Topic 7: Nucleic Acids

Question 1

Outline how Franklin and Wilkins used X-ray crystallography to elucidate the structure of DNA

Answer 1

DNA was crystallised and then targeted with an X-ray (whose beam became diffracted by DNA crystals). The scattering pattern created by the diffracted X-ray was recorded on film. This pattern was then analysed to elucidate the structure of DNA.

Question 2

Describe the organisation of DNA into chromatin within eukaryotic cells

Answer 2

DNA is wrapped around histone proteins to form nucleosomes. Nucleosomes are grouped together (chromatosomes) and then arranged into fibres (chromatin). DNA is usually organised as chromatin within the nucleus, except during cell division (when the chromatin condenses to form chromosomes).

Question 3

Differentiate between euchromatin and heterochromatin

Answer 3

Euchromatin is more loosely packed and corresponds to active segments of DNA (i.e. active genes). Heterochromatin is more densely packaged and corresponds to inactive segments of DNA. Different cells have different segments of DNA packaged as euchromatin and heterochromatin.

Question 4

Outline the structure of the nucleosome (and identify its functions)

Answer 4

A nucleosome consists of DNA and histone proteins. DNA is wrapped around an octamer of histone proteins. Nucleosomes are linked by an interconnecting H1 histone.

Nucleosomes serve two key functions: they help to supercoil DNA, they help to regulate transcription.

Question 5

List five examples of non-coding DNA

Answer 5

Satellite DNA (e.g. short tandem repeats), telomeres, introns, non-coding genes, gene regulatory sequences.

Question 6

Explain the role of tandem repeats in DNA profiling

Answer 6

Short tandem repeats are short repeating segments within satellite DNA. The number of repeats for a particular loci will differ between individuals. The STRs can be excised and separated on a gel to create a distinct DNA profile of a given individual.

Question 7

Outline the methodology and conclusions of the Hershey-Chase experiment

Answer 7

Hershey and Chase demonstrated that DNA was the genetic material by using radioactively labelled viruses. Viruses were prepared with radioactive phosphorus (labels DNA) or radioactive sulphur (labels protein). Viruses then infected bacteria, before the bacteria and virus were separated via centrifugation (bacteria is heavier and forms a pellet, while the smaller virus remains in the supernatant). When radioactive sulphur was used, radioactivity was detected in supernatant (not transferred to bacteria). When radioactive phosphorus was used, radioactivity was detected in pellet (WAS transferred to bacteria)

Question 8

State the direction of DNA replication

Answer 8

Replication occurs in a 5’ to 3’ direction (on the newly synthesised strand)

Question 9

State the role of the following enzymes of the DNA replication process

Answer 9

Helicase: unwinds and separated double stranded DNA

DNA gyrase: relieves torsional strain created by helicase action to prevent supercoiling

SSB proteins: prevents DNA strands from re-annealing

DNA primase: lays down a short RNA primer to provide an initiation point for polymerisation (DNA pol III can only add nucelotides to the 3’-end of an existing nucleotide chain)

DNA polymerase III: extends the nucleotide chain from the primer

DNA polymerase I: removes and replaced RNA primers with DNA nucleotides

DNA ligase: covalently joins Okazaki fragments together (on lagging strand)

Question 10

Outline the difference between leading and lagging strands as the relate to Okazaki fragments

Answer 10

DNA strands are antiparallel, so DNA polymerase III moves in opposite directions on the two strands. On the leading strand, DNA polymerase III moves in the same direction as helicase so synthesis continuous. On the lagging strand, DNA polymerase III moves in the opposite direction to helicase - synthesis is discontinuous. The fragments generated on the lagging strand are called Okazaki fragments.

Question 11

Describe the role of deoxynucleoside triphosphates (dNTPs) in the replication process

Answer 11

Deoxynucleoside triphosphates (dNTPs) align opposite their complementary base partner. DNA pol III cleaves two of the phosphates and uses the energy to form a covalent phosphodiester bond. In this way, DNA pol III synthesises a new DNA strand

Question 12

Explain how dideoxynucleotides are used in DNA sequencing

Answer 12

Dideoxynucleotides (ddNTPs) lack the 3’-hydroxyl group needed to form a phosphodiester bond. This means the inclusion of a ddNTP will terminate the extension of a DNA sequence at that point. Four PCR cycles are set up, each with a different ddNTP (ddA, ddT, ddG or ddC) and a stock of normal bases. Each time the ddNTP is incorporated the sequence stops, generating fragments. When these fragments are separated and then ordered according to length, the DNA sequence is discerned. This process can be automated by using fluorescently labelled ddNTPs that can be detected by machine.

Question 13

Distinguish between the sense and antisense strands

Answer 13

The antisense strand is transcribed into mRNA (complementary to the DNA template). The sense strand is not transcribed into mRNA (it is identical to the DNA template.

Question 14

Identify the three sections of a gene

Answer 14

Promoter - initiation point (where RNA polymerase binds)
Coding Region - sequence that is transcribed
Terminator - termination point

Question 15

Explain the process of transcription including the role of RNA polymerase

Answer 15

RNA polymerase attaches to the promoter (requires the presence of transcription factors)
RNA polymerase unwinds the DNA and separates the strands
Nucleoside triphosphates (NTPs) align opposite complementary bases and are joined by RNA polymerase
This continues until the terminator, at which point the enzyme and transcript dissociate from the DNA

Question 16

List three examples of post-transcriptional modifications

Answer 16

Methyl cap added to the 5’-end (protects against degradation and allows recognition by ribosome)
Polyadenylation occurs at the 3’-end (improves stability and facilitates nuclear export)
Splicing occurs (introns are removed)

Question 17

Distinguish between introns and exons

Answer 17

Introns are intruding sequences within a gene that are removed prior to translation (not expressed). Exons are the expressing sequences translated into protein.

Question 18

Outline how alternative splicing increases the number of different proteins an organism can produce

Answer 18

Exons can be selectively removed via alternative splicing to produce different protein variants from the same gene - this increases the number of proteins an organism can produce relative to the number of genes

Question 19

Outline the role of the promoter and transcription factors in the regulation of transcription

Answer 19

Transcriptional activity is regulated by transcription factors which mediate binding of RNA polymerase to the promoter (initiation point for transcription of a gene sequence). The presence of certain transcription factors may be tissue-specific (explaining why different genes are activated in different tissues) and environmentally sensitive (explaining why expression patterns can change).

Question 20

Discuss the role the environment can play in regulating transcription

Answer 20

Environmental factors can trigger the production of transcription factors. Example, hydrangea flowers change colour according to the pH of the soil.

Question 21

Describe the role regulatory proteins play in moderating gene expression levels

Answer 21

The binding of RNA polymerase to the promoter can be controlled by regulatory proteins. Activators bind to enhancer sequences and make it easier for the enzyme to bind (increases expression). Repressors bind to silencer sequences and make it harder for the enzyme to bind (decreases expression).

Question 22

Outline the role of the nucleosome in the regulation of gene expression

Answer 22

Nucleosomes are involved in the supercoiling of DNA (supercoiled DNA is accessible for transcription). Acetylation of the histones makes the DNA less tightly coiled (increased gene expression). Methylation of the histones makes the DNA more tightly coiled (decreases gene expression).

Question 23

Describe how DNA methylation patterns may differ between identical twins and the significance of this

Answer 23

Direct methylation of DNA also decreases gene expression (prevents binding of transcription factors). Hence genes that are methylated are not expressed, while genes that are not methylated are expressed. Because DNA methylation is influenced by environmental factors, identical twins will develop different methylation patterns over time - this explains why twins may look different despite have identical genomes.

Question 24

Identify the following sites on a tRNA molecule and describe their function

Answer 24

Acceptor stem - carries the amino acid
T arm - associates with the ribosome (via E, P, A sites)
Anticodon - associates with mRNA (via complementary codon)
D arm - associates with tRNA-activating enzyme

Question 25

Describe the process by which amino acids are attached to tRNA molecules

Answer 25

Each tRNA molecule carries a specific amino acid (designated by the codon / anticodon)
Hence, different tRNA molecules associate with specific tRNA-activating enzymes

Question 26

Outline the structure of ribosomes

Answer 26

Ribosomes are composed of a large and a small subunit (made of protein and rRNA). The small subunit binds to mRNA. The large subunit contains three tRNA binding sites (E, P, and A)

Question 27

Differentiate prokaryotic vs eukaryotic ribosomes

Answer 27

Prokaryotic ribosomes are smaller than eukaryotic ribosomes
Prokaryote = 70S ribosomes ; Eukaryote = 80S ribosomes

Question 28

Differentiate bound vs free ribosomes

Answer 28

Free ribosomes are located in the cytoplasm and synthesize proteins for internal cellular use, while bound ribosomes are attached to the ER membrane and synthesize proteins for secretion or membrane incorporation.

Question 29

Explain the initiation process of translation

Answer 29

Small ribosomal subunit binds to mRNA and moves along it in a 5’ - 3’ direction until the START codon (AUG)
Specific tRNA (carrying Met) binds to the START codon (via complementary anticodon)
Large ribosomal subunit binds to the tRNA molecule (via the P-site), completing the ribosome

Question 30

Explain the elongation process of translation

Answer 30

Another tRNA molecule binds to the next codon in the mRNA sequence. The ribosome transfers the amino acid in the P-site to the amino acid in the A-site (via peptide bond)

Question 31

Explain the translocation process of translation

Answer 31

Ribosome moves one codon position along the mRNA sequence (in a 5’ - 3’ direction)
Deacylated tRNA molecule moves to E-site and is released, while tRNA in the A-site moves to the P-site
A new tRNA enters the empty A-site and the elongation and translocation stages are continually repeated

Question 32

Explain the termination process of translation

Answer 32

The ribosome reaches a stop codon and a releasing factor is recruited. This causes the completed polypeptide chain to dissociate and the ribosome to disassemble.

Question 33

Explain the termination process of translation

Answer 33

The ribosome reaches a stop codon and a releasing factor is recruited. This causes the completed polypeptide chain to dissociate and the ribosome to disassemble.

Question 34

Define polysome

Answer 34

A polysome is a group of ribosomes moving along an mRNA sequence simultaneously.

Question 35

Explain the levels of protein structure and indicate their significance

Answer 35

Primary structure: sequence of amino acids, formed via peptide bonds
Secondary structure: folding of polypeptide chain (alpha helices or beta-pleated sheets)
Tertiary structure: overall three dimensional shape of the polypeptide chain
quaternary structure: the presence of multiple polypeptides.