tssm 2023 answers

biology exam tssm unit 3 2023

with answers

Reading time: 15 minutes
Writing time: 1 hour and 30 minutes

Section A

25 marks.

Question 1

The purpose of ribosomal RNA (rRNA) is to

• carry genetic information from the nucleus to the ribosome.
• bring a specific amino acid to the ribosome during transcription.
• bind with proteins to form ribosomes.
• catalyse the addition of free nucleotides in a growing mRNA molecule.

Ribosomal RNA forms part of the structural component of ribosomes. mRNA carries the genetic code from the nucleus, and tRNA carries amino acids.

Question 2

Operator regions regulate transcription in prokaryotes by

• producing repressor proteins that can bind to the promotor region.
• creating a truncated protein when a protein is not required to be expressed.
• removing introns and splicing exons together.
• providing a binding site for the repressor protein.

Regulatory genes produce repressor proteins and attenuation creates truncated proteins. Prokaryotes do not undergo RNA processing. The operator allows the repressor protein to bind, inhibiting transcription from occurring.

Question 3

Gel electrophoresis is a tool used to separate DNA fragments based on their molecular size. Two students ran a gel using the same sample of DNA, the same restriction enzymes, and the same concentration of buffer solution. They both ran the sample for 30 minutes. When they analysed their results, they noticed that in the first students’ sample, the fragments travelled further than the second students’ sample. This would be due to

• the first students’ DNA sample having a stronger negative charge.
• the second student having a stronger current running through the gel.
• the first student having a contaminated sample.
• the second student using the micropipette more accurately than the first student.

As the students had the same DNA sample and the same restriction enzymes, the factor that would affect the position of the bands is the current applied, with a higher current causing the fragments to move through the gel faster than a lower current.

Question 4

In photosynthesis, the light dependent reaction splits water into hydrogen ions and oxygen molecules. The hydrogen ions

• bind to NAD+ and move to the matrix.
• bind to NADPH and move to the cristae.
• bind to NADP+ and move to the stroma.
• are enzymes that catalyse carbon fixation.

The H+ ions from water bind to the co-enzyme NADP+ and move from the grana to the stroma. The matrix and cristae are in the mitochondria, and Rubisco is responsible for carbon fixation.

Question 5

The protein secretory pathway provides a way in which proteins can be exported from the cell. The correct sequence of organelles in which this process occurs is which of the following?

• rough endoplasmic reticulum, ribosome, Golgi body, secretory vesicle, transport vesicle
• ribosome, rough endoplasmic reticulum, secretory vesicle, Golgi body, transport vesicle
• ribosome, rough endoplasmic reticulum, transport vesicle, Golgi body, secretory vesicle
• ribosome, smooth endoplasmic, reticulum, secretory vesicle, Golgi body, transport vesicle

The protein is synthesized at the ribosome, folding occurs at the rough endoplasmic reticulum. The protein is then placed in a transport vesicle and sent to the Golgi body for further modification. Finally, the protein is placed in a secretory vesicle for export from the cell.

Question 6

A plant in a hot environment with ample access to water displays minimal photorespiration. This plant is most likely to be which of the following?

• C3
• C4
• CAM
• a cactus

Photorespiration occurs in C3 plants, whereas C4 and CAM (including cacti) have adaptations to minimize photorespiration. As water is available, the plant is most likely to be C4 as CAM plants are adapted to arid regions.

Question 7

When creating a recombinant plasmid of insulin, the gene for insulin needs to be inserted within the β galactosidase gene. The reason for this is

• the promotor for the β galactosidase gene allows the insulin to be transcribed.
• itconfersantibioticresistancetotherecombinantplasmid.
• the recognition site for the endonuclease is found in the β galactosidase gene.
• β galactosidase is the binding site for RNA polymerase.

As the B galactosidase gene is part of the lac operon, it contains a promoter region. As such, by inserting the gene for insulin within this gene, the insulin will be transcribed and translated. Different endonucleases have different recognition sequences and antibiotic resistance is in another gene.

Question 8

In bacteria, the CRISPR CAS 9 complex occurs naturally to protect from re-exposure to a pathogen. The viral DNA is stored as a

• repeat sequence
• spacer sequence
• guide sequence
• PAM site sequence

Viral DNA is stored as a spacer, with the repeats separating the sequences stored from different viral genomes.

Question 9

A tRNA anticodon has the 3-nucleotide sequence of UUA. The corresponding sequence on the coding strand of DNA would be which of the following?

• AAT
• TTA
• AAU
• UUA

The coding strand is complementary to the template strand from which mRNA is read. If the coding strand is TTA, the template strand would be AAT and the complementary mRNA would be UUA. As the anticodon is complementary to the mRNA, this would be UUA.

Question 10

The amino acids tryptophan and methionine are only encoded for by one codon. This means that these amino acids are

• degenerate
• not degenerate
• catalysts
• co-enzymes

As Methionine and Tryptophan are only coded for one codon, they are not degenerate. Degenerate means that multiple codons will code for one amino acid. Enzymes are catalysts, and co-enzymes increase the affinity of an enzyme to a substrate.

Question 11

A competitive inhibitor

• binds to the allosteric site, allowing the rate of reaction to eventually reach the same rate as an uninhibited reaction.
• creates an attenuated protein.
• regulates the rate of transcription by binding to the operator region.
• binds to the active site, and an increase in substrate minimises the impact of the inhibitor.

A competitive inhibitor binds to the active site of an enzyme, whereas a non-competitive inhibitor binds to the allosteric site. Adding more substrate increases the probability of the substrate binding to the enzyme, rather than the inhibitor, reducing the impact on the rate of reaction.

Question 12

The final H+ acceptor in aerobic respiration is

• oxygen
• FAD+
• NADP+
• NAD+

Oxygen is the final acceptor in the electron transport chain, with NAD accepting H+ ions in glycolysis and the Krebs cycle, and FAD accepting H+ ions in the Krebs cycle only. NADP+ is a H+ acceptor in photosynthesis.

Question 13

A small sample of DNA was extracted from a fossil and PCR was conducted in a laboratory to amplify the sample. The sample, RNA polymerase, free nucleotides and primers were added.
After 30 minutes it was observed that the amount of DNA in the sample remained unchanged. The most likely explanation for this is

• PCR requires a large sample.
• the incorrect enzyme was added.
• insufficient free nucleotides were added.
• primers are not used in PCR.

PCR uses Taq or DNA polymerase, not RNA polymerase, which is why the DNA sample would not amplify. If there was a low number of free nucleotides, it would still be expected that the initial DNA sample would somewhat increase in volume.

Question 14

When considering changes in pH and temperature for an enzyme driven reaction, it is fair to state

• at a pH above and below the optimum, and a temperature below the optimum, the rate of reaction will be low.
• at an optimum pH and temperature, the amount of substrate will remain unchanged.
• at a pH and temperature above the optimum, the rate of reaction will increase.
• at a pH and temperature below the optimum, the rate of reaction will increase.

At a pH level above and below the optimum, and a temperature above the optimum, the rate of reaction decreased as the enzyme denatures. Substrate is converted to product when the enzyme is at the optimal level.

Question 15

What is the total ATP yield in glycolysis?

• 2
• 4
• 26
• 30

The net ATP in glycolysis is 2, however 4 are produced in total. 26 ATP are produced in the electron transport chain, and a total of 30 ATP is produced in aerobic respiration.

Question 16

The optimal temperatures for polymerase chain reaction in order are

• 95 C, 55 C and 82 C
• 55 C, 95 C and 72 C
• 95 C, 55 C and 72 C
• 82 C, 95 C and 55 C

The first stage, denaturation, occurs at 95oC. The second stage, annealing, occurs at 55oC and the final stage, elongation; occurs at 72oC.

Question 17

Post translational folding of proteins initially occurs

• at the smooth endoplasmic reticulum.
• at the rough endoplasmic reticulum.
• within the nucleus.
• at the Golgi complex.

Initial folding occurs at the rough endoplasmic reticulum, with further modification at the Golgi body. Post transcriptional modification occurs within the nucleus, and the smooth endoplasmic reticulum is the site of lipid synthesis.

Question 18

The production of bioethanol requires a series of steps, one of which is enzymatic hydrolysis. The purpose of enzymatic hydrolysis is to

• produce carbon dioxide.
• directly convert glucose into ethanol.
• break stored sugars into monomers.
• break down chitin into monomers.

Enzymatic hydrolysis is the process of using enzymes and the addition of a water molecule to convert chains of stored sugars into monomers. Chitin is found in fungal cell walls, and as such, is not used to produce bioethanol.

Question 19

The labels of the axis on the graph below could be

• x: temperature; y: rate of reaction
• x: substrate concentration; y: rate of reaction
• x: rate of reaction; y: pH
• x: enzyme concentration; y: substrate concentration

The x axis could be either substrate concentration or enzyme concentration, however the rate of reaction is the only option for the y axis. A pH or temperature graph would show a decline following the optimum value.

Question 20

The Cas9 enzyme acts as molecular scissors, cutting through the double strand of DNA. Cas9 cuts at a precise location by

• binding to the target sequence of DNA and cutting 2-5 nucleotides upstream.
• recognising a specific palindromic sequence of DNA and cutting 2-5 nucleotides downstream.
• binding to the tracer sequence of DNA and cutting 2-5 nucleotides downstream.
• binding to the PAM sequence of DNA and cutting 2-5 nucleotides upstream.

The Cas9 enzyme binds to the PAM sequence and cuts 2-5 nucleotides upstream. Endonucleases search for a specific palindromic sequence.

Question 21

RNA processing

• removes introns as they are non-coding regions.
• adds a methyl cap to the 3’ end and a poly A tail to the 5’ end to protect the mRNA molecule from degradation when it leaves the nucleus.
• splices introns together.
• converts mRNA to DNA using reverse transcriptase.

RNA processing removes introns and splices exons together. The methyl cap is added to the 5’ end and the poly A tail is added to the 3’ end.

Note

The following information relates to questions 22 - 25
A student conducted an experiment to observe the rate of cellular respiration in yeast. Prior to undertaking the experiment, they conducted research and identified that yeast cells have mitochondria, therefore can undergo aerobic respiration.

They added a yeast and sugar solution to a beaker and placed an airtight lid on top. They used an ethanol, carbon dioxide and oxygen probe to measure the changes in the solution over a 30-minute period.

Question 22

After 30 minutes it would be expected that

• ethanol levels would decrease.
• carbondioxidelevelswoulddecrease.
• oxygen levels would increase.
• ethanol levels would increase.

Initially aerobic respiration would occur, increasing carbon dioxide and decreasing oxygen levels. When oxygen resources have been depleted, anaerobic respiration would occur, increasing carbon dioxide and ethanol levels.

Question 23

After 30 minutes it would be expected that

• aerobic respiration would be occurring.
• anaerobic respiration would have stopped.
• carbon dioxide is being used as an input of aerobic cellular respiration.
• anaerobic respiration would occur.

As oxygen levels would have been depleted, the yeast would then undergo anaerobic respiration.

Question 24

The change in carbon dioxide levels at 15 minutes would be due to

• carbon dioxide being an input in glycolysis.
• carbon dioxide being produced in the Krebs cycle.
• carbon dioxide being an oxygen acceptor in the electron transport chain.
• carbon dioxide being produced in the cristae.

As oxygen would still be present, aerobic respiration would still be occurring. Carbon dioxide is produced in both glycolysis and the Krebs cycle, and oxygen is the final acceptor in the electron transport chain.

Question 25

A controlled variable in the experiment would be

• the volume of yeast used.
• the oxygen concentration.
• the number of bubbles produced.
• the amount of product produced.

The volume of yeast should be the same for any repetitions of the experiment. Oxygen, ethanol and carbon dioxide levels all changed and were measured.

Section B

50 marks.

Question 1

The trp operon contains a group of genes that regulate the production of the amino acid tryptophan in prokaryotes.

1. Why is it important for the bacterium to produce tryptophan? ]]2

   Tryptophan is an amino acid and amino acids are the monomers of proteins (1). If tryptophan was unavailable, the bacterium may not be able to produce all required proteins OR energy would be wasted transcribing mRNA that could not be translated (1)

2. What role does the attenuator region of the trp operon have in regulating the rate of tryptophan synthesis? ]]3

   The attenuator region contains domains which will control the expression of the trp genes (1). These domains can create loops which either cause a terminator hairpin loop to form – stopping translation (1) or a non-terminator hairpin loop – allowing translation to continue (1)

3. Describe one way in which protein synthesis differs between prokaryotes and eukaryotes. ]]1

   One of:
   Protein synthesis is faster in prokaryotes than eukaryotes as transcription and translation occur simultaneously.
   Eukaryotes have RNA processing whereas prokaryotes do not.
   Eukaryotes have introns and exons whereas prokaryotes have only exons.

4. The amino acid code is described as redundant. What is meant by this term? ]]1

   More than one codon codes for the same amino acid.

Question 2

DNA manipulation involves techniques that can amplify, separate DNA fragments based on molecular size or alter the sequence of nucleotides in an organism.
One technique, creating recombinant plasmids, can introduce DNA from a different species into a bacterial genome.

1. What term is used to describe a genetically modified organism that has genetic material from another species? ]]1

   Transgenic

2. Describe the characteristic of DNA allows for genetic material to be transferred between species? ]]2

   DNA is universal (1) with the same DNA triplet coding for the same amino acid in all organisms (1)

3. Identify two enzymes used in DNA manipulation and describe their role. ]]4

   Two of:
   DNA ligase (1) restores phosphodiester bonds in DNA (1)
   Taq/DNA polymerase (1) catalyses the addition of free nucleotides (1) Endonucleases (1) cut DNA at a specific recognition sequence (1)

Question 3

Wheat is the largest grain crop in Australia, with regions in the west of Victoria, including the Wimmera, Central goldfields and Mallee forming part of Australia’s wheat belt.
As a C3 plant, wheat does not have any adaptations to overcome changes in photosynthetic rate that can occur the high temperatures.

1. What are the names and locations of the two stages of photosynthesis? ]]2

   Light dependent stage in the grana/thylakoid (1) and light independent in the stroma (1)

2. In the Wimmera and Mallee regions, temperatures can reach above 30 degrees Celsius. Wheat ready for harvest can be exposed to these high temperatures. Explain, at a molecular level, what is occurring within the chloroplast at high temperatures in the wheat plant. ]]3

   Chlorophyll is a protein(1) at high temp it will denature(1) and the tertiary structure will irreversibly change(1)

3. A farmer decided to grow maize, a C4 plant that is a starchier version of a typical sweet corn, in place of wheat. Describe how the process of carbon fixation would differ in the wheat, a C3 plant, and the maize plant at high temperatures? ]]3

   Carbon fixation in maize would occur in the bundle sheath, whereas in C3 it occurs in the mesophyll (1). C3 plants produce a 3-carbon molecule whereas C4 plants produce a 4-carbon molecule (1). C4 plants are more efficient than C3 plants at high temperatures (1).

4. CRISPR Cas9 technologies can be used to increase crop yield. Altering Rubisco to increase its affinity to carbon dioxide is one such option. Outline the process that scientists would undertake to modify rubisco using CRISPR. ]]6

   6 of:
   Isolate the gene of interest (1).
   Create a single guide RNA sequence complementary to the target DNA sequence, with a PAM sequence adjacent to the target DNA (1).
   Add the sgRNA to the cell with the Cas9 enzyme (1).
   Cas9 binds to the PAM sequence and unzips the target DNA (1).
   sgRNA binds to the target sequence (1).
   Cas9 moves 3-5 nucleotides upstream of PAM and cuts the sequence (1)
   Sequence can be modified by disrupting, adding or deleting DNA bases (1)

5. Referring to the use of CRISPR technology on wheat, outline an ethical issue and describe how an ethical principle should be considered before making decisions. ]]3

   Respect (1) – does wheat have an intrinsic value? (1). Wheat plants must be respected and their role in the ecosystem considered (1)
   Non-maleficence (1) – Is there harm being done? (1). No other organisms should be harmed when using CRISPR technology (1)
   Justice (1) – consider competing claims (1). Do all farmers have access to the technology? (1) Integrity (1) – Is all data accurately reported (1). Data from trials must be available publicly (1) Beneficence (1) – Is there maximise benefits and minimise risk when considering outcomes? (1). More people/organisms should benefit from the technology with the cost being minimal (1)

Question 4

The production of biofuels, bioethanol and biodiesel, is a growing industry as traditional fossil fuels are becoming scarcer.

1. What is the original source of bioethanol? ]]1

   Plant matter/plants (1)

2. There are 4 stages in the production of bioethanol. Complete the table below to outline the key events that occur at each stage. ]]4

Stage	Key event
Pre-treatment	Biomass is mechanically broken down into smaller pieces
Enzymatic hydrolysis	Enzymes are used to break polymers into monomers
Fermentation	Yeast and sugars are added to allow fermentation, and production of ethanol to occur
Distillation	Molecular sieve used to remove any impurities

3. Describe how changes in temperature and pH can alter the rate of bioethanol production. ]]4

   If the temperature is below the optimum, there is low kinetic energy and the rate of reaction would be lower (1). Above the optimum temperature, enzymes would denature lowering the rate of reaction (1).
   If the pH is above or below the optimum, the enzymes denature, lowering the rate of reaction (1). At the optimal pH and temperature, the rate of reaction is greatest (1).

Question 5

A biochemical pathway is a series of enzyme-mediated reactions where the product of one reaction is used as the substrate in the next.
Below is an example of a common biochemical pathway found within the body:

1. What is the name of the biochemical pathway shown, and where does it occur? Use evidence to support your response. ]]3

   Krebs cycle (1) that occurs in the mitochondrial matrix (1). It can only by Krebs as the coenzyme FAD+ is only found in the Krebs cycle (1).

2. Inhibitors can alter the rate of an enzyme driven chemical reaction. Describe the action of each type of inhibitor listed below and the effect of the rate of reaction if the concentration of substrate increased.
   1. a competitive inhibitor: ]]3

      Competitive inhibitors bind to the active site of an enzyme (1) preventing the substrate from binding (1). Increasing substrate can increase the rate of reaction OR increase in competitive inhibitors can decrease reaction rate (1).

   2. a non-competitive inhibitor: ]]3

      Non-competitive inhibitors bind to the allosteric site of an enzyme (1) causing a conformational change in the shape of the active site – preventing the substrate from binding (1). Adding more substrates will not change the rate of reaction (1)

3. What is the total ATP yield for the biochemical pathway? ]]1

   2 ATP