Exam questions · Biology
Inheritance
- 42 exam questions
- 143 marks
- 63 quick checks
Alleles, Genotype and Phenotype
Just this lesson-
1 State [2 marks]
State what is meant by (a) an allele and (b) a dominant allele.
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Model answer
(a) A different version of the same gene. (b) An allele that shows its effect even when only one copy is present.
Mark scheme
- Different version of a gene — 1 mark
- Shows its effect with one copy — 1 mark
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2 Name [2 marks]
In pea plants, tall (T) is dominant to dwarf (t). Name the genotypes that give a tall plant.
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Model answer
TT and Tt.
Mark scheme
- TT — 1 mark
- Tt — 1 mark
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3 Explain [3 marks]
Explain what is meant by the terms homozygous and heterozygous, using the genotypes TT and Tt as examples.
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Model answer
Homozygous means that the two alleles of the gene are the same, such as TT. Heterozygous means that the two alleles are different, such as Tt.
Mark scheme
- Homozygous: two identical alleles — 1 mark
- Heterozygous: two different alleles — 1 mark
- Correct example of each — 1 mark
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4 Explain [3 marks]
A heterozygous tall plant has the genotype Tt. Explain why it is tall and not dwarf.
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Model answer
The allele for tall (T) is dominant, so it shows its effect even though only one copy is present. The allele for dwarf (t) is recessive, so it only has an effect if there are two copies. The plant therefore has the phenotype tall.
Mark scheme
- T is dominant — 1 mark
- Shows with one copy — 1 mark
- t is recessive and only shows with two copies — 1 mark
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5 Describe [4 marks]
Cystic fibrosis is caused by a recessive allele (f). Describe the genotypes of a person who has cystic fibrosis, a carrier and a person who does not have the allele.
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Model answer
A person with cystic fibrosis has the genotype ff, with two recessive alleles. A carrier has the genotype Ff, with one dominant and one recessive allele, and does not have the disorder. A person who does not have the allele has the genotype FF.
Mark scheme
- ff has cystic fibrosis — 1 mark
- Ff is a carrier — 1 mark
- Carrier does not have the disorder — 1 mark
- FF has no recessive allele — 1 mark
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6 Explain [6 marks]
Explain how the alleles of a gene are passed from parents to their children.
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Model answer
Each parent has two alleles of a gene. In meiosis the pair of chromosomes is separated, so each gamete carries only one allele of the gene. A heterozygous parent produces two types of gamete. At fertilisation, one gamete from the mother fuses with one gamete from the father to form a zygote. The zygote therefore has two alleles again, one from each parent. Which gametes fuse is a matter of chance, so the combination of alleles in the child is random.
Mark scheme
- Each parent has two alleles — 1 mark
- Meiosis separates the alleles, one per gamete — 1 mark
- Heterozygous parent makes two types of gamete — 1 mark
- Fertilisation fuses one gamete from each parent — 1 mark
- The zygote has two alleles again — 1 mark
- Random / chance — 1 mark
Quick check
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1
What is an allele?
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C: A different version of a gene
An allele is a different version of the same gene.
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2
Which genotype is homozygous recessive?
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D: tt
Two lower-case letters, tt, show two copies of the recessive allele.
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3
What is the phenotype of a plant with the genotype Tt, if T (tall) is dominant?
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A: Tall
The dominant allele shows, so the plant is tall.
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4
What does genotype mean?
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B: The alleles it has
The genotype is the combination of alleles an organism has.
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5
How many alleles of a gene are in a gamete?
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A: One
A gamete contains only one allele of each gene.
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6
Which describes a heterozygous genotype?
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C: Bb
Heterozygous means that the two alleles are different, such as Bb.
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7
A person with genotype Ff does not have cystic fibrosis, but has a recessive allele. What is the person called?
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D: A carrier
A person with one recessive allele who does not show the disorder is a carrier.
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8
A plant is tall. Which statement is correct?
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B: Its genotype could be TT or Tt
The plant could be TT or Tt, because both have the dominant phenotype.
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9
Why does the recessive phenotype only appear in the homozygous recessive genotype?
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A: A dominant allele would hide the recessive allele
If there is a dominant allele, it shows its effect and hides the recessive allele, so the recessive phenotype needs two recessive alleles.
Monohybrid Crosses and Punnett Squares
Just this lesson-
1 Complete [3 marks]
A Punnett square is used to show the cross Tt x tt. Complete the genotypes of the four offspring.
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Model answer
Tt, Tt, tt and tt.
Mark scheme
- Two Tt — 1 mark
- Two tt — 1 mark
- Gametes shown correctly — 1 mark
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2 State [2 marks]
Two heterozygous parents (Bb) have a child. State the probability that the child will have the genotype bb, as a fraction and as a percentage.
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Model answer
1 in 4 (\(\frac{1}{4}\)), which is 25%.
Mark scheme
- 1 in 4 — 1 mark
- 25% — 1 mark
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3 Calculate [3 marks]
Two heterozygous tall pea plants (Tt) are crossed and produce 320 seeds. Calculate the number of seeds that would be expected to grow into dwarf plants.
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Model answer
The ratio is 3 tall : 1 dwarf, so \(\frac{1}{4}\) are dwarf. 320 ÷ 4 = 80 dwarf plants.
Mark scheme
- Ratio of 3 : 1 / one quarter dwarf — 1 mark
- 320 divided by 4 — 1 mark
- 80 — 1 mark
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4 Draw [4 marks]
In guinea pigs, the allele for black fur (B) is dominant to the allele for white fur (b). Draw a genetic diagram to show the cross between a heterozygous black guinea pig and a white guinea pig.
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Model answer
Parents Bb and bb. Gametes: B and b from the heterozygous parent, and b and b from the white parent. Offspring: Bb, Bb, bb, bb. This gives 2 black (Bb) and 2 white (bb), a ratio of 1 : 1.
Mark scheme
- Parents' genotypes Bb and bb — 1 mark
- Correct gametes — 1 mark
- Correct offspring genotypes — 1 mark
- Ratio of 1 black : 1 white — 1 mark
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5 Explain [3 marks]
Two carriers of cystic fibrosis already have an affected child. A friend says that their next child cannot be affected. Explain why the friend is wrong.
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Model answer
Each fertilisation is a separate, random event. The chance of a child inheriting two recessive alleles is 1 in 4 every time, and it is not affected by earlier children. So the next child still has a 1 in 4 chance of being affected.
Mark scheme
- Each fertilisation is independent / random — 1 mark
- Chance is 1 in 4 each time — 1 mark
- Not affected by previous children — 1 mark
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6 Explain [6 marks]
A farmer has a black bull (Bb) and a black cow (Bb). Black is dominant to red (b). Use a genetic diagram to predict the offspring of the cross, and explain why the actual results in a herd of 4 calves may not match the prediction.
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Model answer
The parents are Bb and Bb. The gametes from each are B and b. The offspring are BB, Bb, Bb and bb, a genotype ratio of 1 : 2 : 1. BB and Bb are black, and bb is red, so the phenotype ratio is 3 black : 1 red, or a 25% chance of a red calf. In a herd of only 4 calves the results may not match the prediction because fertilisation is random. Each calf has a 1 in 4 chance of being red, and with such a small number there could be none, or two, red calves. A larger number of offspring would be closer to 3 : 1.
Mark scheme
- Correct gametes — 1 mark
- Correct Punnett square genotypes — 1 mark
- Ratio of 3 black to 1 red — 1 mark
- Fertilisation is random / chance — 1 mark
- Small number gives unreliable results — 1 mark
- Larger numbers closer to the ratio — 1 mark
Quick check
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1
What is the genotype ratio from a cross of Tt x Tt?
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B: 1 : 2 : 1
The four boxes are TT, Tt, Tt and tt, so the ratio is 1 : 2 : 1.
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2
What is the phenotype ratio from Tt x Tt, if T is dominant?
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A: 3 : 1
TT and Tt show the dominant phenotype, and tt the recessive, so the ratio is 3 : 1.
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3
What is the chance of a tt offspring from Tt x Tt?
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C: 1 in 4
One of four boxes is tt, so the chance is 1 in 4, or 25%.
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4
What are all the offspring of TT x tt?
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D: Tt
Every gamete from TT is T and every gamete from tt is t, so all offspring are Tt.
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5
Which cross gives a ratio of 1 : 1?
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B: Tt x tt
A heterozygous parent crossed with a homozygous recessive parent gives half Tt and half tt.
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6
100 offspring come from Tt x Tt. How many would you expect to be tt?
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A: 25
One quarter are tt, and \(100 \div 4 = 25\).
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7
In a Punnett square, what is placed along the top and the side?
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C: The gametes
The gametes of each parent go along the top and down the side.
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8
A couple who are both carriers of cystic fibrosis have a 1 in 4 chance of an affected child. Why might they have 3 affected children out of 4?
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B: Fertilisation is random, so small numbers can differ from the prediction
The result is a matter of chance, and small numbers do not have to match the probability.
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9
What is the purpose of a test cross?
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D: To find out if a dominant phenotype is homozygous or heterozygous
Crossing with a homozygous recessive shows the unknown genotype, because only a heterozygous plant can give recessive offspring.
Family Pedigrees
Just this lesson-
1 State [2 marks]
In a pedigree chart, state what is shown by (a) a circle, (b) a shaded symbol.
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Model answer
(a) A female. (b) A person who has the characteristic or disorder.
Mark scheme
- Female — 1 mark
- Has the disorder — 1 mark
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2 Name [2 marks]
A genetic disorder is caused by a recessive allele (n). Name the genotype of a carrier, and the genotype of a person with the disorder.
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Model answer
A carrier is Nn. A person with the disorder is nn.
Mark scheme
- Nn — 1 mark
- nn — 1 mark
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3 Explain [3 marks]
Two parents without cystic fibrosis have a daughter who has the disorder. Explain how this is possible.
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Model answer
Cystic fibrosis is caused by a recessive allele. Both parents are carriers, Ff, so each has one dominant allele that hides the effect of the recessive one. The daughter inherited a recessive allele from each parent, so she is ff and has the disorder.
Mark scheme
- The allele is recessive — 1 mark
- Both parents are carriers (Ff) — 1 mark
- She inherited f from each parent — 1 mark
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4 Calculate [3 marks]
A man has Huntington's disease (Hh) and his partner does not have it (hh). Draw a Punnett square, and state the percentage chance that their child has Huntington's disease.
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Model answer
Gametes from the man are H and h, and from the woman h and h. The offspring are Hh, Hh, hh, hh. Half have Hh, so the chance is 50%.
Mark scheme
- Correct gametes — 1 mark
- Correct genotypes Hh, Hh, hh, hh — 1 mark
- 50% — 1 mark
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5 Explain [4 marks]
In a pedigree for a disorder, two parents who are both affected have a child who is unaffected. Explain whether the disorder is caused by a dominant or a recessive allele.
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Model answer
The disorder is caused by a dominant allele. If it were recessive, both affected parents would be homozygous recessive, and could only have affected children. Since they have an unaffected child, the parents must have a dominant allele that is hidden in the child. The parents are heterozygous, and the child has two recessive alleles.
Mark scheme
- Dominant allele — 1 mark
- If recessive, all children would be affected — 1 mark
- Parents are heterozygous — 1 mark
- Child received the recessive allele from each — 1 mark
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6 Evaluate [6 marks]
People with a family history of Huntington's disease can have a genetic test to find out whether they carry the allele. Evaluate the advantages and disadvantages of taking the test.
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Model answer
An advantage is that people who test positive can make decisions about their future, such as whether to have children, and they can plan for their care. They can also avoid the uncertainty of not knowing. People who test negative are relieved of worry. A disadvantage is that there is currently no cure, so a positive result could cause great distress and depression. The result might affect insurance or employment, and it also has implications for relatives, who might not want to know. Overall the decision is personal, and counselling should be available before and after the test.
Mark scheme
- Can plan or make decisions about children — 1 mark
- Ends uncertainty — 1 mark
- No cure, so distress — 1 mark
- Insurance or employment — 1 mark
- Implications for relatives — 1 mark
- A reasoned conclusion — 1 mark
Quick check
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1
In a pedigree, what does a filled square show?
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B: An affected male
A square is a male and a filled symbol means the person has the disorder.
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2
Two unaffected parents have an affected child. What type of allele causes the condition?
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C: Recessive
The allele is hidden in the parents, so it is recessive.
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3
What is the genotype of a person with cystic fibrosis?
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D: ff
Cystic fibrosis is recessive, so a person with it has two recessive alleles, ff.
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4
Which genotype is a carrier of cystic fibrosis?
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A: Ff
A carrier has one dominant and one recessive allele, Ff.
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5
Huntington's disease is caused by a dominant allele. What is the chance that a child of one affected parent (Hh) and one unaffected parent (hh) is affected?
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B: 50%
Half of the children inherit H from the affected parent, so the chance is 50%.
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6
In a pedigree, what does a horizontal line between a square and a circle show?
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C: Two partners
A horizontal line joins two partners.
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7
Two carriers (Ff) have a child. What is the probability that the child is a carrier?
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A: 50%
The Punnett square gives FF, Ff, Ff and ff, so 2 out of 4 are carriers: 50%.
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8
In a family, every affected person has an affected parent, and the condition appears in every generation. What type of allele is most likely?
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D: Dominant
This pattern points to a dominant allele.
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9
Why can the genotype of an unaffected child of two carriers not be known for certain?
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B: The child could be FF or Ff
The child could be FF or Ff, because both look the same, so the pedigree cannot tell which one it is.
Inheriting Sex
Just this lesson-
1 State [2 marks]
State the sex chromosomes in (a) a human female body cell, and (b) a human male body cell.
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Model answer
(a) XX. (b) XY.
Mark scheme
- XX — 1 mark
- XY — 1 mark
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2 State [2 marks]
State the sex chromosomes that are in a sperm cell, and in an egg cell.
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Model answer
A sperm cell contains either an X or a Y chromosome. An egg cell always contains an X chromosome.
Mark scheme
- Sperm: X or Y — 1 mark
- Egg: X — 1 mark
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3 Explain [3 marks]
Explain why the sex of a baby is decided by the father's gamete and not the mother's.
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Model answer
All of the eggs made by the mother carry an X chromosome. The father makes two types of sperm, one that carries X and one that carries Y. The type of sperm that fertilises the egg decides if the baby is a girl (XX) or a boy (XY).
Mark scheme
- All eggs carry X — 1 mark
- Sperm carry either X or Y — 1 mark
- Which sperm fertilises decides the sex — 1 mark
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4 Draw [4 marks]
Draw a genetic diagram to show how the sex of a child is inherited.
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Model answer
Parents: XX and XY. Gametes: X and X from the mother, X and Y from the father. Offspring: XX, XY, XX, XY, which is 2 girls and 2 boys, a ratio of 1 : 1.
Mark scheme
- Parents XX and XY — 1 mark
- Correct gametes — 1 mark
- Correct offspring XX, XY, XX, XY — 1 mark
- Ratio 1 : 1 / 50% each — 1 mark
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5 Calculate [3 marks]
A couple have three daughters. Calculate the probability that their fourth child is a girl, and the probability that all four children are girls. Give your answers as fractions.
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Model answer
The fourth child has a probability of \(\frac{1}{2}\) of being a girl. The probability of all four being girls is \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{16}\).
Mark scheme
- Probability of a girl is 1/2 — 1 mark
- Multiplies four probabilities — 1 mark
- 1/16 — 1 mark
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6 Evaluate [6 marks]
A newspaper article claims that the mother decides whether a baby is a boy or a girl. Use your knowledge of genetics to evaluate this claim.
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Model answer
The claim is incorrect. A mother is XX, so all of her eggs carry an X chromosome, and she can only pass on an X. The father is XY, so half of his sperm carry an X and half carry a Y. If a sperm with a Y fertilises the egg, the baby is a boy (XY), and if one with an X does, it is a girl (XX). So the sex is decided by the sperm, which is random. The chance of each is 50%, with a ratio of 1 : 1. The mother has no way to decide which sperm fertilises her egg.
Mark scheme
- Claim is incorrect — 1 mark
- All eggs carry X — 1 mark
- Sperm carry X or Y — 1 mark
- Sperm that fertilises decides the sex — 1 mark
- Random / 50% chance each — 1 mark
- Mother has no control — 1 mark
Quick check
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1
Which sex chromosomes does a human male have?
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B: XY
Males have one X and one Y chromosome, XY.
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2
Which sex chromosome do all eggs have?
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A: X
A woman is XX, so every egg has an X chromosome.
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3
What is the probability that a baby is a girl?
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C: 50%
The chances of XX and XY are equal, so a girl has a 1 in 2 chance, or 50%.
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4
Which gamete decides the sex of the baby?
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D: The sperm
The sperm carries either X or Y, so it decides the sex.
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5
What is the sex ratio of the offspring from XX x XY?
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B: 1 : 1
Two of four boxes are XX and two are XY, so the ratio is 1 : 1.
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6
A couple have four boys. What is the chance that the fifth child is a boy?
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A: 1 in 2
Each child is an independent event, so the chance is still 1 in 2.
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7
Which genotype is a baby boy?
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C: XY
A boy has one X chromosome and one Y chromosome.
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8
What does the SRY gene do?
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D: Starts the development of testes
The SRY gene on the Y chromosome starts the development of male features.
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9
What is the probability that three babies in a row are girls?
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B: 1 in 8
Each has a probability of 1/2, so \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\).
Mendel and the Discovery of Inheritance
Just this lesson-
1 State [2 marks]
State two reasons why Mendel used pea plants in his experiments.
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Model answer
Any two from: they grow quickly; they are easy to grow; they have clear differences between varieties; many offspring can be produced; they can be cross-pollinated by hand.
Mark scheme
- First reason — 1 mark
- Second reason — 1 mark
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2 Describe [3 marks]
Describe what Mendel found when he crossed pure-breeding tall pea plants with pure-breeding dwarf pea plants.
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Model answer
All of the F1 offspring were tall. When these were self-pollinated, the F2 offspring were about three-quarters tall and one-quarter dwarf, a ratio of about 3 : 1.
Mark scheme
- F1 all tall — 1 mark
- F2 included dwarf plants — 1 mark
- Ratio of about 3 : 1 — 1 mark
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3 Calculate [3 marks]
In one of his experiments, Mendel counted 5474 round seeds and 1850 wrinkled seeds in the F2 generation. Calculate the percentage of seeds that were round. Give your answer to 1 decimal place.
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Model answer
The total is 5474 + 1850 = 7324. The percentage that was round is 5474 ÷ 7324 × 100 = 74.7%.
Mark scheme
- Total of 7324 — 1 mark
- 5474 divided by the total, times 100 — 1 mark
- 74.7% — 1 mark
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4 Explain [3 marks]
Explain how Mendel's results showed that the idea of blending inheritance was wrong.
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Model answer
In the F1 generation all of the plants were tall, and there were no medium plants, as blending would predict. In the F2 generation the dwarf plants came back unchanged. If the features had blended, the dwarf feature would have been lost.
Mark scheme
- F1 not medium — 1 mark
- Dwarf feature reappeared in F2 — 1 mark
- Unchanged, so it had not been mixed or lost — 1 mark
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5 Explain [4 marks]
Explain why Mendel's work was not recognised until after his death.
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Model answer
Scientists at the time believed in blending inheritance, so his results did not fit their ideas. Chromosomes and meiosis had not been linked to inheritance, so no one knew how his factors could be passed on. His use of mathematics was unusual in biology, and he published in a little-known journal.
Mark scheme
- Went against blending — 1 mark
- No mechanism known — 1 mark
- Chromosomes and meiosis not understood — 1 mark
- Unusual methods / obscure publication — 1 mark
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6 Explain [6 marks]
Evaluate the importance of Mendel's experiments to our understanding of genetics, and explain why scientists in the nineteenth century found it difficult to understand inheritance.
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Model answer
Mendel showed that characteristics are controlled by separate factors, which are passed on in pairs, with one from each parent, and that one can be dominant and hide the other. These ideas were later explained when genes were found on chromosomes and the structure of DNA was discovered, so his work was the basis of genetics. Scientists in the nineteenth century found inheritance difficult to understand because they believed in blending, and could not explain why a feature disappeared and then reappeared. They did not know about chromosomes, meiosis, genes or DNA, so there was no mechanism to explain the results. His work was therefore ignored until it was rediscovered in 1900.
Mark scheme
- Separate factors / pairs — 1 mark
- Dominant and recessive — 1 mark
- Later linked to chromosomes and DNA — 1 mark
- Blending inheritance — 1 mark
- No knowledge of chromosomes, meiosis, genes or DNA — 1 mark
- Rediscovered in 1900 — 1 mark
Quick check
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1
Which organism did Mendel use in his experiments?
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C: Pea plants
Mendel worked with pea plants.
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2
What were all the F1 plants in the cross of pure-breeding tall and dwarf?
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A: Tall
All of the F1 plants were tall, because the tall factor is dominant.
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3
What was the approximate ratio of tall to dwarf in the F2 generation?
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D: 3 : 1
The F2 plants were about three tall to one dwarf.
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4
What did Mendel call what we now call genes?
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B: Hereditary factors
He called them hereditary factors.
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5
What is a pure-breeding plant?
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C: One that always produces offspring like itself
A pure-breeding plant always produces offspring with the same characteristic, because it is homozygous.
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6
Which idea did Mendel's results disprove?
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A: Blending inheritance
Mendel's results showed that features do not blend, because the recessive feature reappeared unchanged.
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7
Why was Mendel's work ignored for many years?
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D: It did not fit the idea of blending and no mechanism was known
There was no known mechanism for his factors, and it conflicted with ideas about blending inheritance.
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8
When was Mendel's work rediscovered?
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B: 1900
It was rediscovered in 1900 by de Vries, Correns and Tschermak.
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9
Mendel counted 750 tall and 250 dwarf plants. What is the ratio of tall to dwarf?
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A: 3 : 1
\(750 \div 250 = 3\), so the ratio is 3 : 1.
Codominance and ABO Blood Groups
Just this lesson-
1 State [2 marks]
State the genotypes that result in (a) blood group AB, and (b) blood group O.
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Model answer
(a) \(I^A I^B\). (b) \(ii\).
Mark scheme
- AB: \(I^A I^B\) — 1 mark
- O: \(ii\) — 1 mark
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2 Explain [2 marks]
Explain what is meant by multiple alleles.
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Model answer
A gene has more than two alleles in a population. For the ABO gene there are three alleles, but each person has only two.
Mark scheme
- More than two alleles — 1 mark
- Example with each person only having two — 1 mark
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3 Explain [3 marks]
Explain why a person with the genotype \(I^A I^B\) has blood group AB.
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Model answer
The alleles \(I^A\) and \(I^B\) are codominant. This means that both alleles show their effect, so the red blood cells have both A antigens and B antigens, making the blood group AB.
Mark scheme
- The alleles are codominant — 1 mark
- Both alleles are expressed — 1 mark
- Both antigens are present — 1 mark
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4 Draw [4 marks]
A man has blood group AB, and a woman has blood group O. Use a genetic diagram to find the possible blood groups of their children.
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Model answer
The man is \(I^A I^B\), and the woman is \(ii\). The man's gametes are \(I^A\) and \(I^B\), and the woman's are all \(i\). The children are \(I^A i\) (group A) or \(I^B i\) (group B), half of each.
Mark scheme
- Correct parent genotypes — 1 mark
- Correct gametes — 1 mark
- Correct offspring genotypes — 1 mark
- Groups A and B in a 1 : 1 ratio — 1 mark
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5 Explain [3 marks]
A couple, one with group A blood and one with group B blood, have a child with group O. Explain how this is possible.
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Model answer
Both parents must be heterozygous, with the genotypes \(I^A i\) and \(I^B i\). Each parent passes on the recessive \(i\) allele, so the child has the genotype \(ii\), which is blood group O.
Mark scheme
- Both parents heterozygous — 1 mark
- Each passes on the allele i — 1 mark
- Child is ii, so group O — 1 mark
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6 Evaluate [6 marks]
Compare dominant and recessive inheritance with codominance, using the ABO blood group system as an example, and say why the ABO system is described as showing multiple alleles.
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Model answer
In complete dominance one allele hides the effect of the other in a heterozygous person, so the heterozygote looks like the homozygous dominant. An example is the \(I^A\) allele with the recessive \(i\): a person with \(I^A i\) has group A and not O. In codominance both alleles show their effect, so the heterozygote has a different phenotype from either homozygote. The genotype \(I^A I^B\) gives group AB, with both A and B antigens. The ABO system shows multiple alleles because the gene has three alleles, \(I^A\), \(I^B\) and \(i\), in the population, even though each person has only two of them.
Mark scheme
- Dominance: one allele hides the other — 1 mark
- Example of dominance, such as \(I^A i\) is A — 1 mark
- Codominance: both alleles shown — 1 mark
- \(I^A I^B\) is AB — 1 mark
- Three alleles in the population — 1 mark
- Each person has only two — 1 mark
Quick check
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1
How many alleles does the ABO blood group gene have in the population?
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C: Three
There are three alleles: \(I^A\), \(I^B\) and \(i\).
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2
Which genotype gives blood group AB?
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B: \(I^A I^B\)
The codominant alleles \(I^A\) and \(I^B\) together give group AB.
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3
Which genotype gives blood group O?
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D: \(ii\)
Group O needs two copies of the recessive allele, \(ii\).
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4
What does codominance mean?
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A: Both alleles show their effect
In codominance both alleles show their effect in the heterozygous person.
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5
Which blood groups can a child of an AB parent and an O parent have?
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B: A or B
The AB parent passes on \(I^A\) or \(I^B\), and the O parent passes on \(i\), so the child is A or B.
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6
A person has the genotype \(I^A i\). What is their blood group?
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A: A
\(I^A\) is dominant to \(i\), so the person is group A.
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7
A group A parent and a group B parent have a group O child. What are the parents' genotypes?
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C: \(I^A i\) and \(I^B i\)
The child is \(ii\), so each parent has an \(i\) allele: \(I^A i\) and \(I^B i\).
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8
A red cow and a white bull produce roan calves. What is this an example of?
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D: Codominance
The calf shows both red and white hairs, so both alleles show their effect: codominance.
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9
Why can a group AB person not have a group O child?
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B: An AB parent has no i allele to pass on
A group AB person is \(I^A I^B\), so can only pass on \(I^A\) or \(I^B\), and the child cannot be \(ii\).
Sex-Linked Genetic Disorders (Higher tier)
Just this lesson-
1 State [2 marks]
State what is meant by a sex-linked disorder.
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Model answer
A disorder caused by an allele on a sex chromosome, usually the X chromosome, so that it is inherited differently in males and females.
Mark scheme
- Allele on a sex chromosome / X chromosome — 1 mark
- Different pattern of inheritance in males and females — 1 mark
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2 Explain [3 marks]
Explain why a male is more likely than a female to have a sex-linked recessive disorder.
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Model answer
A male has only one X chromosome, so he has only one allele for the gene. If it is the recessive allele, it will show its effect. A female has two X chromosomes, so a dominant allele on the other X would hide the recessive one, and she needs two recessive alleles to be affected.
Mark scheme
- Males have only one X — 1 mark
- A single recessive allele shows — 1 mark
- Females need two recessive alleles — 1 mark
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3 Complete [3 marks]
Haemophilia is caused by a recessive allele, \(h\), on the X chromosome. State the genotype of (a) a man with haemophilia, (b) a female carrier, and (c) a man without haemophilia.
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Model answer
(a) \(X^h Y\). (b) \(X^H X^h\). (c) \(X^H Y\).
Mark scheme
- \(X^h Y\) — 1 mark
- \(X^H X^h\) — 1 mark
- \(X^H Y\) — 1 mark
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4 Draw [4 marks]
A woman who is a carrier of colour blindness has a child with a man who has normal vision. Draw a genetic diagram to show the possible children.
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Model answer
The mother is \(X^B X^b\) and the father is \(X^B Y\). The gametes are \(X^B\) and \(X^b\) from the mother, and \(X^B\) and \(Y\) from the father. The children are \(X^B X^B\) (normal girl), \(X^B X^b\) (carrier girl), \(X^B Y\) (normal boy) and \(X^b Y\) (colour-blind boy).
Mark scheme
- Parents' genotypes — 1 mark
- Gametes — 1 mark
- All four offspring genotypes — 1 mark
- Phenotypes identified — 1 mark
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5 Calculate [3 marks]
A woman is a carrier of haemophilia and her partner is unaffected. Calculate the percentage of their sons expected to have haemophilia, and the percentage of their daughters expected to have haemophilia.
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Model answer
Half of the sons, 50%, are expected to have haemophilia, because the sons are \(X^H Y\) or \(X^h Y\). None of the daughters, 0%, will have haemophilia, as they all receive the father's normal \(X^H\).
Mark scheme
- 50% of sons — 1 mark
- 0% of daughters — 1 mark
- Reasoning about the father's X — 1 mark
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6 Explain [6 marks]
A man with haemophilia has a daughter and a son. Explain, using a genetic diagram, whether the children are likely to have haemophilia, and why haemophilia cannot be passed from a father to his son.
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Model answer
The father is \(X^h Y\) and the mother is \(X^H X^H\). The father makes \(X^h\) and \(Y\) sperm, and the mother makes only \(X^H\) eggs. The daughters are all \(X^H X^h\), so they are carriers and do not have haemophilia. The sons are all \(X^H Y\), so they are normal. A father passes the Y chromosome to his sons, and the allele is on the X chromosome, so he cannot pass it to them. His sons get their X from their mother, who has normal alleles. A carrier daughter could, however, pass the allele to her own children.
Mark scheme
- Parents' genotypes — 1 mark
- Correct gametes — 1 mark
- Daughters are carriers — 1 mark
- Sons are unaffected — 1 mark
- Sons get the Y from the father — 1 mark
- Allele is on the X chromosome — 1 mark
Quick check
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1
On which chromosome is the haemophilia allele found?
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B: The X chromosome
Haemophilia is caused by a recessive allele on the X chromosome.
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2
Why are sex-linked recessive disorders more common in males?
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C: Males only have one X chromosome
A male has only one X chromosome, so one recessive allele is enough to cause the disorder.
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3
What is the genotype of a female carrier of haemophilia?
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A: \(X^H X^h\)
A carrier has one normal and one haemophilia allele: \(X^H X^h\).
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4
What is the genotype of a man with haemophilia?
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D: \(X^h Y\)
He has the haemophilia allele on his only X chromosome, and a Y: \(X^h Y\).
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5
A carrier woman and an unaffected man have a son. What is the chance that he has haemophilia?
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B: 50%
He gets his X from his mother, which is \(X^H\) or \(X^h\) with equal chance, so 50%.
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6
What are the daughters of a man with haemophilia and a woman with two normal alleles?
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C: All carriers
Each daughter receives his \(X^h\) and her mother's \(X^H\), so all are carriers.
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7
Why can a father not pass haemophilia on to his sons?
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D: He passes the Y chromosome to his sons, and the allele is on his X
A father passes his Y chromosome to his sons, and the haemophilia allele is on the X chromosome.
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8
Which of these is a sex-linked recessive disorder?
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A: Red-green colour blindness
Red-green colour blindness is caused by recessive alleles on the X chromosome.
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9
A woman who has colour blindness has a son. What is the chance that he is colour blind?
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B: 100%
She has the genotype \(X^b X^b\), so every son receives an \(X^b\) from her, and he is colour blind (\(X^b Y\)).