ALL 2

Question 1

Select the INCORRECT pair relating to the specificity of ligands and receptors:

Answer 1

Adrenaline and muscarinic acetylcholine receptor

Incorrect because adrenaline binds to adrenergic receptors, not muscarinic acetylcholine receptors.

Question 2

Glucagon and the GlucaGen HypoKit:

Answer 2

Glucagon uses a G protein-coupled receptor to produce signal transduction.

Correct: Glucagon receptors are GPCRs.

Question 3

Second messenger NOT used by GPCRs:

Answer 3

Adenylate cyclase

Adenylate cyclase is an enzyme, not a second messenger.

Question 4

INCORRECT statement regarding receptor proteins:

Answer 4

Receptors release the ligand as a changed product.

Incorrect: Receptors typically do not chemically change ligands.

Question 5

Good chemical starting point for discovering a GPCR antagonist:

Answer 5

The endogenous ligand for their target receptor.

This would be a good starting point for discovering an antagonist.

Question 6

Flow of genetic information in a cell:

Answer 6

DNA -> mRNA -> Protein.

Question 7

Translation is the synthesis of:

Answer 7

Protein from an mRNA template.

Question 8

Key elements of a gene:

Answer 8

Promoter region, exons, transcription factor binding sites, introns.

Question 9

Alternate forms of a gene are called:

Answer 9

Alleles.

Question 10

True statement about the inheritance of alleles:

Answer 10

One allele comes from each parent.

Question 11

Parents’ genotypes if two straight-haired people have a curly-haired child:

Answer 11

Heterozygous.

Question 12

Least likely genetic variant to have severe consequences:

Answer 12

A variant that changes a codon to another codon for the same amino acid.

Question 13

How a person’s genotype can be determined:

Answer 13

All of the above.

All listed assays can be used for genotyping.

Question 14

Who will show symptoms of the genetic disease (from gel image):

Answer 14

Person 2.

Person 2 has the homozygous recessive genotype.

Question 15

Why PCR is commonly used in genetic testing:

Answer 15

It generates a large number of copies of a specific DNA region.

Question 16

Sickle Cell Anaemia is:

Answer 16

A monogenic disease caused by the E6V haemoglobin genetic variant.

Question 17

A disease or disorder can be:

Answer 17

All of the above.

Diseases can have varying degrees of genetic and environmental causes.

Question 18

Genetic variant that could cause a severe phenotype:

Answer 18

Anywhere in a reaction pathway.

Question 19

The effect of a genetic variant on a phenotype can be modified by the environment if:

Answer 19

The genetic variant does not severely change the protein’s function.

Question 20

Best explanation for why older people have a higher chance of getting cancer:

Answer 20

Because their DNA has been damaged over time.

Question 21

A complex disease is:

Answer 21

Caused by multiple, complicated gene-environment interactions.

Question 22

Best host to produce functional “PepZ” (requiring gamma-carboxylation):

Answer 22

Chinese Hamster Ovary Cells.

Question 23

NOT an advantage of using eukaryotic cells for therapeutic protein production:

Answer 23

Can make proteins from any tissue.

Eukaryotic cells cannot make all proteins.

Question 24

INCORRECT statement about recombinant expression vectors:

Answer 24

They are double-stranded pieces of linear DNA.

They are circular DNA.

Question 25

Protein type and recombinant production systems NOT correctly paired:

Answer 25

Glycosylated protein and bacteria. ## Footnote Bacteria do not glycosylate proteins.

Question 26

Restriction enzymes cut:

Answer 26

DNA at specific sites.

Question 27

Correct statement about human insulin:

Answer 27

The A and B chains of human insulin can be synthesized recombinantly without the requirement for in vivo post-translational modifications.

Question 28

Recombinant insulin is produced in:

Answer 28

E.coli.

Question 29

Erythropoietin (EPO):

Answer 29

Can be used to treat anaemic cancer patients.

Question 30

Genes containing introns cannot be expressed in bacteria because:

Answer 30

Bacteria do not have the cellular machinery required for splicing out introns.

Question 31

Statement about EPO and insulin production in E. coli:

Answer 31

EPO produced in E. coli is not functional because it requires glycosylation.

Question 32

C. Can be used to treat anaemic cancer patients.

Answer 32

D. Bacteria do not have the cellular machinery required for splicing out introns. ## Footnote Example sentence: Bacteria lack the necessary enzymes for intron splicing.

Question 33

Statement about EPO and insulin production in E. coli:

Answer 33

A. EPO produced in E. coli is not functional because it requires glycosylation. ## Footnote Example sentence: EPO needs glycosylation for proper function.

Question 34

Restricting expression of recombinant proteins to specific tissues in animals can be controlled by:

Answer 34

B. A mammalian cell-type specific promoter. ## Footnote Example sentence: Specific promoters can target expression to certain tissues.

Question 35

Major benefit of gene therapy:

Answer 35

B. It can treat genetic disorders at their source. ## Footnote Example sentence: Gene therapy addresses the root cause of genetic disorders.

Question 36

Statement about dilution of solution A (12 mL of solution A mixed with 24 mL of solution B):

Answer 36

C. The undiluted solution A had a concentration of 90 mg/mL. ## Footnote Example sentence: The original concentration of solution A was 90 mg/mL.

Question 37

Volume delivered by the pipette:

Answer 37

C. 350 μL. ## Footnote Example sentence: The pipette dispensed 350 microliters of liquid.

Question 38

50 mmol/μL equals:

Answer 38

A. 0.05 mol/mL. ## Footnote Example sentence: 50 millimoles per microliter is equivalent to 0.05 moles per milliliter.

Question 39

A 1 in 12 dilution of an enzyme means:

Answer 39

A. 1 part enzyme diluted with 11 parts of buffer. ## Footnote Example sentence: A 1 in 12 dilution involves adding 1 part enzyme to 11 parts buffer.

Question 40

To dilute a solution from 90 mg/mL to 15 mg/mL:

Answer 40

B. 1 part solution to 6 parts water. ## Footnote Example sentence: Diluting from 90 mg/mL to 15 mg/mL requires mixing 1 part solution with 6 parts water.

Question 41

Structural component of starch responsible for blue-black colour in iodine test:

Answer 41

A. Amylose. ## Footnote Example sentence: Amylose is the starch component that gives a blue-black color with iodine.

Question 42

High heat (80°C) affects bonds stabilising protein:

Answer 42

B. Secondary, tertiary, and quaternary structure. ## Footnote Example sentence: Protein structure is affected at 80°C, impacting secondary, tertiary, and quaternary bonds.

Question 43

Beer's Law in spectrophotometry indicates:

Answer 43

C. Concentration of the solution. ## Footnote Example sentence: Beer's Law relates absorbance to solution concentration.

Question 44

Lambert's Law in spectrophotometry indicates:

Answer 44

C. Decreases logarithmically with increasing pathlength. ## Footnote Example sentence: Lambert's Law describes the logarithmic decrease in absorbance with longer pathlengths.

Question 45

In the Beer-Lambert Law, the term 'l' denotes:

Answer 45

C. Light-path length. ## Footnote Example sentence: The variable 'l' in the Beer-Lambert Law represents the distance light travels through the sample.

Question 46

Wavelength at which a molecule absorbs light maximally:

Answer 46

D. Can be determined by constructing an absorbance spectrum. ## Footnote Example sentence: Absorbance spectra help identify the wavelength of maximum light absorption for a molecule.

Question 47

INCORRECT statement about spectrophotometry:

Answer 47

C. The wavelength of a standard cuvette is 1 cm. ## Footnote Example sentence: The path length, not the wavelength, of a standard cuvette in spectrophotometry is typically 1 cm.

Question 48

Correct statement about enzyme assays and absorbance:

Answer 48

A. When using enzyme assays to analyse complex solutions for specific molecules, it is assumed that the reaction goes to completion. ## Footnote Example sentence: Enzyme assays assume complete reaction to analyze complex mixtures for specific molecules.

Question 49

Dilution of solution A (80 mL of solution A added to 720 mL of water):

Answer 49

B. 1 in 10 dilution. ## Footnote Example sentence: Diluting solution A by adding 720 mL of water to 80 mL results in a 1 in 10 dilution.

Question 50

Reason for the decrease in gradient of a progress curve:

Answer 50

B. Substrate concentration decreases over the reaction period. ## Footnote Example sentence: The decline in gradient on a progress curve is due to decreasing substrate concentration during the reaction.

Question 51

Good standard curve (absorbance plotted against concentration):

Answer 51

B. I and III (passes through the origin and is a straight line). ## Footnote Example sentence: A standard curve that intersects the origin and is linear is considered good for absorbance versus concentration.

Question 52

Alcohol dehydrogenase catalyses the conversion of:

Answer 52

A. Ethanol to acetaldehyde. ## Footnote Example sentence: Alcohol dehydrogenase converts ethanol into acetaldehyde.

Question 53

INCORRECT statement about C-blue solution absorbance:

Answer 53

A. The concentration of the 1 in 6 diluted C-blue sample is 15 mmol/L. ## Footnote Example sentence: The correct concentration for the 1 in 6 diluted C-blue sample is not 15 mmol/L.

Question 54

Correct concentration of diluted solution Y based on absorbance:

Answer 54

C. 11.4 μmol/L. ## Footnote Example sentence: The concentration of diluted solution Y, determined by absorbance, is 11.4 micromoles per liter.

Question 55

Initial velocity of assay using 200 μL of ADH:

Answer 55

A. 0.075 ΔA/s. ## Footnote Example sentence: The initial velocity of the assay with 200 μL of ADH is 0.075 absorbance units per second.

Question 56

KM for enzyme from Lineweaver-Burk graph:

Answer 56

C. 2.5 mmol/L. ## Footnote Example sentence: The KM value for the enzyme, obtained from the Lineweaver-Burk plot, is 2.5 millimoles per liter.

Question 57

Vmax for enzyme from Lineweaver-Burk graph:

Answer 57

B. 0.2 ΔA/min. ## Footnote Example sentence: The Vmax value for the enzyme, derived from the Lineweaver-Burk graph, is 0.2 absorbance units per minute.