Case study (4 Marks)

Question 14 Marks

Given below is the pattern of temperature in a person suffering from a non-viral disease transmitted by mosquitoes. Study the graph and answer the questions that follow:

A. Explain the factor(s) responsible for this pattern of temperature.
B. How does this pathogen multiply in the human body?
C. How is this infection transmitted to humans?
OR
D. Which stages of the life cycle of this pathogen are completed in the mosquito's gut?

Answer

A. The rupture of RBCs associated with the release of toxic substance haemozoin is responsible for the chills and fever/recurring every 3 - 4 days.
B.
- The parasite reproduces asexually in liver cells, bursting the cell and releasing into the blood.
- Parasite further reproduces asexually in red blood cells. Released parasite infects new red blood cells. Sexual stages (gametocytes) develop in red blood cells.
C. The infection is caused by the bite of the female Anopheles mosquito which introduces the sporozoites in the human body.
D. Fertilisation and development take place in the mosquito's gut.

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Question 24 Marks

Assuming that within a population of beetles where Hardy Weinberg conditions are met, the colour black (B) is dominant over the colour red (b). 40% of all beetles are red (bb).
Given this information, answer the questions below:
A. What is the frequency of red beetles?
B. Calculate is the percentage of beetles in the population that are heterozygous.
C. What is the frequency of homozygous dominant individuals?
OR
D. Assuming that Hardy Wienberg conditions are met in the beetle population consisting of 1500 beetles. How many beetles would you expect to be black and red in colour respectively?

Answer

A. Given is bb = q² = 0.4 .
To determine q, which is the frequency of the recessive allele in the population, simply take the square root of q² which works out to be 0.632 (i.e. 0.632 x 0.632 = 0.4 ).
So, q = 0.63
B. As we know that p + q = 1
then p must be 1-0.63 = 0.37.
The percentage of beetles in the population that are heterozygous would be 2pq. 2(0.37)(0.63) = 0.47
C. Given is BB = p² = (0.37)² = 0.14
OR
D. As, 1500 is the total population
40% is red population,
hence the number of beetles with red colour will be
1500 x 0.4 = 600
If total population is 1500 and red is 600 then black would be 1500- 600=900

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BIOLOGY Case study (4 Marks)

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