L11: Human Genome & Gene Expression

Question 1

Mitochondrial vs Nuclear Genome

Answer 1

Mitochondrial Genome: 93% highly conserved
Nuclear Genome: 1.5% conserved coding areas

Question 2

Human Nuclear DNA may be categorized as follows:

Answer 2

Single sequence DNA or low-copy DNA (about 45% of the total) –coding sequence only account for ~1.5% of the total genome, remainder is DNA from introns
Intermediate repeated DNA, present at between 10^2 and 10^5 copies per genome (about 45% of the total)
Highly repetitive DNA, present at up to 10^6 copies per genome (about 10% of the total)

Question 3

Human Genes

Answer 3

~30-35,000 genes in total
Some genes (histone genes) do not contain introns
α-globin gene is 0.8 kb in size and contains three introns, which account for 30% of the total genomic locus
dystrophin (defective in DMD) is 2.4 Mb in size and contains 79 introns, which account for 99.4% of the total genomic locus
The mean size of genes is 27 kb

Question 4

Genes <10kb
Genes <100kb
Genes >100kb

Answer 4

Genes <10kb: Insulin, B-globin, Sialidase
Genes <100kb: ApoB, LDL receptor, PAH
Genes >100kb: CFTR, dystrophin

Question 5

Gene Families

Answer 5

Thought to have arisen by a process of gene duplication and divergence from an ancestral gene.
Because there are multiple copies of genes in a family, sometimes loss of function in one of them may be tolerated –leads to evolution of pseudogenes

Question 6

Pseudogenes

Answer 6

Defective gene copies found in gene families:
- Unprocessed pseudogenes in a gene cluster (tandem gene duplication, inappropriate termination codons)
- Truncated genes and internal gene fragments (e.g. class I HLA)
- Non-processed pseudogenes in a dispersed gene family (e.g. NF-1 gene fragments over 7 chromosomes)
- Processed pseudogenes in a dispersed polypeptide-encoding gene family (cDNA+oligo dA/dT; copied by retrotransposition)
- Processed pseudogenes in an RNA-encoding gene family (retrotransposition from RNA polymerase III transcripts)

Question 7

Non-protein coding genes functions

Answer 7

The bulk of cellular RNA consists of rRNA and tRNA required for mRNA translation

Additional functions of noncoding RNAs:
1) Small nuclear RNA genes (snRNA) involved in spliceosome function
2) Small nucleolar RNA genes (snoRNA) that control site specific base modifications in rRNA and U6 snRNA
3) The 7SL RNA molecule forms part of signal recognition particle required for translocation of pr across ER
4) MicroRNAs (miRNAs) ~22 nucleotides long can function as antisense regulators of genes to inhibit translation

Question 8

Genes

Answer 8

DNA seq that contributes to phenotype of organism in a way that depends on its sequence

Question 9

Gene Expression

Answer 9

Transcription in humans, as in all eukaryote cells, is catalyzed by 3 types of RNA polymerase:
1. RNA polymerase I: transcribes rRNA
2. RNA polymerase II: transcribes nuclear genes that encode proteins
3. RNA polymerase III: transcribes tRNA genes, and small # of functional RNA molecules such as 5S RNA and nuclear RNAs that mediate splicing

Question 10

Promoter

Answer 10

Typical eukaryotic gene contains promoter region, which extends for ~200 bp upstream of transcription start site at nucleotide +1 contain 1/more of the following elements: enhancers, response elements and silencers
Most human promoters contain element called TATA box centred around position 30 (upstream) relative to transcription start site
TATA box binds general transcription factors, designated TFIIX
In pol II transcription, process is initiated by binding of TFIID mediated by another protein called TATA-binding protein (TBP). This is followed by assembly of initiation complex with pol II and other general transcription factors.

Question 11

Regulatory Elements within the Promoter

Answer 11

GC box (GGGCGG) binds transcription factor SP1
CAAT box (GGCCAATCT) binds transcription factors CTF and NF1
Oct box (ATTGCAT) binds transcription factors Oct-1 and Oct-2
Enhancers are operationally distinguished from promoters by 3 criteria:
- located considerable distance from transcription start site
- located upstream or downstream
- action is not dependent on orientation –can be in sense/antisense

Question 12

Eukaryotic Gene Structure

Answer 12

Regulatory regions upstream
Promoter
mRNA start
5’ leader
Exons & introns (w/ internal regulatory regions)
3’ trailer
mRNA end
Regulatory regions downstream

Question 13

Gene Regulation

Answer 13

DNA looping bring pr bound to enhancers or silencers into direct contact w pr bound to promoter
Can increase/decrease transcription; can increase/decrease presence of pr

Question 14

Response Elements

Answer 14

Induce genes in response to signals
Examples: cAMP response element (CRE), serum response element (SRE) and heat-shock response element (HRE)
- Response elements may be found within promoter region, closely upstream of promoter or in more distant locations
- Negative control of gene expression as a result of binding of certain pr to elements known as SILENCERS
- Some genes, termed housekeeping genes, are expressed in most tissues most of the time, and are responsible for functions likely to be necessary in any cell

Question 15

cAMP Response

Answer 15

1) Hormone binds to receptor
2) Conformational change in G protein
3) Adenylate cyclase converts ATP to cAMP
4) Elevated cAMP activates protein kinase A
5) PKA enters nucleus & phosphorylates CREB

Question 16

5’ cap formation

Answer 16

As soon as transcription starts, 5’ end of nascent mRNA molecule is covalently modified to form a structure known as a cap.
This is formed by addition of 7-methylguanosine to 5’ nucleotide of nascent mRNA in a 5’-5’ triphosphate linkage. The cap is essential for initiation of translation (Protects 5’ end of RNA)

Question 17

Termination of Transcription

Answer 17

Appearance of sequence AAUAAA in nascent RNA molecule causes transcription to terminate a few hundred base pairs downstream

Question 18

Polyadenylation

Answer 18

The RNA molecule is then cleaved ~20 bp downstream of AAUAAA signal. The 3’ end is also covalently modified by addition of 100-200 adenine residues to form structure => poly-A tail

Question 19

One Gene-More than 1 Protein

Answer 19

Some genes have alternative transcription and processing of gene
1. Use of alternate promoters can generate tissue specific isoforms
2. Human genes are prone to alt splicing and alt polyadenylation
3. RNA editing→substitution by deamination from very selected cytosine or adenine residues. e.g., C→U editing in APOB gene

Question 20

Tissue-specific RNA editing during processing of the human apolipoprotein B gene (APOB)

Answer 20

Liver:
- MTP goes into ApoB100 into VLDL
- VLDL—LDL—peripheral –HDL—liver and bile to be excreted

In intestine, a cytosine deaminase converts cytosine to uridine
- MTP load triglycerides –APOB48 and goes into chylomicron –goes only to liver

Question 21

CpG Islands

Answer 21

DNA can be mod by methylation of cytosine to form 5-methylcytosine; occurs at dinucleotide CpG
Dinucleotide CpG occurs less frequently in genome b/c: Accidental deamination of cytosine produces uracil, whereas deamination of 5-methylcytosine produces thymine
Uracil is not found in DNA, it is efficiently excised and replaced with cytosine by an enzyme called glycolase
Thymine, a natural constituent of DNA, is now mispaired with a G residue although this will also be repaired, the process is not perfectly efficient

Question 22

DNA Methylation

Answer 22

Recognition of a methylated DNA segment
- Hpa II and MSP I -both have same recognition site
- If have probe, can cut up genome, and blot with probe –Hpa will not cut methylated areas (if promoter was cut with Hpa, not methylated; if not cut with Hpa but was cut with Msp, then it is methylated)
DNA methylation is generally associated with repression of transcription
CpG islands are about 1 kb in length and tend to extend over promoters of expressed genes
About 56% of human genes are estimated to be associated with such sequences.

Question 23

Changes in DNA methylation during mammalian development

Answer 23

Germ lines (ovaries/testes)
Primordial germ cells –lose methylation completely, once develop into oocyte/sperm, increase methylation
Sperm methylated more than egg
Sperm completely unmethylated once fertilized (steep decline compared to egg)
After blastocyst, 100% maternal regulators during development –methylated occurs and peaks at somatic cells

Question 24

Studying Chromatin Conformation

Answer 24

Take DNA and cross link proteins to DNA (thru formaldehyde) to isolate chromatin
- Allow antibodies to bind to transcription factors –immuno-precipitated
Reverse crosslink, degrade protein, purify DNA frag –broken down pr was bound to TFs, proves it exists on that gene