Lecture 15

Question 1

Give two reasons why gene regulation is more complicated in eukaryotes as compared to bacteria.

Answer 1

-typically encode more transcription factors than bacteria

-regulatory sequences can be in the “long range” rather than just in the promoter

-alteration of chromatin states is a mechanism of gene regulation in eukaryotes

Question 2

What two types of regulation are present within multicellular eukaryotes that are not resent in
bacteria? Explain what is meant by each.

Answer 2

1. temporal regulations: certain genes are turned on during different checkpoints of development
2. Spatial: turning on/off certain genes in specific tissue

Question 3

Compare and contrast the characteristics of the core promoter with that of an enhancer sequence.
How are they similar and how are they different?

Answer 3

The core promoter binds generalized transcription factors and must be upstream of a gene

Silencers and enhancers however promote or prevent transcription via the interaction of proteins with the core promoter, they can be located anywhere in the gene.

Can provide on/off regulation and dimmer regulation.

Question 4

Explain how different patterns of SHH expression are accomplished in the brain and in the limbs.

Answer 4

SHH gene has a LONG RANGE enhancer to promote gene expression in the limbs and a closer enhancer to promote gene expression in the brain .

In the limb there are no specific brain transcription factors we express SHH in the limb specific way, vice verse for expressing brain specric patter of SHH.

Question 5

Propose a plausible outcome for using CRISPR to replace the cobra SHH enhancer with the
mouse SHH enhancer within a cobra.

Answer 5

this will results in the cobra having an SHH enhancer that upregulated limb development so when the SHH is activated forto allow for transcription it is likely that limb development in cobras would occur

Question 6

Summarize the general argument for why morphological evolution is presumed to more often be
driven by regulatory rather than coding changes. Illustrate this argument using the example of
humans and chimps.

Answer 6

human and chimps have 99% commonalities of AAs however we have vastly different morphological attributes due to there being

different AMOUNTS of proteins being turned on at different TIMES, in different TISSUES

Question 7

A mutation in the beta-globin LCR causes Thalassemia. Propose a plausible mechanism of action
of this mutation.

Answer 7

A locus control region controls many different genes and said genes enhacer proteins, however each gene produces its on mrna as opposed to polycistronic mrna.

mutation to the beta-globin LCR changes the 1:1 stoichiometry of alpha and beta globin, not lethal due to modularity but still has an effect caus thalssemia.

fetus have different globins to steal oxygen from the mothers blood

Question 8

Define modularity in the context of enhancers and silencers. Suppose that enhancers were not
modular, how would this change the consequences of mutations in these in these regions?

Answer 8

modularity is present in enhancers and silencers because if a protein can’t bind to specific enhance or silencers that will not affect how other silencers or enhancers in the gene bind.

If enhancers were not modular, the natural section would select against mutations to these enhancer sequences, leading to species that are very diverged evolutionarily having the same sequences.

Question 9

Why are some non-coding enhancer elements so highly conserved?

Answer 9

natural selection selects agains enhancer regions ever having LOF mutations as this individual will likely die due to gene expression being completely down-regulated when it shouldn’t

Question 10

How can the property of modularity give rise to different species using different binding sites and
different trans-acting proteins to regulate a gene, but the pattern of gene expression remains similar
between these species?

Answer 10

Binding site turnover results in enhancer modules differing across species due to a mutation in the binding site causing the binding site to be in the same location but changing the protein which binds.

THis results in the enhancer sequences across species using different binding sites and proteins but having the same out put which preserves modularity as there are different inputs but the same gene expression across species.

Question 11

Explain the mechanism of activation of the Gal genes in yeast. Your answer should explain the
roles of UASG Gal80, Gal4, and Gal3 both in the presence and in the absence of galactose

Answer 11

For galactose use in the body you need gal 4 to bind UpstreamActivatorSequence which is the enhancer sequence.

In the absence of galactose:

gal4 binds gal 80 instead of UAS preventing the 4 genes necessary for galactose utilization to be expressed.

Gal 3 is within the cytoplasm.

In the presence of galactose:

gal 3 will migrate to the nucleus and bind gal 80.

this will release gal 4 from gal 80 allowing it to bind the UAS and activate the necessary genes.

Question 12

Explain the mechanism of repression of the Gal pathway in yeast. Your answer should explain the
roles/location if Mig1 both in the presence and absence of glucose.

Answer 12

Instead of there being an inducer as seen in bacteria w/ lac operon you have silencers.

When glucose is present(preferred carbon source over galactose)

MIg1 will bind a silencer between UAS and Gal genes to prevent activation of gal pathway

When glucose is absent :

enzyme phosphorylates MIG1 –> exported from nucleus so it does not obscure UAS

Question 13

Compare and contrast the gal pathway with the lac operon. How are they similar and how are
they different?

Question 14

How are enhancers directed to only act to increase the transcription of specific target genes?

Answer 14

insulator blocks communication between enhancers and non-target promoters and will allow enhancers nearby the promoter that are not blocked by the insulator to be activated

Question 15

What is position effect variegation? Are these mutations altering coding sequence? Explain

Answer 15

A structural mutation moved the gene that controlled the drosophilas eye’s pigment to a heterochromatic region. The mutation are not altering coding sequence they epigenetic and are affect expression of the gene based on how wrapped the DNA is around the histone in that region

Question 16

Identify each of the following histone marks as either generally heterochromatic or euchromatic:
H3K27me3, H3K9ac, H3K9me3, H3K4me

Answer 16

H3K27me3: is a repressive mark is facultative heterochromatic

H3K9me3: repressive heterochromatic mark

h3k9ac: euchromatic mark

H3k4me: Euchromatic mark

Question 17

Histone deacetylases remove acetyl groups from histones. Would we expect greater activity of
these enzymes in a particular genomic region to increase or decrease transcriptional activity of that
region and why?

Answer 17

We would expect transcriptional activity to decrease as removing the euchromatic mark will tighten the DNA wrapped around the nucleosome making the gene less transcriptionally active

Question 18

Contrast nucleosome sliding with nucleosome ejection.

Answer 18

nucleosome sliding: is where the nucleosome remains bound however the regulatory sequence is exposed

nucleosome rejection: is where the nucleosome is removed and placed at another sequence to expose regulatory sequence

Question 19

Explain the regulation of PHO5 in yeast in as much detail as possible.

Answer 19

PHO5 encodes a phosphatase in yeast that is repressed in the presence of excess phosphate

PHO4 is present in the cytoplasm in high phosphate condition and the TATA box is bound on the -1 nucleosome preventing the gene from being turned on

two enhancers:

one is blocked by a -2 nucleosome and one is bound by PHO 2

in low phosphate conditions it will localize in the nucleus and turn on PHO5
pho4 binds pho 2

nuA4 aceylyzes histones openening oup chromaitn
TATA box is exposed

Question 20

What are lncRNAs and what is their role in X inactivation

Answer 20

Xist is a gene that encodes a lncrna located on the inactivated X and not located on the active X.

the lncrna will coat the inactivated X chromosome and recruit histone deacetylases and methylases which condenses the chromatin and turns x chromosome into barr body

Question 21

Define allele specific expression and genomic imprinting

Answer 21

Most genes express equal amount of the maternal and parental copy

however SOME have maternal /paternal specific expression

there is an epigenetic mark that is placed on the gene by the father causing the genen to only express the maternal allele

this is done by a sex-specific pattern of DNA methylation

if the zygote is male, it will add this epigenetic mark; if it is female, it will not.

Question 22

Explain how IGF2 expression from exclusively the paternal allele is accomplished

Answer 22

On the maternal chromosome there is no methylation/imprinting on the imprinting control region(ICR) which results in an insulator being bound and the enhancer is blocked from expressing IGF2 however it is able to activate the closer gene H19

On the paternal chromosome, there is methylation along the ICR region AND H19, so while an insulator protein cannot bind the ICR region, since H19 is blocked IGF2 gene can still be activated by enhancer

Without genetic imprinting development cannot occur

HOW DID THIS GENETIC IMPRINTING DEVELOP?

The male wants all offspring to express as much of the paternal resources (IGF2 ) as possible, however, the female wants offspring to have an equal amount of resources

Question 23

Contrast the production of siRNAs and miRNAs

Answer 23

Bidirectional promotors that produce (siRNA- small interfering RNA) is cut up by dicer to produce 21-24 bp ds sequences and can thus turn off genes that have complementary sequences to that RNA

mRNA( micro RNA) - can fold back on themselves similar to stem loops

Question 24

What is the role of dicer in RNAi?

Answer 24

it is a “molecular ruler” processes dsRNA that is transported to the cytoplasm but cutting it up into 21-25 bp siRNA or processing mRNA such that these “digestable” sequences can be used as a guide to find complementary sequences in the nucleus

Question 25

What are two possible outcomes of siRNA or miRNA binding to RISC and then binding a
complementary mRNA that would lower expression post transcriptionally?

Answer 25

RISC will denature one of the strands of the RNA molecule

RISC will use this ssRNA from either (siRNA, miRNA, or dsRNA from a virus) as a GUIDE to find complementary strands.

1. If a complementary strand is found it can be DESTROYED or prevented from being translated by binding it

Question 26

Explain the RITS mechanism for how RNAi can alter transcription

Answer 26

Similar to RISC however

RITS: is bound to the siRNA and carries it to the nucleus to find complementary RNAS at the SOURCE

RITS doesn’t destroy or prevent translation it recruits histone modify enzymes to close chromatin and spread heterochromatin

allows silencing of part of the genome that don’t need to be transcribed

RISC: will find complementary dsRNAs that are transported to the cytoplasm

Question 27

What is the likely reason that RNAi evolved? What is one piece of evidence that supports this?

Answer 27

RNAi was evolved to destroy transposable elements that render alleles null. Mutation to RNA can result in activation of normally silent transposable elements.