You test-cross plants you suspect to be AaBb dihybrids and look at 100 offspring, with the following results: Phenotype AB 22 Ab 28 aB 29 ab 21 100 total1. You wish to test the hypothesis that the data are best explained as a dihybrid test cross AaBb x aabb. For a chi-square test, what expected number would you use for the aB class?2. Calculate the chi-square value for these data given the model that the results are from a dihybrid test cross of two genes on separate chromosomes.3. Using the chi-square table provided, determine the probability that the model would give the observed data. Degrees P Values
of freeom | 099 | 090 | 050 | 0.10 | 0.05 | 0.01 | 0.001
1 - 002 .45 2.71 3.84 6.64 10.83
2 0.02 0.21 1.39 4.61 5.99 9.21 13.82
3 0.11 0.58 2.37 6.25 7.81 11.35 16.25
4 0.30 1.03 3.36 7.78 9.49 13.28 18.47
5 0.55 1.61 4.35 9.24 11.07 15.09 20.524. Given this probability, would you accept or reject your model?

From the test cross and X² test, I assume the individual is dihybrid and genes are in separate chromosomes. 1)25 aB. 2)X²=2. 3)P₀.₀₅=7.81. 4)Accept the model.

What is a test cross?

A test cross is a cross we use to determine the genotype of individuals whose phenotype we already recognize.

The cross occurs between the individual of an unknown genotype and a recessive h0m0zygous individual. We want to know if the unknown genotype is heter0zyg0us or h0m0zyg0us dominant for a single gene.

Considering only one gene involved in the cross,

if the individual is heter0zyg0us, the progeny will be 50% heter0zyg0te and 50% h0m0zyg0te expressing the recessive trait.

The phenotypic and genotypic ratio is 1:1

if the individual is h0m0zyg0us dominant, the whole progeny will be heter0zyg0us.

Considering two genes that assort independently are test crossed,

If the individual is a dihybrid -heter0zyg0us for both genes-, the progeny will express equal phenotypic and genotypic ratio, 1:1:1:1.

If more than one gene are involved, we can also determine if these genes assort independently, or if they are linked.

If genes are located in different chromosomes or are far from each other in the same chromosome, they assort independently, and their phenotypic and genotypic ratios are 1:1:1:1

If these genes are linked, after the test cross, the distribution of phenotypes among the progeny differs from 1:1:1:1.

When phenotypes appear in different proportions than the expected ones, we can assume that these genes are linked.

However, the difference between the observed and the expected data can be by random chance. This means that genes could assort independently, but by chance, the observed ratios vary from the expected ones.

To analyze these situations and determine is the difference is by chance or because genes are linked, we need to perform a chi-square test.

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Now, let us answer the question.

Available data:

Test cross

F1 = 100 individuals

AB 22
Ab 28
aB 29
ab 21

1) What expected number would you use for the aB class?

To answer this question, we assume genes assort independently, so if this is the case, genotypic ratio should be 1:1:1:1, or 25% AB : 25% aB : 25% Ab : 25% ab

100% ------------- 100 individuals

25% -------------X = (25% x 100) / 100% = 25 individuals AB

25% -------------X = (25% x 100) / 100% = 25 individuals aB

25% -------------X = (25% x 100) / 100% = 25 individuals Ab

25% -------------X = (25% x 100) / 100% = 25 individuals ab

So, accordig to our calculations, the expected number for the aB class is 25 individuals.

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2) Calculate the chi-square value.

Model: dihybrid test cross of two genes on separate chromosomes -assort independently-.

AB Ab aB ab

Observed, O 22 28 29 21

Expected, E 25 25 25 25

(O-E)² / E 0.36 0.36 0.64 0.64

X² = Σ(Obs-Exp)²/Exp = 0.36 + 0.36 + 0.64 + 0.64 = 2

X² = 2

The chi-square value is X² = 2

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3) Determine the probability that the model would give the observed data.

Freedom degrees = n - 1 Where n is the number of classes ⇒ 4.

Freedom degrees = 4 - 1 = 3

Significance level = 0.05

Conventionally, we use 0.05 = 5% as the significance level.

Critical value (table value) = 7.81

Considering a significance level of 0.05 and 3 freedom degrees, the critical value equals 7.81.

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4) would you accept or reject your model?

X² = 2
P₀.₀₅ = 7.81

X² < P₀.₀₅ ⇒ 2 < 7.81

p value is greater than X², so there is not enough evidence to reject the null hypothesis. The ungenotype might be a dihybrid, and there might be independent assortment.

I would accept the model.

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