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Describe the role of DNA polymerase in the semi-conservative replication of DNA - AQA - A-Level Biology - Question 5 - 2020 - Paper 1
Question 5
Describe the role of DNA polymerase in the semi-conservative replication of DNA. DNA polymerase plays a crucial role in the semi-conservative replication of DNA by ... show full transcript
Worked Solution & Example Answer:Describe the role of DNA polymerase in the semi-conservative replication of DNA - AQA - A-Level Biology - Question 5 - 2020 - Paper 1
Step 1
Step 2
It took less time for 25% of cells with cyclin D to be undergoing DNA replication than for 25% of cells without cyclin D. Use Figure 5 to calculate this time difference as a percentage decrease. Show your working.
Answer
From Figure 5,
- Time taken for cells without cyclin D = 6 hours.
- Time taken for cells with cyclin D = 5.5 hours.
Percentage decrease can be calculated using the formula: rac{ ext{Initial} - ext{Final}}{ ext{Initial}} imes 100
So, substituting the values: rac{6 - 5.5}{6} imes 100 = rac{0.5}{6} imes 100 \\ = 8.33 ext{%}
Thus, the time difference as a percentage decrease is 8.33%.
Step 3
Describe how an enzyme can be phosphorylated.
Answer
An enzyme can be phosphorylated through the addition of a phosphate group (PO4) that can be transferred from a donor molecule, typically ATP or ADP. This phosphorylation can lead to a conformational change in the enzyme, resulting in its activation or inhibition.
Step 4
Use Figure 5 to suggest why higher than normal concentrations of cyclin D could result in a tumour.
Answer
Higher concentrations of cyclin D may lead to a shortened interphase, causing cells to replicate DNA faster than normal. This rapid division can result in uncontrolled cell division, potentially leading to the formation of tumours as cells accumulate mutations or undergo malignant transformation.