4 (a) Figure 5 shows how alcohol consumption increases the risk of developing liver cancer - Edexcel - GCSE Biology Combined Science - Question 4 - 2021 - Paper 1
Question 4
4 (a) Figure 5 shows how alcohol consumption increases the risk of developing liver cancer.
Someone who does not drink alcohol has a 1.0 risk of developing liver can... show full transcript
Worked Solution & Example Answer:4 (a) Figure 5 shows how alcohol consumption increases the risk of developing liver cancer - Edexcel - GCSE Biology Combined Science - Question 4 - 2021 - Paper 1
Step 1
Calculate the risk of developing liver cancer for someone who consumes 4 units of alcohol a day.
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Answer
To determine the risk of developing liver cancer for someone who consumes 4 units of alcohol:
Convert units of alcohol to grams:
4extunitsimes8extgrams/unit=32extgramsofalcohol
Based on the graph in Figure 5, since a consumption of 32 grams corresponds to a risk factor of 1.2, the risk of developing liver cancer is:
extRisk=1.2
Step 2
Describe how cancer develops in the liver.
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Answer
Cancer development in the liver involves several key processes:
Mutations in DNA: Changes or mutations occur in the DNA sequence of liver cells.
Uncontrolled cell division: These mutations lead to rapid and uncontrolled cell division.
Formation of tumors: The uncontrolled division results in the formation of a tumor, which is a mass of abnormal cells.
Step 3
Give two safety precautions needed when blood is removed from this person.
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When removing blood from a person with haemochromatosis, two key safety precautions are:
Wear gloves: This protects against exposure to blood.
Use a sterile needle: This minimizes the risk of infection.
Step 4
State and explain the genotype of female Z.
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The genotype of female Z is heterozygous. This means that she has one dominant allele and one recessive allele for the haemochromatosis trait.
Explanation: One offspring must have inherited the recessive allele for haemochromatosis, indicating that at least one parent is a carrier. Therefore, female Z cannot be homozygous recessive.