Lecture 1+2 - Introduction/Gene Analysis Techniques

Question 1

What key discovery did Avery, Macleod and McCarty make?

Answer 1

They identified DNA as the transforming principle in 1944.

Question 2

In Avery, Macleod and McCarty’s experiment what did the ‘smooth’ strain result in?

Answer 2

The ‘smooth’ strain caused disease and death in the mice.

Question 3

In Avery, Macleod and McCarty’s experiment what did the ‘rough’ strain result in?

Answer 3

The ‘rough’ strain had no effect on the mice.

Question 4

Briefly outline Avery, Macleod and McCarty’s experiment in 1944.

Answer 4

They showed that DNA was the transforming principle not proteins as previously thought. They injected mice with strains of pneumonia, some that were ‘smooth’ which resulted in disease and death and ‘rough’ strains which had no effect.

If heat-killed smooth (deadly) strains were injected into the mice then there was no effect, however if heat-killed smooth was injected with live ‘rough’ (no effect) the mouse died.

After extracting the DNA, live smooth strains were found. The S strain extract somehow had “transformed” the R strain bacteria to S form.

Question 5

Who’s original observations was Avery, MacLeod and McCarty’s experiment based on?

Answer 5

Griffith’s initial observations in 1928.

Question 6

Who showed that nucleic acid is the genetic material in 1952?

Answer 6

Hershey and Chase

Question 7

In Hershey and Chase’s experiment, what media labelled the DNA of the phage?

Answer 7

T2 phages were inserted into 32P media which labelled the DNA of the phage.

Question 8

In Hershey and Chase’s experiment, what media labelled the protein of the phage?

Answer 8

T2 phages were inserted into 35S media which labelled the protein of the phage.

Question 9

Briefly outline Hershey and Chase’s experiment.

Answer 9

One group of T2 phages were inserted into a 32P media which labelled their DNA, a second group was inserted into a 35S media which labelled their proteins.

These labelled phage were infected into an unlabelled media. The DNA labelled phage produced unlabelled ghosts but bacteria labelled with 32P.

The protein labelled phage produced ghosts labelled with 35S but unlabelled bacteria. This showed that it is DNA that is the genetic material.

Question 10

Describe the life cycle of T2 phage.

Answer 10

Phage have DNA or RNA present in a protein coat. These phage attach to a bacterial cell and insert their genetic material leaving an empty ‘ghost’. The bacteria cell will then produce new phage particles.

Question 11

Which two bases are Purines?

Answer 11

Adenine and Guanine are both Purines.

Question 12

Which two bases are Pyrimidines?

Answer 12

Cytosine, Thymine and Uracil are all Pyrimidines.

Question 13

How can you distinguish between a Purine and a Pyrimidine?

Answer 13

A Purine ring consists of a pyrimidine ring fused to an imidazole ring. Whereas a pyrimidine has just one nitrogenous ring.

Question 14

By which bonds are nucleotides joined together?

Answer 14

Phosphodiester bonds join nucleotides together.

Question 15

Which end is extended in DNA synthesis?

Answer 15

The 3’ end is extended, DNA synthesis is always 5’ - 3’.

Question 16

Who proposed the anti-parallel double helix for DNA?

Answer 16

Watson and Crick in 1953.

Question 17

What is supercoiled DNA?

Answer 17

Supercoiling of DNA refers to the twisting of the molecules. It can only be introduced into or released from a closed circular DNA molecule by breaking at least one phosphodiester bond.

Question 18

What is used to describe the level of supercoiling within a DNA molecule?

Answer 18

The Linking number (Lk) refers to the number of double helical turns in the original linear molecule. The linking difference gives an indication of the overall levels of supercoiling within a molecule.

Question 19

Most DNA is typically positively supercoiled, true or false?

Answer 19

FALSE, most DNA is negatively supercoiled.

Question 20

What nucleosomes are present in the DNA core?

Answer 20

All but H1, therefore H2A, H2B, H3, H3, H4, H4

Question 21

What is the distance between the DNA strands?

Answer 21

2nm

Question 22

Who proved the existence of semi-conservative replication?

Answer 22

Meselson & Stahl gave proof of semi-conservative replication.

Question 23

In Meselson and Stahl’s experiment what medium was E. coli grown in initially?

Answer 23

Many generations of E. coli were grown in a medium containing the heavy isotope of nitrogen, 15N. The mobility band of the bacteria was then noted.

Question 24

After the E. coli had been grown in 15N what was done next in Meselson and Stahl’s experiment?

Answer 24

The mobility band of the DNA was noted and then the bacteria were transferred to a media containing the normal isotope of nitrogen, 14N.

Question 25

After the E. coli had been transferred to 14N what was observed in Meselson and Stahl’s experiment?

Answer 25

After one generation,the isolated DNA from the bacteria ran through the density gradient with a mobility different form that seen previously. After two generations, two DNA species were observed.

One had the same mobility seen after one generation the other ran with the same mobility as DNA isolated from unlabelled media.

Later generations showed that the intensity of the latter band increased and the intensity of the intermediate density decreased.

Question 26

What is a gene?

Answer 26

A gene is a unit of heredity composed of nucleic acid, it is a discrete part of the chromosome that determines a particular characteristic.

Question 27

What is a polycistronic message?

Answer 27

An mRNA found in prokaryotes that encodes more than one protein.

Question 28

In prokaryotes the binding of what triggers transcription?

Answer 28

The proteins RNA polymerase (promoter) and enhancer elements such as cAMP.

Question 29

What do repressor proteins do and where are they found?

Answer 29

The operon may contain the binding sites for repressor protein, which when bound to DNA, prevent polymerase binding and therefore stop transcription.

Question 30

Where are enhancer elements located in eukaryotic genes?

Answer 30

Enhancer elements may be present in eukaryotic genes and are located several thousand base pairs upstream of the gene which are required for its full expression.

Question 31

What is the major difference between prokaryotic and eukaryotic genes?

Answer 31

In eukaryotes introns are present which split the coding exon sequences. The introns are removed by splicing before the transcript is translated.

Question 32

In prokaryotes are genes produced individually or all at once?

Answer 32

In prokaryotes the genes required to code for a specific function, for example an enzymatic pathway are all transcribed together, as a polycistronic message forming multiple proteins.

Question 33

What is the lac operon?

Answer 33

The lac operon is the assortment of three adjacent genes and their regulatory regions. Lactose is only metabolised when the structural genes are expressed and the protein products produced.

lacZ-lacY-lacA =
B-galactosidase - Lactose permease - Thiogalactoside transacetylase.

Question 34

What DNA sequence elements does the lac operon have?

Answer 34

Operator (lacO) - the binding site for the repressor
Promoter (lacP) - the binding site for the RNA polymerase
Repressor (lacI) gene - encodes for the Lac repressor protein. Binds to DNA at the operator and blocks the transcription of the structural genes by RNA polymerase bound at the promoter.
Pi - the promoter for lacI
CAP binding site for cAMP/CAP complex.

Question 35

If lactose is absent what occurs?

Answer 35

RNA polymerase binds to the promoter but is prevented from transcribing the structural genes by the Lac repressor. which is bound to the operator site.

Question 36

What occurs when lactose is present?

Answer 36

When lactose enters the cell, it binds to the lac repressor and causes a conformational change that inhibits its ability to bind DNA. Therefore the polymerase bound at P will transcribe the lac structural genes (lacZ, lacY and lacA).

lacZ » ß-galctosidas
lacY » Lactose permease
lacA » Thiogalactosideas transacetylase

Full activation of the lac genes only occurs in the absence of glucose when the CAP binds to the operon and aids the binding of RNA polymerase to P.

Question 37

In eukaryotes where can the activator protein bind?

Answer 37

The activator protein can bind through its DNA binding domain (DBD) in a nucleosome free region or on DNA already bound within a nucleosome.

Question 38

What is the structure of a mature mRNA molecule in eukaryotes?

Answer 38

After transcription a 7-methyl guanosine cap is added to the 5’-end of the message.

The ribose sugar of the first (and sometimes second) nucleotide is methylated at the 2’ position. The 3’-end of the transcript is polyadenylated with the addition of 100-200 A residues.

Question 39

What is polyadenylation?

Answer 39

Polyadenylation is the addition of a poly(A) tail to a messenger RNA.

The process of polyadenylation begins as the transcription of a gene finishes, or terminates. The 3’-most segment of the newly made pre-mRNA is first cleaved off by a set of proteins; these proteins then synthesize the poly(A) tail at the RNA’s 3’ end.

In some genes, these proteins may add a poly(A) tail at any one of several possible sites. Therefore, polyadenylation can produce more than one transcript from a single gene (alternative polyadenylation), similar to alternative splicing.

Question 40

How many chromosomes do humans have?

Answer 40

23 chromosomes.

Question 41

How many chromosomes do mice have?

Answer 41

21 chromosomes.

Question 42

Briefly explain alternative splicing.

Answer 42

Alternative splicing can generate different forms of mRNA from identical pre-mRNA molecules, so that expression of one gene can give rise to a number of proteins, with similar or different functions.

Alternative splicing increases the number of proteins that can be made from each gene. As a result, the number of proteins that an organism can make—its proteome—is not the same as the number of genes in the genome, and protein diversity can exceed gene number by an order of magnitude.

Question 43

What does the 7-methyl guanosine cap do?

Answer 43

It improves mRNA stability.

Question 44

Which strand is the ‘sense’ strand?

Answer 44

5’ - 3’ is the sense strand.

Question 45

Which strand is the ‘antisense’ strand?

Answer 45

3’ - 5’ is the antisense strand.

Question 46

What does phosphatase treatment do?

Answer 46

It can be used to remove phosphates from the free 5’ ends of cut DNA.

Question 47

Can plasmids be used as vectors?

Answer 47

Yes, and they usually carry antibiotic resistance.

Question 48

What do activators do?

Answer 48

They have multiple roles in switching on gene expression in eukaryotes.

The activator binds through its DNA binding domain (DBD), the chromatin of the strand is modified and remodelled and RNA polymerase II holoenzyme is recruited.