Inheritance (Edexcel International AS Biology)

Exam Questions

2 hours11 questions
1a
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1 mark

A person with diabetes has a blood glucose level that can be too high.

When the blood glucose level of a person without diabetes becomes too high, the liver stores glucose as a polysaccharide.

Which polysaccharide does the liver store?

  A Amylopectin
  B Cellulose
  C Glycogen
  D Starch
1b
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1 mark

Blood glucose levels can become high following the digestion of carbohydrates.

Which of the following can be digested to release glucose?

  A Both fructose and sucrose
  B Both fructose and galactose
  C Both galactose and lactose
  D Both lactose and sucrose
1c
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4 marks

Diabetes is a risk factor for cardiovascular disease.

(i)

One estimate is that there are 415 million people with diabetes in the world and that 46% of these people are undiagnosed.

Calculate the number of people who have undiagnosed diabetes.

(1)

(ii)

There are two types of diabetes, Type I and Type II.

The treatment for Type I diabetes is different from the treatment for Type II diabetes, so it is important for a correct diagnosis to be made.

Diagnosis can be difficult, particularly in people aged between 20 and 40 years old.

A genetic screening method is now available for the diagnosis of diabetes.

Explain why doctors are more likely to screen individuals once they develop diabetes than use methods such as prenatal testing.

(3)

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2a
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1 mark

Most Bengal tigers are orange with black stripes but there is a very small number of Bengal tigers that are white with black stripes.

The photograph shows a white Bengal tiger with black stripes.

white-tiger

SusuMa, CC0, via Wikimedia Commons

White tiger offspring are produced by two Bengal tigers that each carry at least one recessive allele for a gene affecting coat colour.

The pedigree diagram shows the phenotypes in one family of tigers, bred in captivity.

q2-2-unit-1-oct-2020-edexcel-ial-biology

The phenotype is affected by the genotype.

State what is meant by the term genotype.

2b
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1 mark

State the probability that the next tiger born to these two parents will be female.

2c
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3 marks

Determine the expected phenotypic ratio of orange tigers to white tigers born to the parents shown in this pedigree diagram.

Use a genetic diagram to support your answer.

2d
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1 mark

The incidence of white tigers in the wild is 1 in 10 000 Bengal tigers.

There are approximately 6000 Bengal tigers in captivity, 200 of which are white.

Calculate the incidence of white tigers in captivity.

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3a
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1 mark

Phenylthiocarbamide (PTC) is a chemical that has a very bitter taste to some individuals (tasters).

The ability to taste PTC is determined by a gene that codes for a bitter-taste receptor on the tongue.

The pedigree diagram shows some of the tasters and non-tasters in a family.

q4-unit-1-june-2021-edexcel-ial-biology

Complete the diagram to show the following information:

    • Individual 7 as a male taster
    • Individual 8 as a male non-taster
    • Individual 9 as a female taster.
3b
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4 marks

Describe the difference between each of the following pairs of terms, using the information in the pedigree diagram to illustrate your answer.

(i)

Gene and allele

(2)

(ii)

Genotype and phenotype

(2)

3c
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2 marks

Explain which is the dominant allele.

Use the information in the pedigree diagram to support your answer.

3d
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2 marks

Explain why this gene is unlikely to be located on the X chromosome.

Use the information in the pedigree diagram to support your answer.

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4a
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3 marks

People with cystic fibrosis produce very thick, sticky mucus.

Cystic fibrosis is caused by mutations in a gene coding for the CFTR protein.

Explain why a mutation in this gene results in the production of very thick, sticky mucus.

4b
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5 marks

The diagram shows some health problems associated with cystic fibrosis, in a female.

q5b-unit-1-june-2021-edexcel-ial-biology
Explain why very thick, sticky mucus results in these health problems.

4c
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3 marks

The graph shows the correlation between the concentration of chloride ions in sweat and the level of function of the CFTR protein.

q5c-unit-1-june-2021-edexcel-ial-biology
Individuals diagnosed with cystic fibrosis have a level of function of the CFTR protein of 18 % or less.

(i)

Which is the change in concentration of chloride ions in the sweat of an individual when the level of function of CFTR protein decreased from 100 % to 18 %?

(1)

  A 15 mmol dm–3
  B 35 mmol dm–3 
  C 40 mmol dm–3
  D 80 mmol dm–3

(ii)

Cystic fibrosis results from different mutations in the CFTR gene.

Explain how the graph provides evidence that cystic fibrosis results from different mutations in the CFTR gene.

(2)

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5a
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1 mark

A number of diseases are associated with lifestyle risk factors.

Some of these risk factors cause mutations.

Mutations can give rise to cancer.

State the meaning of the term mutation.

5b
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7 marks

Exposure to ultraviolet light is associated with the development of skin cancer.

The graphs show the incidence of skin cancer in males and females in one country in the Far East, from 2000 to 2005.

q7b-unit-1-june-2021-edexcel-ial-biology

(i)

The graphs show some correlations.

State the meaning of the term correlation.

(1)

(ii)

Describe the correlations shown by these graphs.

(3)

(iii)

Suggest a reason for each of the correlations shown by these graphs.

(3)

5c
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6 marks

The table shows some information from a study of the incidence of emphysema in smokers and non-smokers.

Information Males Females
Number of individuals in the study 25 25
Mean age when diagnosed / years 53.1 54.2
Range of ages when diagnosed / years 32 to 77 34 to 68
Number of smokers 5 6
Number of non-smokers 20 19
Number of smokers with emphysema 1 6
Number of non-smokers with emphysema 0 0


Criticise the design of this study.

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6a4 marks

The five stages of cancer are used to describe the size and spread of a cancer in a human body.

The mitotic index of a tissue can be used to determine the stage of cancer.

A higher mitotic index is usually linked to a later stage of cancer.

The mitotic index is calculated using the formula 

screenshot-2022-12-15-15-31-31

Tests were performed on three patients, P, Q and R, who had cancer.

The table shows the number of cells that were counted in each stage of the cell cycle per mmof tissue, taken from the same organ in each patient.

Patient Interphase Prophase Metaphase Anaphase Telophase
P 16 1 3 0 0
Q 14 2 1 1 2
R 11 2 2 3 2

(i)

State what is meant by the term tissue.

(1)

(ii)

Using the data in the table, determine the stages of cancer in patients P and R.

(3)

6b3 marks

The graph shows the probability of survival for different stages of a cancer, after diagnosis.

q5b-unit-2-oct-2020-edexcel-ial-biology

Compare and contrast the probabilities of survival for the different stages of this cancer, as shown by the graph.

6c2 marks

Anticancer drugs have to undergo double‐blind trials before they are used to treat patients.

Describe how a placebo is used in a double‐blind trial.

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7a
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1 mark

Mutations can give rise to cancer.

What is a mutation?

  A a change in the amino acid sequence in DNA
  B a change in the amino acid sequence in a protein
  C a change in the base sequence in DNA 
  D a change in the base sequence in a protein
7b
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1 mark

Name two types of mutation.

7c
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3 marks

The graph shows the number of cases of one type of cancer in a human population.

q1c-unit-1-oct-2021-edexcel-ial-biology
Describe the effect of age and sex on the number of cases of cancer.

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8a
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2 marks

Genetic screening can be used to test for aneuploidy.

Aneuploidy is the presence of an abnormal number of chromosomes in a cell.

Aneuploidy can affect the miscarriage rate of implanted embryos.

Following screening, only embryos with the correct number of chromosomes are implanted into the female.

The table shows the miscarriage rate of two groups of implanted embryos:

  • embryos not screened for aneuploidy
  • embryos screened and shown not to have aneuploidy.
Age range
of women at
implantation/years
Miscarriage rate (%)
Implanted embryos
not screened for
aneuploidy
Implanted embryos
screened and shown
not to have aneuploidy
<35 12.0 11.2
35 to 37 16.8 13.0
38 to 40 25.0 13.6
41 to 42 37.9 16.3
>42 58.8 17.2

Explain how this data shows that there is a correlation between the age of the women and the miscarriage rate.

8b
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5 marks
(i)
Explain the conclusions that can be made from these data about the causes of miscarriage.

(2)

(ii)
Explain why conclusions made using these data may not be valid.

(3)

8c
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3 marks

Discuss the implications of screening embryos for aneuploidy before implantation.

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9a
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1 mark

Cystic fibrosis is an inherited recessive disease caused by mutations in a gene on chromosome 7.

Give the meaning of the term gene.

9b
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3 marks

Explain how mutations result in cystic fibrosis.

9c
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3 marks

Population carrier screening (PCS) is one type of genetic screening.

This involves screening people who want a child to see if they are carriers of genetic disorders.

In one country, the number of babies born with cystic fibrosis went down following the introduction of PCS.

Suggest why the number of babies born with cystic fibrosis went down.

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10a
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6 marks

The drawing shows a speckled chicken. These chickens have a mixture of black and white feathers.q8-unit-1-oct-2021-edexcel-ial-biology

The colour of the feathers of a chicken is an example of codominance.

One parent of this speckled chicken had white feathers and the other parent had black feathers.

Describe the difference between each of the following pairs of terms, using feather colour to illustrate your answer.

(i)

Gene and allele

(3)

(ii)

Genotype and phenotype

(3)

10b
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3 marks

A black chicken was mated with a speckled chicken. They had 25 chicks.

Determine the expected number of speckled chicks.

You must use a genetic diagram.

10c
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6 marks

In an experiment, several pairs of speckled chickens were mated together.

They produced 480 chicks.

The table shows the expected number of speckled chicks, white chicks and black chicks. It also shows the actual number of each type of chick.

Steps in the calculation
for the statistics test
Colour of feathers of chicks
Speckled White Black
Observed number (O) 243 125 112
Expected number (E) 240 120 120
open parentheses straight O minus straight E close parentheses      
open parentheses straight O minus straight E close parentheses squared over straight E      


This table can be used in a statistics test.

(i)

Name the statistics test being used to analyse these data.

(1)

(ii)

Complete this table to show the missing values.

(2)

(iii)     Calculate

 sum for blank of open parentheses straight O minus straight E close parentheses squared over straight E

(1)

(iv)

Explain how a critical value table could be used to accept or reject a null hypothesis for this experiment.

(2)

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11a
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5 marks

Sickle cell disease is caused by a gene mutation affecting the β‐globin chain of haemoglobin.

The mutation occurs in the seventh triplet code of this gene.

This mutation results in the amino acid Glu being replaced with the amino acid Val.

The table shows the sequence of bases in the first part of the DNA in a person who does not have sickle cell disease. It also shows the corresponding sequence of amino acids in the β‐globin chain.

Position of triplet code first second third fourth fifth sixth seventh eighth ninth
 DNA AUG GUG CAC CUG ACU CCU GAG GAG AAG
 β‐globin chain (start) Val His Leu Thr Pro Glu Glu Lys

(i)

Give the seventh triplet code in the gene for the β‐globin chain in a person who has sickle cell disease.

(1)

(ii)

Name the type of mutation that causes sickle cell disease.

(1)

(iii)

The amino acid Glu is hydrophilic (polar) and the amino acid Val is hydrophobic (non‐polar).

Suggest why this mutation causes haemoglobin molecules to stick together.

(3)
11b
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2 marks

In 2020, about 140 million babies were born in the world.

About 305 800 babies are born with sickle cell disease each year.

Estimate the ratio of babies born with the disease to babies not born with the disease.

11c
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8 marks

The red blood cells in a person with sickle cell disease are sickle shaped and less elastic. They also have a shorter lifespan than healthy red blood cells.

These sickle shaped red blood cells cannot carry as much oxygen as healthy red blood cells and they get stuck in the capillaries.

The graph shows oxygen dissociation curves for groups of people with sickle cell disease and those without the disease.

The shaded areas represent the range of values for each group of people.

q8c-unit-1-january-2022-edexcel-ial-biology

(i)

A p50 value is the partial pressure of oxygen that results in 50% saturation of haemoglobin.

Calculate the largest difference in the p50 values between a person with sickle cell disease and a person without this disease.

Give your answer to an appropriate number of decimal places.

(2)

(ii)

Identify two conclusions, other than the difference in p50 values, that can be made from this graph.

(2)

(iii)

The lifespan of a red blood cell in a person with sickle cell disease is 11 days.

This is 9.17% of the lifespan of a healthy red blood cell.

Calculate the lifespan of a healthy red blood cell.

Give your answer to the nearest day.

(1)

(iv)

Sickle cell disease can result in death.

Explain why the changes in the structure of haemoglobin and the shape of the red blood cells could result in death in a person with sickle cell disease.

(3)

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