DNA Replication, Transcription and Translation

Question 1

Semi-Conservative Replication of DNA

Answer 1

When a cell prepares to divide, the two strands of the double helix separate and each of the original strands serve as templates for the creation of a new strand.

The new strands are formed by adding nucleotides, one by one, and linking them together. The result is two DNA molecules, both composed of an original strand and a newly synthesised strand, hence semi-conservative.

Complementary bases are added to the new strands.

Question 2

How does complementary base pairing work?

Answer 2

Complementary bases are added to the new strands. Only a nucleotide carrying a base that is complementary to the next base on the template strande can successfully be added to the new strand.

1. This is because complementary bases form hydrogen bonds with each other, stabilising the structure. If a nucleotide with the wrong base started to be inserted, hydrogen bonding between bases would not occur and the nucleotide would not be added to the chain.
2. Also because, when T and C are added there is too much space between the strands, when A and G would be pair bases, then there would be not enough space between the strands (purines - big molecues, pyrimidines - small molecules).

Question 3

Number of hydrogen bonds between complementary bases

Answer 3

C and G – 3 hydrogen bonds

A and T – 2 hydrogen bonds

Question 4

Meselson and Stahl and DNA replication

Answer 4

Meselson and Stahl cultured E. coli bacteria for many generations in a medium where the only nitrogen source was ¹⁵N, so the nitrogen in the bases of the bacterial DNA was ¹⁵N. They then transferred the bacteria abruptly to a medium with the less dense ¹⁴N isotope. Samples were taken at each stage.

A solution of caesium chloride was spun in an ultracentrifuge at 45,000 revolutions per minute for 24 hours. Caesium ions are heavy so tend to sink, establishing a gradient with the greatest caesium concentration and therefore density at the bottom. Any substance centrifuged with the caesium chloride solution becomes concentrated at the level of its density. Meselson and Stahl spun samples of DNA. The DNA shows up as a dark band in UV light.

After one generation the DNA was intermediate in density between ¹⁴N and ¹⁵N, as expected with one old and one new strand. After two generations there were two equal bands, one still ¹⁴N/¹⁵N and one at ¹⁴N density. In the following generations the less dense ¹⁴N band became stronger and the ¹⁴N/¹⁵N band weaker.

Question 5

Helicase

Answer 5

Helicase (enzyme) unwinds the double helix and separates the two strands by breaking hydrogen bonds between the complementary bases using ATP.

• Enzyme formation is donut-shaped (six-globular polypeptides), with one strand of the DNA molecule passing through the center of the donut.
• Double-stranded DNA cannot be split into two strands while it is still helical. Helicase therefore causes the unwinding of the helix at the same time as it separates the strands.

Question 6

DNA polymerase

Answer 6

DNA polymerase (enzyme) links nucleotides together to form a new strand (5’ to 3’), using the pre-existing strand as a template.

DNA polymerase brings nucleotides into the position where hydrogen bonds could form, but unless this happens and a complementary base pair is formed, the nucleotide breaks away again - one at the time. Hence, a very high degree of fidelity exists — few mistakes are made during DNA replication.

Question 7

lagging vs leading strand

Answer 7

Because of the antiparallel structure of DNA, the two strands have to be replicated in different ways.

On the leading strand DNA polymerases can move in the same direction as the replication fork so replication continuous.

On the lagging strand DNA polymerases have to move in the opposite direction to the replication fork, so replication is discontinuous.

Question 8

Outline of DNA replication (+general facts)

Answer 8

• replication occurs constantly during interphase
• occurs in a semi-conservative manner (one old strand, one new strand)

1. DNA gyrase moves in advance of helicase and relieves strains in the DNA molecule that are created when the double helix is uncoiled. It weakens and breaks the hydrogen bonds.
2. Helicase uncoils and breaks the hydrogen bonds between nucleotides, separating the two template strands.
3. Single-stranded binding proteins keep the strands apart long enough to allow the template strand to be copied.
4. RNA primase puts down an RNA primer, which initiates DNA polymerase.
5. DNA polymerase III adds complimentary nucleotides from 5’ to 3’ using deoxynucleoside triphosphate (dNTP).
6. DNA polymerase I removes the RNA primer and replaces it with DNA. A nick is left in the sugar-phosphate backbone of the molecule where two nucleotides are still unconnected.
7. Because DNA polymerase III can’t copy from 3’ to 5’, the RNA primase needs to go ahead and put down RNA primer, which RNA primase can then follow from 5’ to 3’ - Okazaki fragments.
8. DNA ligase seals up the nick by making another sugar-phosphate bond.

Question 9

Outline of Transcription (+splicing)

Answer 9

Transcription is the synthesis of mRNA copied from the DNA base sequences by RNA polymerase.

1. The enzyme RNA polymerase binds to a site on the antisense DNA at the start of a gene - promoter.
2. RNA polymerase moves along the gene separating DNA into single strands and pairing up RNA nucleotides with complementary bases on one strand of the DNA. Uracil pairs with adenine. 5’ to 3’
3. RNA polymerase forms covalent bonds between the RNA nucleotides.
4. The RNA separates from the DNA and the double helix reforms.
5. Transcription stops at the end of the gene, terminal, and the completed mRNA molecule is released, which carries the information needed to synthesise a polypetide in translation.
6. Splicing takes place, which removes introns (non-coding) from the exons (coding and expressive), forming mature mRNA.
7. The mature mRNA is taken to the ribosomes at rER.

Question 10

benefits of splicing

Answer 10

Splicing of mRNA increases the number of different proteins an organism can produce.

Question 11

Sense and antisense strand (+product of transcription)

Answer 11

The product of transcription is a molecule of RNA with a base sequence that is complementary to the template strand of DNA - the sense strand.

So, to make an RNA copy of the base sequence of one strand of a DNA molecule, the other strand is transcribed - the antisense strand.

Question 12

Translation

Answer 12

Translation is synthesis of polypeptides on ribosomes, using the base sequence of an RNA molecule produced in transcription.

Question 13

composition of subunits of ribosomes and their purpose

Answer 13

The two subunits of a ribosome are composed of rRNAandproteins.

Small subunit is where the mRNA is fed through.

Large subunit makes peptide bonds between amino acids, to link them together into a polypeptide and has tRNA binding sites.

Question 14

features of the genetic code

Answer 14

triplet code (three bases = one codon) code for 20 amino acids

Degenerate—having more than one base triplet to code for one amino acid.

Universal—found in all living organisms

Question 15

Messenger RNA and the genetic code

Answer 15

The amino acid sequence of polypeptides is determined by mRNA according to the genetic code.

Question 16

types of RNA

Answer 16

• mRNA (messenger) - produced in transcription and used in translation
• tRNA (transfer) - involved in decoding the base sequence of mRNA into amino acids during translation
• rRNA (ribosomal) - is part of the structure of ribosomes

Question 17

promoters

Answer 17

A specific region of non-coding DNA that initiates transcription of a particular gene.

Question 18

Epigenetics

Answer 18

When the environment has an impact on gene expression, which can be passed on to children.

Question 19

Codons and the genetic code

Answer 19

The genetic code is a triplet code - three bases code for one amino acid. Three bases are one codon. With four nucleotide bases, there are 64 different codons (4³); more than enough to code for the twenty amino acids in proteins. All 64 are used; there are two or more codonds for most amino acids.

These condons are universal for all organisms, with few exceptions, and code for the same amino acids.

Question 20

Codons & Anticodons

Answer 20

Translation depends on complementary base pairing between codons on mRNA and anticodons on tRNA.

Three components work together to synthesise polypeptides by translation:

• mRNA has a sequence of codons that specifies the amino acid sequence of the polypeptide
• tRNA molecules have an anticodon of three bases that binds to a complementary codon on mRNA and they carry the amino acid corresponding to that codon
• ribosomes act as the binding site for mRNA and tRNAs and also catalyse the assembly of polypeptide

Question 21

Outline of Translation

Answer 21

1. An mRNA binds to the small subunit of the ribosome on rough endoplasmic reticulum.
2. A molecule of tRNA with an anticodon complementary to the start codon to be translated on the mRNA binds to the ribosome.
3. A second tRNA with an anticodon complementary to the first codon on the mRNA then binds. A maximum of two tRNAs can be bound at the same time.
4. The ribosome transfers the amino acid carried by the first tRNA to the amino acid on the second tRNA, by making a new peptide bond. The second tRNA is then carrying a chain of two amino acids — a dipeptide.
5. The ribosome moves along the mRNA so the first tRNA is released, the second becomes the first, from 5’ to 3’.
6. Another tRNA binds with an anticodon complementary to the next codon on the mRNA.
7. Stages 4,5 and 6 are repeated again and again, with one amino added to the chain each time the cycle is repeated. The process continues along the mRNA until a stop codon is reached, when the completed polypeptide is released. (If a large polypeptide chain is produced, the mRNA can move accross multiple ribosomes at once.)
8. Folding of the polypeptide completes the conformation of the amino acids.

Question 22

an example of the universality of the genetic code

Answer 22

Production of human insulin in bacteria is an example of the universality of the genetic code allowing gene transfer between species.

The insulin produced in E. coli, can be used for diabetes treatment. The amino acid sequence of insulin that is produced in these organsims is identical to the sequence produced in humans, because organisms use the same genetic code so each codon in the mRNA is translated into the same amino acid when insulin is made.

Despite some occasional differences in the amino acid sequence between animal and human insulin, they all bind to the human insulin receptor and cause lowering of blood glucose concentration.

There are, however, some exceptions in organisms like yeast.

Question 23

Difference between replication and transcription

Answer 23

1. Replication is the duplication of two-strands of DNA. Transcription is the formation of single, identical RNA from the two-stranded DNA.
2. There are different proteins involved in replication and transcription (tRNA, mRNA, etc.)
3. In replication, the end result is two daughter cells, while in transcription, the end result is a protein molecule.
4. In transcription, DNA serves as the template for RNA synthesis.
   http: //www.diffen.com/difference/Replication_vs_Transcription

Question 24

peptides and polypeptides

Answer 24

Peptides short chains of amino acid monomers linked by peptide bonds.

Hence, peptides fall under the broad chemical classes of biological oligomers and polymers, alongside nucleic acids, oligosaccharides and polysaccharides, etc.

Question 25

a peptide bond

Answer 25

A peptide bond is a covalent bond formed between two amino acids.

Question 26

location of transcription and translation in prokaryotes and eukaryotes

Answer 26

Eukaryotes

Transcription occurs in the nucleus and then the mRNA leaves through the membrane pores to ribosomes on the rER.

Prokaryotes

Transcription occurs in the nucleoid and translation can happen right after or during transcription because there is no nuclear membrane.

Question 27

different tRNA binding sites in ribosomes (large sub-unit)

Answer 27

A site - tRNA arrives and binds to the condons with their anticodons

P site - the tRNA from A moves to P and gives of its attached amino acid to the tRNA at site A.

E site - mRNA and tRNA from P exists the ribosome