Exam 3

Question 1

Differentiate Euploidy and Aneuploidy.

Answer 1

A euploid organism has multiples of the basic chromosome set (genome); ex. 2n, 3n, 4n

An aneuploid organism’s chromosome number differs from the wild type by part of a chromosome set; ex. 2n + 1, 2n - 1

Question 2

What are the three major types of chromosome mutations?

Answer 2

1) Relocation of genetic material, ie. translocation and inversion
2) Loss of genetic material, ie. deletion
3) Gain of genetic material, ie. duplication and extra chromosomes

Question 3

Why are triploids most often sterile?

Answer 3

Due to unpaired chromosomes at meiosis, gametes will either receive one or two chromosomes for each chromosomal type. It is unlikely that they will receive the same number of each chromosome type, and so they are most often aneuploid, and inviable.

Question 4

In what organisms is polyploidy common, and what are its effects?

Answer 4

Plants, as shown by much higher incidence of even numbers of chromosomes above haploid number 12, indicating doubling.

Polyploid plants are often larger than diploid plants, but remain proportional.

Question 5

What is nondisjunction?

Answer 5

A failure of normal segregation of homologous chromosomes or chromatids during meiosis I or II; two chromosomes incorrectly go to one pole or the other.

Question 6

What is the difference between primary and secondary nondisjunction?

Answer 6

Primary nondisjunction occurs in meiosis I (anaphase); products will be two O gametes, and two 2n gametes with different chromatids if parent is heterozygous.

Secondary nondisjunction occurs in meiosis II (anaphase); products will be one O gamete, one 2n gamete with the same chromatids, and two normal 1n gametes with the other chromatid.

Question 7

Why does the chance of a child with Down’s syndrome increase with maternal age?

Answer 7

Oocytes arrest in meiotic prophase I and resumes at each menstrual period; with age, the bivalents must stay together for longer and longer, and likelihood of mutagenesis events breaking the bivalent increases. This leads to increased primary nondisjunction.

Question 8

Explain what is meant by a “gene-dosage effect” in regards to human aneuploidy.

Answer 8

In general, the amount of transcript produced by a gene is directly proportional to the number of copies of that gene in a cell. If there is an extra (or missing) chromosome, this leads to uneven gene dosage, interfering with gene function. Because of this, human aneuploidy is almost always lethal.

Question 9

Which cases of human sex chromosome aneuploidy are fertile?

Answer 9

XYY and XXX are fertile and have no obvious phenotype.

Question 10

What is the result of an XO genotype? What type of nondisjunction could lead to this genotype?

Answer 10

Turner syndrome - female with underdeveloped gonadal structures, sterile

Could occur due to primary or secondary nondisjunction in either parent.

Question 11

What is the result of an XXY genotype? What type of nondisjunction could lead to this genotype?

Answer 11

Klinefelter syndrome - sterile male with underdeveloped gonadal structures, IQ impairment, and some breast development

Could occur due to primary nondisjunction in the father, or primary or secondary nondisjunction in the mother.

Question 12

Can a single meiotic nondisjunction event give rise to both a Klinefelter child and a child with Turner syndrome?

If so, would it be a primary nondisjunction, a secondary nondisjunction, or is either possible?

Answer 12

Yes, it could. During a primary nondisjunction event.

Question 13

What rules govern sex determination in drosophila?

Answer 13

One X chromosome –> male
Two X chromosomes –> female

absence of X is lethal, three X chromosomes is lethal

Question 14

What rules govern sex determination in humans?

Answer 14

Y chromosome present –> male
Y chromosome absent –> female

absence of X is lethal

Question 15

What are three funky ways in which sex can be determined in various species?

Answer 15

1) Bees and Wasps –> females are diploid, males are haploid
2) Lizards and Alligators –> females in cool temperature development, males in warm temperature development
3) Anemone –> young adult are male, older adults are female

Question 16

A woman with Turner syndrome is also colorblind (an X-linked recessive phenotype). Both her parents have normal vision.

How is it possible for her to be colorblind if neither parent was?
Can you determine which parent experienced nondisjunction?
Did the nondisjunction event happen at meiosis I or II?

Answer 16

a) She only has one X chromosome, so whatever genotype it is is dominant. One of her parents must be a carrier.
b) Her mother must be the carrier, because her father would express the phenotype is he was. Thus, she must have received the X from her mother, and the O from her father.
c) Can’t determine.

Question 17

How do chromosomal inversions and translocations come about?

Answer 17

1) Double stranded breaks (x-ray zapping, etc.)

2) Crossing over between repetitive sequences in the genome (like those produced by transposable elements)

Question 18

What is the difference between a paracentric and pericentric inversion?

Answer 18

Paracentric doesn’t include the centromere, pericentric does.

Question 19

What is the problem with dicentric and acentric fragments?

Answer 19

They either lack a centromere and won’t be pulled to either pole in anaphase, or they have two centromeres and will be pulled to both poles simultaneously.

Question 20

What does a chromosome require in order to replicate properly?

Answer 20

exactly 1 centromere, and 2 telomeres

Question 21

Why is the inversion loop produced?

Answer 21

When chromosomes line up in meiosis, the inverted chromosome twists to properly pair with the same genes on the non-inverted strand, creating an inversion loop. Crossing over in an inversion loop yields acentric and dicentric fragments, as well as duplications and deletions.

Question 22

Why is TM2 a bad gene to have?

Answer 22

It has multiple inversions, and is homozygous lethal; Gametes that are products of recombination are inviable, and heterozygotes carrying TM2 might be sicker than wildtype.

Question 23

Describe the basic transcriptional regulation process in prokaryotes?

Answer 23

Activators and repressors bind to sites in the vicinity of the promoter to control accessibility of DNA to RNA polymerase.

Question 24

What are the structural genes in the lac operon?

Answer 24

Z: ß-Galactosidase
Y: Permease
A: Transacetylase

Question 25

What are the regulatory components of the lac operon?

Answer 25

I: the lac repressor protein encoding gene
P: the promoter that initiates the transcription of the structural genes
O: the lac operator site to which the lac repressor protein binds
crp: the CAP encoding gene

Question 26

Describe the mechanism by which the presence of lactose in an environment allows for the synthesis of lactose processing proteins.

Answer 26

1) In presence of lactose, lactose functions as an inducer, and binds to an allosteric site on the repressor protein
2) Repressor protein can no longer bind to the operator sequence, and falls off
3) The RNA polymerase is then able to transcribe the lac structural genes.

Question 27

What is meant by constitutive mutations?

Answer 27

Constitutive mutations cause the lac structural genes to be expressed regardless of environmental factors –> transcription is no longer inducible.

Question 28

What is the effect of an I-/I+ phenotype?

Answer 28

I- mutations stop the transcription of the repressor protein, but the repressor is trans-acting, so the I+ will be the dominant phenotype, and transcription is still inducible.

Question 29

What is the effect of an Is/I- phenotype?

Answer 29

Is mutations lead to production of repressors that cannot be bound by inducers. Thus, the Is repressors would bind both of the operator sequences, leading to the halt of all production of lac structural genes.

Question 30

What is the effect of an O+/Oc phenotype?

Answer 30

The Oc mutation is cis-acting, meaning is only affects one strand. However, no repressor will be able to bind that operator sequence, so lac structural genes will always be transcribed at low levels, despite the other strand’s operator sequence still functioning in the negative feedback loop.

Question 31

How does the presence of glucose affect the operation of the lac operon?

Answer 31

In the presence of glucose, cAMP is present at very low levels. In the absence of glucose, cAMP is present at high levels, and can bind to CAP which then binds to a sequence in the lac operon, allowing transcription.

Question 32

How would a crp- mutation affect transcription of lac structural genes?

Answer 32

A crp- mutation would halt synthesis of CAP. cAMP would not be able to bind CAP, even if it were present in high levels as in low-glucose conditions. The lac operon would be inducible but would only produce very low (basal) levels of proteins.

Question 33

Describe the mechanism by which trp operon gene expression is governed by levels of tryptophan in the environment.

Answer 33

1) Simple negative feedback - trp repressor binds tryptophan when present in significant amounts, and then can bind operator sequence to switch off tryptophan production
2) Attenuation - post transcription; a leader sequence between the trp operator and the trpE gene is transcribed and translated; contains two trp codons, so if trp is abundant in the cell, enough trp tRNAs will be present for the ribosome to move past both codons quickly, leading to a formation of a step and loop in the mRNA that favors the halt of transcription. But, when trp is not present, the ribosome stalls at the trp codons, favoring the continuation of transcription.

Question 34

How are transcriptional regulation in eukaryotes and prokaryotes the same?

Answer 34

Both utilize control of transcription through DNA binding proteins (ie. activators and repressors, transcription factors)

Question 35

What mechanisms of transcriptional regulation differ between eukaryotes and prokaryotes?

Answer 35

Prokaryotes - operons, attenuation
Eukaryotes - chromatin structure, differential splicing, mRNA processing, differential polyadenylation, differential transport of nucleus to cytoplasm, RNA interface carried out by microRNAs

Question 36

What is the large-scale functional difference between prokaryotic and eukaryotic transcriptional regulation?

Answer 36

Bacterial genes are “on” in the ground state, for the most part, while eukaryotic genes are “off.” Eukaryotes require the presence of transcription factors and enhancers to initiate and continue transcription.

Question 37

What are promoter-proximal elements?

Answer 37

Sequences (TATA, CAAT, GC-rich) found near the promoter that interact with TFs to allow the binding of RNA polymerase and the initiation of transcription; these alone will produce only basal levels of transcription.

Question 38

What are enhancers/upstream activating sequences?

Answer 38

Cis-acting regulatory sequences that be located far away from promoters, but interact with TFs to make the environment more optimal for RNA polymerase, increasing transcription above basal levels.

Question 39

Describe the mechanism by which Gal4 regulates genes through UASs.

Answer 39

Gal 4 is a sequence specific DNA binding protein (aka TF) that binds to many UASs. In the absence of galactose, the activation domain is bound by Gal 80. In presence of galactose, Gal80 undergoes a conformational change, and allows Gal 4 to enhance transcription of target genes.

Question 40

What are the two domains of the Gal4 protein and what are their functions?

Answer 40

1) Activation domain - activates transcription

2) DNA-binding domain - binds to the correct DNA sequence

Question 41

What is the structure of chromatin?

Answer 41

A nucleosome is composed of DNA wrapped around eight histones. Nucleosomes exist in coiled chains that are condensed to different degrees and make up chromatin fibers.

Question 42

How do histones relate to transcriptional regulation?

Answer 42

Histones are heavily modified post-translation; protruding amino terminal ends (histone tails) can become acetylated or methylated through covalent bonding, influencing how tightly DNA is bound and packed.

Question 43

How might a readily transcribable gene’s chromatin structure differ from a non-transcribed gene’s?

Answer 43

DNA methylation is associated with inactive genes and leads to closed chromatin.

Histone modifications, including acetylation and some methylation events, lead to open chromatin that can be transcribed more efficiently.

Question 44

Differentiate heterochromatin from euchromatin.

Answer 44

Euchromatin is DNA actively in use and is acetylated; heterochromatin is inactive, DNA methylated.

Question 45

How might histone modifications be inherited?

Answer 45

When the replisome copies the parental strands, it also disassembles the nucleosomes in the parental strands and reassembles them in both parental and daughter strands. Old histones provide a template for the modification of new histones to match.

Question 46

What is gene imprinting?

Answer 46

A phenomenon in which certain autosomal genes are only expressed if they are inherited from either the mother or father. Maternal imprinting if gene from mother is inactive, vise versa.

Question 47

How does imprinting work on a molecular level as in ifg2 and H19 in mice?

Answer 47

In females, the imprinting control region is unmethylated, and can bind to s CTCF dimer, forming an insulator that blocks enhancer activation of lgf2, such that only H19 is expressed.

In males, the imprinting control region is methylated, preventing binding of CTCF dimer, and also preventing transcription of H19, so only lgf2 is expressed via the action of the downstream enhancer.

Question 48

What is a Barr body?

Answer 48

In females, the inactivated X chromosome that can be seen in the nucleus as a darkly staining, highly condensed, heterochromatic structure.

Question 49

What is a model for the inactivation of one X-chromosome in females?

Answer 49

1) X-inactivation center locus (Xic) of X to be inactivated produces 17kb noncoding RNA Xist.
2) Xist binds Xi-coating protein, and transfers it to the DNA strand.

Question 50

How might genetic silencing proceed?

Answer 50

Histone methylases bind to histone complex, and recruit DNA methylases, etc.

Question 51

Why does Handel et al. suggest we need to rethink twin studies?

Answer 51

Previously, it has been assumed that discordance in twins is due to environmental factors. Now, we must also consider randomly and environmentally determined gene methylation patterns that diverge heavily as twins age.

Question 52

What environmental factors might affect the epigenome?

Answer 52

Stress, smoking, sex hormones, and dramatically poor nutrition

Question 53

What was the question and the main findings of Dolinoy et al?

Answer 53

Dolinoy et al. looked at the effect of maternal BPA exposure on variation in offspring coat color. They found that offspring coat color was lighter when mothers had been exposed to BPA, and this lightness of coat was associated with decreased methylation of CpG sites. Importantly, methylation in different tissue layers did not differ, indicating that the effect of BPA probably occurs very early in fetal development.

Question 54

What are the basic requirements of evolution?

Answer 54

1) Must be genetic variation within a population
2) Variation must be heritable; individuals who reproduce effectively will contribute more to the genetic makeup of the population
3) Some alleles confer advantages in survival and reproduction, and some confer disadvantages

Question 55

What forces tend to decrease variation in a population?

Answer 55

1) Directional selection
2) Genetic drift
3) Selection against heterozygotes

Question 56

What forces tend to increase variation in a population?

Answer 56

1) Balanced polymorphism
2) Migration
3) Mutation

Question 57

Why are large populations more genetically diverse than small populations?

Answer 57

In small populations, alleles are easily lost to genetic drift over generations, while in larger populations, the effects of genetic drift are minimized.

Question 58

What is a selective sweep?

Answer 58

When one advantageous mutation is selected for, the genetic material around it is also selected for via recombination

Question 59

How can neutral mutations be used to track how long ago two species diverged?

Answer 59

Neutral mutations accumulate steadily, and aren’t selected for, or against. The number of different neutral mutations might be an indicator of how long ago two species began mutating separately.

Question 60

What do differences in number of amino acid substitutions tell us about two species?

Answer 60

This tells us how quickly, and how much, a species is diverging; more neutral substitutions indicates more sensitivity to mutation, ie. quicker change. This is due to fewer highly conserved sequences that are essential to function.

Question 61

What does the ratio of synonymous to nonsynonymous mutations in a gene tell you?

Answer 61

Whether the gene has undergone selection. If the number of nonsynonymous mutations is lower than synonymous, the gene has undergone purifying selection. If the ratio is equal to 1, the gene is probably not a functional gene (ie. fingerprints).

Question 62

How do gene duplications come about?

Answer 62

1) transposons/retroviruses
2) nondisjunction events
3) small insertions from crossing over

Question 63

What can happen to a duplicate gene?

Answer 63

1) Gene inactivation (pseudogenization)
2) Evolution of a new function (neofunctionalization)
3) Function of gene is divided between the duplicates (subfunctionalization)

Question 64

What are some key features about the life cycle of HIV?

Answer 64

• a retrovirus with two copies of the RNA genome per particle
• has three open reading frames, with 3 genes coding for inner proteins, outer proteins, and replication proteins
• interacts with protein receptor on T-cells, virus fuses with cell, reverse transcriptase generates DNA from RNA and jumps into the genome –> many copies of the virus in the genome and translation of viral proteins

Question 65

What are mechanisms by which HIV evolves at such a fast rate?

Answer 65

1) Fast replication rate with error prone reverse transcriptase –> more mutations, faster
2) Under selective pressure by immune system and drug therapies
3) Recombination between two RNA genomes per molecule, and between different strains present in one person