CAIE Biology A2 17.2.5 Hardy-Weinberg Principle

[Maximum number: 7]

Cystic fibrosis is an autosomal recessive genetic disease. People with cystic fibrosis have a homozygous recessive genotype.

In 2020:
- there were 10800 people with cystic fibrosis in the UK
- the UK population was estimated to be 67100000 people.
A proportion of people in the UK population are heterozygous for the gene that causes cystic fibrosis and do not have symptoms of the disease.
Use the Hardy-Weinberg principle to calculate the number of people in the UK population who are expected to be heterozygous for the gene that causes cystic fibrosis.
The two equations for the Hardy-Weinberg principle are provided.
equation 1

p+q=1

equation 2

p^{2}+2 p q+q^{2}=1

p= frequency of the dominant allele
q= frequency of the recessive allele
$p^{2}=$ frequency of the homozygous dominant genotype
2 p q= frequency of the heterozygous genotype
$q^{2}=$ frequency of the homozygous recessive genotype
The first stage of the calculation has been completed for you.

q^{2}=\frac{10800}{67100000}

2 p q=
number of people in the UK expected to be heterozygous for the gene =
(ii) The Hardy-Weinberg principle provides a useful estimate of the number of people in the UK who are heterozygous for cystic fibrosis. However, the estimate is lower than the actual number. This underestimation occurs because not all the conditions of the Hardy-Weinberg principle apply.

In the UK in 2020, the mean life expectancy of:
- people with cystic fibrosis was approximately 50 years
- all people was approximately 80 years.

Explain how this information accounts for the underestimation of the number of people in the UK that are heterozygous for cystic fibrosis.
(c) A screening programme for cystic fibrosis was introduced in 2007 for all children born in the UK. Children are tested within seven days of their birth. Children identified from the screening programme as being at high risk of having cystic fibrosis can have a genetic test to confirm whether they have the disease.
(i) Table 3.1 shows the median predicted life expectancy for people born in the UK who have cystic fibrosis. Predictions are shown for people born in 2008, 2012, 2016 and 2020.

Table 3.1

Describe the trend shown in Table 3.1 and outline how early screening for cystic fibrosis may have contributed to this trend.
(ii) In many countries, a genetic test for cystic fibrosis is available to adults who do not have cystic fibrosis but have a family member who either has cystic fibrosis or is heterozygous for the gene that causes cystic fibrosis.

These adults include partners, parents, offspring, brothers and sisters of the family member. The aim is to find out if any of these adults are heterozygous for the gene that causes cystic fibrosis.

Discuss the ethical and social considerations of making a genetic test for cystic fibrosis available to these adults.

4 Holstein Friesian cattle are a breed of cattle used by dairy farmers in many countries of the world for the high milk yield of their cows.

Fig. 4.1 shows Holstein Friesian cattle.

Fig. 4.1

Milk yield in Holstein Friesian cattle is affected by heat stress. Heat stress occurs when homeostatic mechanisms are not enough to keep the body temperature down to normal levels.

One of the factors that contributes to heat stress is air temperature.

Fig. 4.2 shows:
- the mean daily air temperature in Central Europe
- the mean monthly milk yield per cow of Holstein Friesian cattle in Central Europe.

Fig. 4.2

(a) With reference to Fig. 4.2, describe the trends in air temperature and milk yield from April to August.
(b) Many dairy farmers in tropical regions use cattle breeds that are tolerant to heat stress (heat-tolerant cattle). These heat-tolerant cattle:
- can tolerate higher air temperatures than Holstein Friesian cattle before heat stress occurs
- have milder symptoms of heat stress than Holstein Friesian cattle for the same high air temperatures.

Where heat stress does not occur, heat-tolerant cattle produce a lower milk yield than Holstein Friesian cattle under the same conditions.

Scientists compared DNA sequences of Holstein Friesian cattle and heat-tolerant cattle for a number of genes known to have an effect on body temperature.

Twenty genes were found that had alleles associated only with heat-tolerant cattle.
With reference to the information provided, including the data in Fig. 4.2:
- state the type (pattern) of phenotypic variation shown by milk yield in cattle
- identify factors that cause phenotypic variation in milk yield in cattle.

In each case, give a reason for your choice.
type (pattern) of phenotypic variation and reason for choice
factors that cause phenotypic variation and reason for each choice
(c) The scientists found that one of the genes studied, P R L R, has a dominant allele known as SLICK. The SLICK allele was identified in Senepol cattle, a heat-tolerant breed, and is not found in Holstein Friesian cattle.

Cattle with the SLICK allele have short hair due to reduced hair growth.
Scientists have used selective breeding to introduce the SLICK allele into Holstein Friesian cattle. The milk yields of normal Holstein Friesian cattle and Holstein Friesian cattle with the SLICK allele are shown in Fig. 4.3, during:
- March, when the mean daily air temperature is $5^{\circ} \mathrm{C}$ .
- September, when the mean daily air temperature is $14^{\circ} \mathrm{C}$ .

Fig. 4.3

With reference to Fig. 4.3, describe the effect of the SLICK allele on milk yield in Holstein Friesian cattle.
(d) The SLICK allele differs from the recessive allele by a single nucleotide deletion. This results in a frameshift mutation and introduces a premature stop codon in the P R L R gene.

Scientists can use gene editing to replicate this mutation in Holstein Friesian cattle. This provides a way to introduce the SLICK allele into Holstein Friesian cattle without selective breeding.

Compare gene editing and selective breeding for introducing the SLICK allele into Holstein Friesian cattle.

Include similarities and differences in your answer.

5 Complete the following paragraphs using the most appropriate word or words.

The theory of evolution describes a process that can lead to the formation of new species from pre-existing species over .

DNA sequence data of different species can be compared to show evolutionary relationships. Two species that have a more recent common ancestor share more in the DNA nucleotide sequences of their genomes than two species that are more distantly related.

Mitochondrial DNA can also be used in the study of evolutionary relationships. Mitochondrial DNA is inherited only from the female gamete, and its nucleotide sequence is unaffected by during the production of gametes.

DNA sequence data can be stored in large biological , allowing faster comparison of the nucleotide sequences of genomes using computer software. DNA sequence data can also be used to predict the sequences of proteins produced by a species.

A can be used to detect many different mRNA molecules at the same time in studies that compare gene expression between different species.

6 A respirometer is a piece of apparatus that can be used to measure the rate of respiration of living tissue such as germinating peas.

A simple respirometer is shown in Fig. 6.1.

Fig. 6.1

A student carried out an investigation to determine the effect of temperature on the rate of respiration of germinating peas.
- The student set up the respirometer as shown in Fig. 6.1 and placed the respirometer in a water-bath at $10^{\circ} \mathrm{C}$ .
- After five minutes, the student used the syringe to adjust the position of the coloured liquid in the right-hand side of the U-shaped tube so that it lined up with 0 cm on the ruler. The student immediately started a timer.
- The germinating peas used up oxygen, causing the coloured liquid in the U-shaped tube to move.
- The student measured the distance moved by the coloured liquid after 20 minutes.
- The student repeated the experiment at temperatures of $20^{\circ} \mathrm{C}, 30^{\circ} \mathrm{C}, 40^{\circ} \mathrm{C}$ and $50^{\circ} \mathrm{C}$ .
(a) State the function of the potassium hydroxide solution used in the investigation.
(b) Suggest how the validity of the results could be assessed.
(c) Explain why the respirometer was left in the water-bath for five minutes before starting the experiment.
(d) The rate of movement of the coloured liquid in the U-shaped tube, calculated from the results, is shown in Table 6.1.

Table 6.1

Plot a graph of the results shown in Table 6.1 on the grid in Fig. 6.2. Draw a curved line of best fit.

Fig. 6.2

(e) The rate of movement of the coloured liquid is related to the rate of respiration.

Explain the effect of temperature on the rate of respiration shown in Table 6.1 and Fig. 6.2.

3(b)(i) Calculation:
$q = \sqrt{\frac{10800}{67100000}} = \sqrt{0.000161} = 0.0127$ p = 1 - 0.0127 = 0.987 $2pq = 2 \times 0.9873 \times 0.0127 = 0.0251$ $67100000 \times 0.0251 = 1680000$

Answer:
number of heterozygous people = 1680000.

q = \sqrt{\frac{10800}{67100000}} = 0.0127

Must be to 3 significant figures or more.
Accept 1675000 to 1685000.
Reject 1700000.
Max 2 if decimal places are used in the final answer.
Allow max 2 for working if the final answer is incorrect.

3(b)(ii) Any three from:
1. reference to natural selection ;
2. selection pressure against cystic fibrosis (CF) ;
3. people with CF do not survive as long / AW ;
4. heterozygotes are calculated from a smaller CF population / AW ;
5. some people with CF who have died will have passed on the CF allele ;

3(c)(i) Any three from:
1. median predicted life expectancy increased over time ;
2. data quote, e.g. from 2008 to 2020, increase from 38.8 to 50.6 years ;
increase of 11.8 years ;
percentage increase of 30.4% ;
3. treatment can be started early ;
4. better treatment / improved care between 2008 and 2020 ;

3(c)(ii) Any three from:
1. reduces worry / AW if result is negative ;
ora
2. can make informed reproductive decisions / AW ;
3. cost / availability of test ;
4. plan for care of child with CF ;
5. idea of further genetic testing, e.g. counselling / testing of embryo / testing of partners ;
6. idea that test is not 100% accurate ;
7. problems related to stigma / discrimination / insurance / confidentiality ;

4(a):
1. temperature increases and milk yield decreases ;
2. comparative data quote, e.g. April = 10 $^{\circ}\mathrm{C}$ and 792-798 kg cow $^{-1}$ ;
August = 18 $^{\circ}\mathrm{C}$ and 711-716 kg cow $^{-1}$ ;

4(b):
1. continuous, because milk yield does not fall into distinct classes / shows a range / intermediates / no distinct categories / AW ;
2. environment, because milk yield is affected by air temperature ;
3. genetic / polygenic, because many / 20 genes have an effect on milk yield, or heat-tolerant cattle have lower milk yield than Holstein Friesian cattle ;

4(c) Any three from:
1. SLICK allele increases milk yield in both months / at both temperatures ;
2. in March / at $5^{\circ}\mathrm{C}$ , there is a small difference between milk yields ;
3. in September / at $14^{\circ}\mathrm{C}$ , there is a larger difference between milk yields ;
4. cattle with SLICK allele maintain milk yield in both months / at both temperatures ;
5. comparative data quote to support marking points 1-4:

month / temperature	Holstein without SLICK / kg cow $^{-1}$	Holstein with SLICK / kg cow $^{-1}$
March / $5^{\circ}\mathrm{C}$	780-785	795-800
September / $14^{\circ}\mathrm{C}$	690-700	770-780

4(d) Any six from:

point	gene editing	selective breeding
1	first offspring have SLICK allele	not necessarily in first offspring
2	can have negative / unknown effects	can have negative / unknown effects
3	presence of SLICK allele can be confirmed by genetic testing / phenotypic observation	presence of SLICK allele can be confirmed by genetic testing / phenotypic observation
4	done by humans / artificial methods	done by humans / artificial methods
5	only PRLR / one gene is affected	affects many genes
6	all offspring have SLICK allele	some offspring have SLICK allele
7	immediate / one generation	performed over many generations
8	performed on embryos / zygotes / cells	mating / artificial insemination
9	technical requirements, e.g. laboratory, training, molecular method	mating / performed on the farm / physiological method
10	maintains heterozygosity / genetic variation / desirable characteristics	loss of heterozygosity / genetic variation / desirable characteristics
11	regulatory approval	no regulatory approval
12	no outbreeding required	outbreeding required

5:
- many generations / many years / a long time / AW ;
- mutations / similarities ;
- meiosis / crossing over / recombination ;
- databases ;
- amino acid ;
- microarray ;

6(a):
absorb carbon dioxide produced by peas ;

6(b) Any two from:
1. idea of repeatability ;
2. check to see how consistent the results are / the more consistent the results, the more valid the results / carry out a named statistical test or measure / calculate standard deviation / calculate standard error ;
3. a control respirometer / described, e.g. with glass beads instead of peas ;
4. reference to checking variables other than temperature, e.g. volume of germinating peas / volume or concentration of KOH ;

6(c):
acclimatisation / equilibration / AW ;

6(d):
- 5 points plotted accurately ;
- smooth curve drawn linking all plotted points / curve of best fit ;
Reject extrapolation.

6(e) Any three from:
1. reference to enzymes ;
2. as temperature increases, kinetic energy increases ;
3. rate increases because more enzyme-substrate complexes are formed in the same time, or because there is a higher proportion / frequency of successful / effective collisions ;
4. rate decreases after $30^{\circ}\mathrm{C}$ due to denaturation of enzymes ;
5. further detail, e.g. active site shape change / reference to optimum temperature.

CAIE A-Level Biology A2 17.2.5 Hardy Weinberg Principle Questions

Exam points

CAIE A-Level Biology A2 17.2.5 Hardy Weinberg Principle Questions question 1