GAMSAT Section 3 Practice Questions

Graphs

Unit 1

Horizontal gene transfer has played a key role in prokaryotic evolution. It is hypothesised that it may also play a key role in the cellular processes of eukaryotes. An experiment was designed to predict the extent of the genes of different unicellular eukaryotes in encoding for different metabolic enzymes. The results of the experiment show the total number of enzymes that were identified as being present in each organism group.

Figure 1. Gene transfer in unicellular eukaryotes. Whitaker, J. W., McConkey, G. A., & Westhead, D. R. (2009). The transferome of metabolic genes explored: analysis of the horizontal transfer of enzyme encoding genes in unicellular eukaryotes. Genome biology10(4), R36.
Q1/2 – Which genus has the highest percentage of enzymes predicted to be ancestral genes?
A. Saccharomyces
The question asks you to find the ‘highest percentage of enzymes predicted to be ancestral (orange) genes’. The two candidates are Saccharomyces and Phytophthora. Since Phytophthora has twice the non-ancestral enzyme count as Saccharomyces, it would need to have at least twice the ancestral enzyme count to have the highest percentage. This is not the case, so the answer is A. If you’re struggling with this, let’s use some simple numbers as an example. If Saccharomyces has 10 non-ancestral and 90 ancestral, it is 90% ancestral. If Phytophthora has twice as much non-ancestral (20), it would need at least twice as much ancestral (200) to have an equal or higher percentage ancestral enzyme count.
B. Ostreococcus
Try again buddy 🙂
C. Diatoms
Try again!
D. Phytophthora
Try again! 🙂
Q2/2 – Which genus has the highest percentage of enzymes predicted to occur from endosymbiotic gene transfer (EGT) and horizontal gene transfer (HGT)?
A. Saccharomyces
Try again 🙂
B. Ostreococcus
Try again 🙂
C. Diatoms
You needed to identify the variable EGT was in fluoro green, locate this on the graph and see that Diatoms had the highest proportion..
D. Phytophthora
Try again! 🙂

Unit 2

Mrs Acer Racer is a protein scientist who is looking to optimise conditions when producing proteins to improve the yield of her crops. (Change 1st sentence to something about what P.pastoris is and why this is useful). By changing settings one at a time whilst keeping other variables fixed, it is possible to find optimum conditions to improve P. pastoris ability to produce and secrete functional single-chain antibody fragments.

The effect of pH, temperature and methanol concentration on the yield of the antibody TS1-218 was recorded. A predictive model was developed that measured absorbance (ABS) values from ELISA assays as a response for the level of TS1-218 production. Results showed that higher starting pH and lower temperatures significantly increased the yield of TS1-218.

Figure 2. Response surface plot showing the influence of temperature and pH on the yield of TS1-218 reflected by the absorbance signal from the ELISA assays. Jafari, R., Sundström, B.E. & Holm, P. Optimization of production of the anti-keratin 8 single-chain Fv TS1-218 in Pichia pastoris using design of experiments. Microb Cell Fact 10, 34 (2011).
Q1/3 – What pH and temperature is approximately best for TS1-218 expression?
A. 7.5, 16°C
Try again buddy!
B. 6.5, 20°C
Try again buddy 🙂
C. 6.5, 11°C
Try again!
D. 7.0, 11°C
In the question stem, it is stated that “higher starting pH and lower temperatures significantly increased the yield of TS1-218.” It also mentions that absorbance (ABS) values is a response for the level of TS1-218 production, which means that as you go vertically upwards in the y-axis, there will be greater yield. This is indicated by the dark coloured circle. The reason it is not C) 6.5, 11°C is because higher starting pH (D has a ph of 7) significantly increases yield.
Q2/3 – What is the peak absorption values predicted for TS1-218?
A. Equal to 1.2
You are required to extrapolate from the graph in 3D. The dark area does not reach above the 1.2 line, and looks to peak at approximately 1.2. 
B. Above 1.2
Try again 🙂
C. Less than 1.2
Try again.
D. None of the above
Try again! 🙂
Q3/3 – What is suggested by the results of this graph?
A. That absorbance was measured at 405 nm and the absorbance at 650 nm was subtracted 
Unrelated! 
B. That P. pastoris is highly capable of producing and secreting TS1-218 single-chain antibody fragments at temperatures as low as  11°C
Deducing the wrong answers was important here. A is wrong as it is unrelated and C is wrong because the question stem already states that ‘higher starting pH and lower temperatures’ increased yield. D is wrong as the caption of this graph is about the contribution of pH and temperature to the yield of TS1-218, rather than about the relationship between pH and temperature. 
C. That decreasing the initial pH of the culture medium and increasing temperature can result in higher absorbance values in the ELISA essays and higher TS1-218 concentration 
C is wrong because the question stem already states that ‘higher starting pH and lower temperatures’ increased yield.
D. That pH and temperature are directly correlated to each other  
D is wrong as the caption of this graph is about the contribution of pH and temperature to the yield of TS1-218, rather than about the relationship between pH and temperature. 

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Worked Answers

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UNIT 1 – Approximating proportions, visualising proportions

Q1 – A. The question asks you to find the ‘highest percentage of enzymes predicted to be ancestral (orange) genes’. The two candidates are Saccharomyces and Phytophthora. Since Phytophthora has twice the non-ancestral enzyme count as Saccharomyces, it would need to have at least twice the ancestral enzyme count to have the highest percentage. This is not the case, so the answer is A.

If you’re struggling with this, let’s use some simple numbers as an example. If Saccharomyces has 10 non-ancestral and 90 ancestral, it is 90% ancestral. If Phytophthora has twice as much non-ancestral (20), it would need at least twice as much ancestral (200) to have an equal or higher percentage ancestral enzyme count.

Q2 – C. You needed to identify the variable EGT was in fluoro green, locate this on the graph and see that Diatoms had the highest proportion.

UNIT 2 – Interpreting graphs, extrapolating information

Q3 – D. In the question stem, it is stated that “higher starting pH and lower temperatures significantly increased the yield of TS1-218.” It also mentions that absorbance (ABS) values is a response for the level of TS1-218 production, which means that as you go vertically upwards in the y-axis, there will be greater yield. This is indicated by the dark coloured circle. The reason it is not C) 6.5, 11°C is because higher starting pH (D has a ph of 7) significantly increases yield.

Q4 – A. You are required to extrapolate from the graph in 3D. The dark area does not reach above the 1.2 line, and looks to peak at approximately 1.2. 

Q5 – B. Deducing the wrong answers was important here. A is wrong as it is unrelated and C is wrong because the question stem already states that ‘higher starting pH and lower temperatures’ increased yield. D is wrong as the caption of this graph is about the contribution of pH and temperature to the yield of TS1-218, rather than about the relationship between pH and temperature. 

 

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