Monohybrid Crosses and Punnett Squares

1. 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
2. 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
3. 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
4. 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
5. 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
6. 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

1. 1

   What is the genotype ratio from a cross of Tt x Tt?

   1. A3 : 1
   2. B1 : 2 : 1
   3. C1 : 1
   4. D2 : 2
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   B: 1 : 2 : 1

   The four boxes are TT, Tt, Tt and tt, so the ratio is 1 : 2 : 1.

2. 2

   What is the phenotype ratio from Tt x Tt, if T is dominant?

   1. A3 : 1
   2. B1 : 1
   3. C1 : 2 : 1
   4. D4 : 0
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   A: 3 : 1

   TT and Tt show the dominant phenotype, and tt the recessive, so the ratio is 3 : 1.

3. 3

   What is the chance of a tt offspring from Tt x Tt?

   1. A1 in 2
   2. B3 in 4
   3. C1 in 4
   4. D1 in 3
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   C: 1 in 4

   One of four boxes is tt, so the chance is 1 in 4, or 25%.

4. 4

   What are all the offspring of TT x tt?

   1. ATT
   2. Btt
   3. CHalf Tt and half tt
   4. DTt
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   D: Tt

   Every gamete from TT is T and every gamete from tt is t, so all offspring are Tt.

5. 5

   Which cross gives a ratio of 1 : 1?

   1. ATT x tt
   2. BTt x tt
   3. CTt x Tt
   4. DTT x TT
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   B: Tt x tt

   A heterozygous parent crossed with a homozygous recessive parent gives half Tt and half tt.

6. 6

   100 offspring come from Tt x Tt. How many would you expect to be tt?

   1. A25
   2. B50
   3. C75
   4. D100
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   A: 25

   One quarter are tt, and \(100 \div 4 = 25\).

7. 7

   In a Punnett square, what is placed along the top and the side?

   1. AThe offspring
   2. BThe phenotypes
   3. CThe gametes
   4. DThe chromosomes
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   C: The gametes

   The gametes of each parent go along the top and down the side.

8. 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?

   1. AThe probabilities are wrong
   2. BFertilisation is random, so small numbers can differ from the prediction
   3. CThe recessive allele changes
   4. DThey cannot be carriers
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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.

9. 9

   What is the purpose of a test cross?

   1. ATo make a plant dwarf
   2. BTo find the number of chromosomes
   3. CTo change the alleles
   4. DTo find out if a dominant phenotype is homozygous or heterozygous
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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.