Unit 3 AOS 1 - What is the role of proteins and nucleic acids in maintaining life?

Question 1

Describe the general structure of an amino acid (3 marks)

Answer 1

Describe the general structure of an amino acid (3 marks)
- Alpha carbon and hydrogen
- Amine group
- Carboxyl group
- Variable R group

Question 2

Describe the process of transcription (3 marks)

Answer 2

• RNA polymerase attaches to the promoter region and unzips DNA
• As RNA polymerase reads the template strand, it assigns complementary nucleotides and replaces thymine with uracil.
• Once RNA polymerase reaches a termination sequence, the mRNA is released and transcription ends

Question 3

Describe the process of RNA processing (3 marks)

Answer 3

• Methyl-G cap added at the 5’ end of pre-mRNA
• Poly-A tail added at the 3’ end of pre-mRNA
• Introns spliced and exons joined together

Question 4

Describe the process of translation (3 marks)

Answer 4

• mRNA attaches to a ribosome
• The ribosome reads the mRNA a codon at a time
• tRNA carrying a specific amino acid travels to the ribosome
• If the tRNA anticodon is complementary to the mRNA codon than the amino acid is added to the polypeptide chain via a condensation reaction

Question 5

Describe how an enzyme produces a product with an substrate (3 marks)

Answer 5

• An enzyme has an induced fit with a substrate.
• The enzyme’s active site is complementary to the substrate, and the two form together to create an enzyme-substrate complex.
• After the two have a reaction, the substrate is broken down into products that are more useful and the enzyme is recycled into more uses.

Question 6

Describe how a competitive inhibitor affects enzyme function (2 marks)

Answer 6

• A competitive inhibitor has a similar structure to the substrate, meaning it is also complementary to the active site of an enzyme.
• When the competitive inhibitor binds with the enzyme, no reaction occurs and it decreases enzyme activity

Question 7

Describe how a non-competitive inhibitor affects enzyme function (2 marks)

Answer 7

• A non-competitive inhibitor binds with an enzyme at a site other than the active site called the allosteric site
• The enzyme undergoes conformational changes in the active site of the enzyme, meaning the substrate is no longer complementary

Question 8

Describe how co-enzymes assist enzyme functions (2 marks)

Answer 8

• Coenzymes are organic, non-protein molecules that assist substrates by helping them fit the active site of target enzymes better
• They release energy during a reaction and are recycled afterwards

Question 9

Explain how DNA differs from RNA (3 marks)

Answer 9

• DNA is doubled stranded where RNA is single stranded
• DNA has a deoxyribose sugar where RNA has a ribose sugar
• DNA has thymine where RNA has uracil

Question 10

Describe the process of repression in the trp operon when trp levels are high and conversely when they are low (6 marks)

Answer 10

• When intracellular tryptophan levels are high, there is available tryptophan to bind to a repressor protein.
• When a trp is binded with this repressor protein, it undergoes a conformational change that allows it to bind to the operator region.
• This blocks RNA polymerase from transcribing the structural genes and thus does not lead to more tryptophan being produced
• When intracellular tryptophan levels are low, there is no available tryptophan to bind to a repressor protein
• This means that the repressor protein does not bind to the operator region
• This means RNA polymerase can transcribe the structural genes and this leads to more tryptophan being produced in the cell

Question 11

Describe the process of attenuation as a form of gene regulation in the trp operon when tryptophan levels change (6 marks)

Answer 11

• When intracellular levels of tryptophan levels are high, there is a sufficient amount of tryptophan bound tRNA.
• When the ribosome translates the two adjacent trp codons along the leader mRNA, tRNA adds trp into the polypeptide chain
• The ribosome pauses at the STOP codon, allowing a terminator hairpin to form, causing RNA polymerase to detach from the operon and prevent the transcription of the structural genes
• When intracellular levels of tryptophan levels are low, there is little tryptophan bound tRNA
• When the ribosome attempts to translate the leader mRNA there are two adjacent trp codons, but since there is not sufficient trp tRNA, the ribosome pauses
• This causes an anti-terminator hairpin to form, and the RNA polymerase does not detach from the operon and transcription of the structural genes occur.

Question 12

Describe the pathway proteins take from being produced to leaving the cell (4 marks)

Answer 12

• Proteins are synthesized at the ribosome
• Proteins are folded and transported at the rough endoplasmic reticulum
• Proteins in transport vesicles are transported to the golgi apparatus
• Golgi apparatus modifies and packages proteins into secretory vesicles
• Secretory vesicles fuse with the plasma membrane and export proteins via exocytosis

Question 13

Describe the process of the Polymerase Chain Reaction (4 marks)

Answer 13

• First, denaturation of DNA occurs. At 95 degrees Celsius, the hydrogen bonds between nitrogenous bases of the double stranded DNA break and DNA becomes single stranded
• Second, annealing occurs at 55 degrees Celsius. Two different primers join on the separated strands of DNA at their complementary sections
• Finally, extension occurs at 72 degrees. This is the optimal temperature of Taq polymerase and the new strand of DNA is built using complementary nucleotides
• Repeat steps until desired amount of DNA is reached

Question 14

Describe the CRISPR-Cas9 defence system in bacteria (6 marks)

Answer 14

• a bacteriophage inserts its DNA into a bacteria cell
• the bacteria cell recognises the PAM sequence as foreign and stores its DNA into a spacer sequence in CRISPR
• gRNA is transcribed from the CRISPR sequence
• gRNA and Cas9 form a complex which seeks out bacteriophage DNA at the PAM site.
• Cas9 unzips the bacteriophage DNA, and if the gRNA is complementary to the DNA it hinds with the sequence
• Cas9 cuts the viral DNA upstream of the PAM sequence disabling the viral gene and protecting the bacteria

Question 15

Describe how scientists can insert genes into bacteria to synthesise proteins using recombinant plasmids. (5 marks)

Answer 15

• A plasmid with antibiotics resistance and a gfp gene, and the gene of interest are cut with the same restriction endonuclease leaving complementary sticky ends
• They are combined in a mixture with DNA ligase to attempt to glue them together
• Bacteria are heat-shocked to make the plasma membrane were permeable allowing some plasmids to enter.
• Scientists will place the bacteria into a agar plate containing antibiotics. The bacteria that survive and do not glow because the restriction site cuts the gfp gene, are the transformed bacteria
• These bacteria are isolated and the proteins extracted and purified

Question 16

Describe the steps taken by scientists to deactivate BCL11A (4 marks)

Answer 16

• sgRNA is create by scientists that is complementary to a section of the BCL11A gene
• a Cas9 enzyme with an appropriate PAM sequence is obtained
• the sgRNA is combined with Cas9 to create a complex
• the complex is injected into bone marrow stem cells
• the Cas9 locates the PAM sequence and checks upstream if the sgRNA is complementary to the target BCL11A gene
• If so, the DNA is cut, and as the cell attempts to repair it, the random nucleotides added disable the gene

Question 17

The mutation in HBB causes a change from GLU to VAL in the beta-chain of haemoglobin. Identify the changes in the protein structure that result from the mutations? (2 marks)

Answer 17

• Primary structure, because the base sequence of amino acids in the polypeptide chain is changed
• Tertiary structure, because the normal folding and interaction in the beta-chain are disrupted leading to the normal round shape changing to a sickle cell shape