino acids

Answer 297

- In prokaryotes, initiation complex includes: a) Initiator tRNA charged with N-formylmethionine b) Small ribosomal subunit c) mRNA strand - Ribosome binding sequence (RBS) of mRNA positions small subunit correctly - Large subunit now added - Initiator tRNA bound to P site with A site empty

Answer 298

a) Initiator tRNA charged with N-formylmethionine b) Small ribosomal subunit c) mRNA strand

Answer 299

- Initiations in eukaryotes similar except a) Initiating amino acid is methionine b) More complicated initiation complex c) Lack of an RBS – small subunit binds to 5′ cap of mRNA

Answer 300

- Elongation adds amino acids a) 2nd charged tRNA can bind to empty A site b) Requires elongation factor called EF-Tu to bind to tRNA and GTP c) Peptide bond can then form d) Addition of successive amino acids occurs as a cycle - Matching tRNA anticodon with mRNA codon - Peptide bond formation - Translocation of ribosome

Answer 301

- There are fewer tRNAs than codons - Wobble pairing allows less stringent pairing between the 3′ base of the codon and the 5′ base of the anticodon - This allows fewer tRNAs to accommodate all codons

Answer 302

- Elongation continues until the ribosome encounters a stop codon - Stop codons are recognized by release factors which release the polypeptide from the ribosome

Answer 303

- In eukaryotes, translation may occur in the cytoplasm or the rough endoplasmic reticulum (RER) - Signal sequences at the beginning of the polypeptide sequence bind to the signal recognition particle (SRP) - The signal sequence and SRP are recognized by RER receptor proteins - Docking holds ribosome to RER - Beginning of the protein-trafficking pathway

Answer 304

- Point mutations alter a single base - Base substitution – substitute one base for another a) Silent mutation – same amino acid inserted b) Missense mutation – changes amino acid inserted - Transitions – pyrimidine for pyrimidine or purine for purine - Transversions – pyrimidine for purine or vice versa c) Nonsense mutations – changed to stop codon

Answer 305

alter a single base

Answer 306

substitute one base for another

Answer 307

same amino acid inserted

Answer 308

changes amino acid inserted

Answer 309

pyrimidine for pyrimidine or purine for purine

Answer 310

pyrimidine for purine or vice versa

Answer 311

changed to stop codon

Answer 312

- Addition or deletion of a single base - Much more profound consequences - Alter reading frame downstream - Triplet repeat expansion mutation a) Huntington disease b) Repeat unit is expanded in the disease allele relative to the normal

Answer 313

Change the structure of a chromosome a) Deletions – part of chromosome is lost b) Duplication – part of chromosome is copied c) Inversion – part of chromosome in reverse order d) Translocation – part of chromosome is moved to a new location

Answer 314

part of chromosome in reverse order

Answer 315

part of chromosome is moved to a new location

Answer 316

- Mutations are the starting point for evolution - Too much change, however, is harmful to the individual with a greatly altered genome - Balance must exist between amount of new variation and health of species

Answer 317

- Can now be measured directly - One study suggests about 70 new single nucleotide variations per birth - Small insertions and deletions of less than 50 bases occur more rarely - Larger deletions or duplications, called copy number variation, occur at an even lower rate

Answer 318

- Controlling gene expression is often accomplished by controlling transcription initiation - Regulatory proteins bind to DNA a) May block or stimulate transcription - Prokaryotic organisms regulate gene expression in response to their environment - Eukaryotic cells regulate gene expression to maintain homeostasis in the organism

Answer 319

- Gene expression is often controlled by regulatory proteins binding to specific DNA sequences a) Regulatory proteins gain access to the bases of DNA at the major groove b) Regulatory proteins possess DNA-binding motifs

Answer 320

- Regions of regulatory proteins which bind to DNA a) Helix-turn-helix motif – two α-helical segments linked by a nonhelical segment - Homeodomain motif b) Zinc finger motif – several forms, use zinc atoms to coordinate DNA binding c) Leucine zipper motif – dimerization motif in which region in one subunit interacts with a similar region on another subunit

Answer 321

- Control of transcription initiation a) Positive control – increases frequency of initiation of transcription - Activators enhance binding of RNA polymerase to promoter b) Negative control – decreases frequency - Repressors bind to operators in DNA c) Effector molecules can act on both repressors and activators

Answer 322

two α-helical segments linked by a nonhelical segment

Answer 323

- Zinc finger motif | - Leucine zipper motif

Answer 324

several forms, use zinc atoms to coordinate DNA binding

Answer 325

dimerization motif in which region in one subunit interacts with a similar region on another subunit

Answer 326

– increases frequency of initiation of transcription | - Activators enhance binding of RNA polymerase to promoter

Answer 327

decreases frequency - Repressors bind to operators in DNA - Effector molecules can act on both repressors and activators

Answer 328

can act on both repressors and activators

Answer 329

bind to operators in DNA

Answer 330

- Prokaryotic cells often respond to their environment by changes in gene expression - Genes involved in the same metabolic pathway are organized in operons - Induction – enzymes for a certain pathway are produced in response to a substrate - Repression – capable of making an enzyme but does not

Answer 331

enzymes for a certain pathway are produced in response to a substrate

Answer 332

capable of making an enzyme but does not

Answer 333

- Contains genes for the use of lactose as an energy source - b-galactosidase (lacZ), permease (lacY), and transacetylase (lacA) - Gene for the lac repressor (lacI) is linked to the rest of the lac operon

Answer 334

- The lac operon is negatively regulated by a repressor protein a) lac repressor binds to the operator to block transcription b) In the presence of lactose, an inducer molecule (allolactose) binds to the repressor protein c) Repressor can no longer bind to operator d) Transcription proceeds - Even in the absence of lactose, the lac operon is expressed at a very low level

Answer 335

- Preferential use of glucose in the presence of other sugars a) Glucose used first, then lactose b) Mechanism involves activator protein that stimulates transcription c) Catabolic activator protein (CAP) is an allosteric protein with cAMP as effector d) Level of cAMP in cells is reduced in the presence of glucose so that no stimulation of transcription from CAP-responsive operons takes place - Inducer exclusion – presence of glucose inhibits the transport of lactose into the cell

Answer 336

an allosteric protein with cAMP as effector

Answer 337

presence of glucose inhibits the transport of lactose into the cell

Answer 338

- Genes for the biosynthesis of tryptophan - The operon is not expressed when the cell contains sufficient amounts of tryptophan - The operon is expressed when levels of tryptophan are low - trp repressor is a helix-turn-helix protein that binds to the operator site located adjacent to the trp promoter

Answer 339

a helix-turn-helix protein that binds to the operator site located adjacent to the trp promoter

Answer 340

- The trp operon is negatively regulated by the trp repressor protein a) trp repressor binds to the operator to block transcription b) Binding of repressor to the operator requires a corepressor which is tryptophan (operon is repressed) c) When tryptophan levels fall, the repressor cannot bind to the operator (operon is derepressed)

Answer 341

operator to block transcription

Answer 342

- Control of transcription more complex - Major differences from prokaryotes a) Eukaryotes have DNA organized into chromatin - Complicates protein-DNA interaction b) Eukaryotic transcription occurs in nucleus - Amount of DNA involved in regulating eukaryotic genes much larger

Answer 343

- General transcription factors a) Necessary for the assembly of a transcription apparatus and recruitment of RNA polymerase II to a promoter b) TFIID recognizes TATA box sequences c) After TFIID binds, TFIIE, TFIIF, TFIIA, TFIIB, and TFIIH bind, along with many transcription-associated factors (TAFs) d) This initiation complex can initiate synthesis at a basal level

Answer 344

- Specific transcription factors a) Act in a tissue- or time-dependent manner to stimulate higher levels of transcription than the basal level b) Each factor consists of a DNA-binding domain and a separate activating domain that interacts with the transcription apparatus; these domains are independent in the protein

Exam 4 Flashcards

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