Biology · Simulation and virtual lab Free
Genetic crosses: Punnett squares, ratios and chi-square
Choose two parents, fill the Punnett square, breed the offspring and test whether chance gives the expected ratio.
- You will learn to
- Use Punnett squares to predict genotypes and phenotypes of monohybrid and dihybrid crosses.
- Explain dominance, codominance and multiple alleles (ABO).
- Compare bred and expected numbers and test the fit with chi-square.
- You will measure
- Numbers of each phenotype bred; expected numbers; χ².
- Time needed
- About 30–40 min
- Equipment
- Nothing extra. Real version: two bags of coloured beads or bottle tops (one per parent) to draw gametes at random.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To use Punnett squares to predict the genotypes and phenotypes of offspring in monohybrid and dihybrid crosses, to compare the predicted ratios with simulated breeding results, and to test the fit using the chi-square test.
What to do
- Choose pea plant height, parents Tt × Tt. Read the gametes and the four boxes of the Punnett square.
- Breed 100 offspring and tap Record. Breed again three times. Are the numbers always 75 : 25?
- Breed 400 offspring. Is the ratio closer to 3 : 1? Look at χ² and the graph.
- Try Tt × tt (a test cross) and TT × tt. Write the expected ratio before you breed.
- Choose cattle coat colour: cross RR × WW, then two roan cattle (RW × RW). Choose blood groups: Iᴬi × Iᴮi.
- Choose pea seeds (dihybrid): cross RrYy × RrYy and check the 9 : 3 : 3 : 1 ratio with χ².
Results
Questions
What is the difference between genotype and phenotype?
Genotype is the pair of alleles an organism has (e.g. Tt); phenotype is the characteristic that shows (e.g. tall).
Why does Tt × Tt give a 3 : 1 ratio?
Each parent makes T and t gametes in equal numbers; the four equally likely combinations are TT, Tt, tT and tt, and three of them contain the dominant T allele.
What is a test cross and why is it used?
Crossing an organism showing the dominant phenotype with a homozygous recessive (tt). If any offspring are recessive, the unknown parent must be heterozygous.
Two roan cattle (RW) are crossed. What ratio of coat colours is expected?
1 red (RR) : 2 roan (RW) : 1 white (WW), because neither allele is dominant (codominance).
Can parents with blood groups A and B have a child with blood group O?
Yes, if both are heterozygous (Iᴬi × Iᴮi): one quarter of the children are expected to be ii, group O.
Why are the bred numbers rarely exactly the expected numbers?
Which gametes join is a matter of chance, so results vary. Bigger samples usually come closer to the expected ratio.
For teachers: expected results and how the model works
Each offspring gets one gamete chosen at random from each parent, so the counts follow the binomial distribution. Expected ratios: TT × tt all tall; Tt × Tt 3 : 1; Tt × tt 1 : 1; RR × WW all roan; RW × RW 1 : 2 : 1; Iᴬi × Iᴮi 1 A : 1 B : 1 AB : 1 O; RrYy × RrYy 9 : 3 : 3 : 1; RrYy × rryy 1 : 1 : 1 : 1. χ² = Σ (O − E)² ÷ E with critical values at the 5 % level: 3.84 (1 degree of freedom), 5.99 (2), 7.81 (3). About 1 run in 20 will be "does not fit" even when the theory is right: a good discussion point. ABO alleles are written Iᴬ, Iᴮ and i. Only 100 offspring are drawn; all are counted.
Exam practice Exam set
National-exam style questions on this simulation (18 marks). Try each one on paper first, then open the marking scheme.
In peas, round seed (R) is dominant to wrinkled (r). A heterozygous plant is crossed with a wrinkled plant. (a) Draw a genetic diagram. (b) Give the expected ratio. (c) 52 round and 48 wrinkled were obtained: do the results fit? (7 marks)
Marking scheme: (a) Parents Rr × rr (1); gametes R, r and r (1); offspring Rr and rr (1). (b) 1 round : 1 wrinkled (1). (c) Expected 50 : 50 (1); χ² = (2² ÷ 50) + (2² ÷ 50) = 0.16 (1); less than 3.84, so the results fit (1).
Explain codominance using an example. (3 marks)
Marking scheme: Both alleles are expressed in the heterozygote (1); e.g. cattle RR red and WW white give RW roan (1), with red and white hairs mixed (1).
A man of blood group AB marries a woman of group O. List the possible blood groups of their children and explain why none can be AB or O. (4 marks)
Marking scheme: Iᴬ Iᴮ × ii (1); children Iᴬi (group A) or Iᴮi (group B), half each (1); every child gets i from the mother, so cannot be AB (1), and gets Iᴬ or Iᴮ from the father, so cannot be O (1).
In a dihybrid cross between two RrYy pea plants, 320 offspring were produced. How many of each phenotype are expected? (4 marks)
Marking scheme: 9 : 3 : 3 : 1 (1); round yellow 180 (1); round green 60 and wrinkled yellow 60 (1); wrinkled green 20 (1).
Donat Sciences and MathsBiology · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Genetic crosses: Punnett squares, ratios and chi-square
Aim: To use Punnett squares to predict the genotypes and phenotypes of offspring in monohybrid and dihybrid crosses, to compare the predicted ratios with simulated breeding results, and to test the fit using the chi-square test.
Apparatus (real laboratory): two bags of coloured beads or bottle tops (one per parent) to draw gametes at random.
What to do (with the simulation on www.donatstem.co.rw)
- Choose pea plant height, parents Tt × Tt. Read the gametes and the four boxes of the Punnett square.
- Breed 100 offspring and tap Record. Breed again three times. Are the numbers always 75 : 25?
- Breed 400 offspring. Is the ratio closer to 3 : 1? Look at χ² and the graph.
- Try Tt × tt (a test cross) and TT × tt. Write the expected ratio before you breed.
- Choose cattle coat colour: cross RR × WW, then two roan cattle (RW × RW). Choose blood groups: Iᴬi × Iᴮi.
- Choose pea seeds (dihybrid): cross RrYy × RrYy and check the 9 : 3 : 3 : 1 ratio with χ².
Results
| # | Cross | Expected ratio | Numbers obtained | Expected numbers | χ² | Fits? (yes/no) |
|---|
| 1 | | | | | | |
| 2 | | | | | | |
| 3 | | | | | | |
| 4 | | | | | | |
| 5 | | | | | | |
| 6 | | | | | | |
| 7 | | | | | | |
| 8 | | | | | | |
Graph
Plot the number obtained against the expected number for each type of offspring. Draw the line y = x and comment.
Questions
- What is the difference between genotype and phenotype? [2]
- Why does Tt × Tt give a 3 : 1 ratio? [2]
- What is a test cross and why is it used? [2]
- Two roan cattle (RW) are crossed. What ratio of coat colours is expected? [2]
- Can parents with blood groups A and B have a child with blood group O? [2]
- Why are the bred numbers rarely exactly the expected numbers? [2]
- In peas, round seed (R) is dominant to wrinkled (r). A heterozygous plant is crossed with a wrinkled plant. (a) Draw a genetic diagram. (b) Give the expected ratio. (c) 52 round and 48 wrinkled were obtained: do the results fit? [7]
- Explain codominance using an example. [3]
- A man of blood group AB marries a woman of group O. List the possible blood groups of their children and explain why none can be AB or O. [4]
- In a dihybrid cross between two RrYy pea plants, 320 offspring were produced. How many of each phenotype are expected? [4]
Donat Sciences and MathsBiology · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Genetic crosses: Punnett squares, ratios and chi-square
Expected results
Each offspring gets one gamete chosen at random from each parent, so the counts follow the binomial distribution. Expected ratios: TT × tt all tall; Tt × Tt 3 : 1; Tt × tt 1 : 1; RR × WW all roan; RW × RW 1 : 2 : 1; Iᴬi × Iᴮi 1 A : 1 B : 1 AB : 1 O; RrYy × RrYy 9 : 3 : 3 : 1; RrYy × rryy 1 : 1 : 1 : 1. χ² = Σ (O − E)² ÷ E with critical values at the 5 % level: 3.84 (1 degree of freedom), 5.99 (2), 7.81 (3). About 1 run in 20 will be "does not fit" even when the theory is right: a good discussion point. ABO alleles are written Iᴬ, Iᴮ and i. Only 100 offspring are drawn; all are counted.
Answers
- What is the difference between genotype and phenotype? [2]
Genotype is the pair of alleles an organism has (e.g. Tt); phenotype is the characteristic that shows (e.g. tall).
- Why does Tt × Tt give a 3 : 1 ratio? [2]
Each parent makes T and t gametes in equal numbers; the four equally likely combinations are TT, Tt, tT and tt, and three of them contain the dominant T allele.
- What is a test cross and why is it used? [2]
Crossing an organism showing the dominant phenotype with a homozygous recessive (tt). If any offspring are recessive, the unknown parent must be heterozygous.
- Two roan cattle (RW) are crossed. What ratio of coat colours is expected? [2]
1 red (RR) : 2 roan (RW) : 1 white (WW), because neither allele is dominant (codominance).
- Can parents with blood groups A and B have a child with blood group O? [2]
Yes, if both are heterozygous (Iᴬi × Iᴮi): one quarter of the children are expected to be ii, group O.
- Why are the bred numbers rarely exactly the expected numbers? [2]
Which gametes join is a matter of chance, so results vary. Bigger samples usually come closer to the expected ratio.
- In peas, round seed (R) is dominant to wrinkled (r). A heterozygous plant is crossed with a wrinkled plant. (a) Draw a genetic diagram. (b) Give the expected ratio. (c) 52 round and 48 wrinkled were obtained: do the results fit? [7]
(a) Parents Rr × rr (1); gametes R, r and r (1); offspring Rr and rr (1). (b) 1 round : 1 wrinkled (1). (c) Expected 50 : 50 (1); χ² = (2² ÷ 50) + (2² ÷ 50) = 0.16 (1); less than 3.84, so the results fit (1).
- Explain codominance using an example. [3]
Both alleles are expressed in the heterozygote (1); e.g. cattle RR red and WW white give RW roan (1), with red and white hairs mixed (1).
- A man of blood group AB marries a woman of group O. List the possible blood groups of their children and explain why none can be AB or O. [4]
Iᴬ Iᴮ × ii (1); children Iᴬi (group A) or Iᴮi (group B), half each (1); every child gets i from the mother, so cannot be AB (1), and gets Iᴬ or Iᴮ from the father, so cannot be O (1).
- In a dihybrid cross between two RrYy pea plants, 320 offspring were produced. How many of each phenotype are expected? [4]
9 : 3 : 3 : 1 (1); round yellow 180 (1); round green 60 and wrinkled yellow 60 (1); wrinkled green 20 (1).