Question

After crossing two organisms that have the phenotype A, you get offspring in two groups: 37 of phenotype A and 18 of phenotype B. Propose the most likely type of inheritance for the traits, write the genotypes of parents and the offspring, and strengthen your arguments by doing a chi-square test. Show calculations in your work... (GENETICS PROBLEM)

Answer 1

Let the alleles involved in determine the two different phenotypes A and B be X and x

Let allele X be dominant over allele x (recessive). Presence of at least one dominant X allele produces the A phenotype whereas the presence of 2 x alleles only produces phenotype B.

Thus homozygous dominant individual will have the genotype XX and Phenotype A.

Heterozygous individuals will have the genotype Xx and Phenotype A.

Homozygous recessive individuals will have the genotype xx and Phenotype B.

In the above cross, it is mentioned that crossing of two Phenotype A individuals gives 37 offspring of Phenotype A and 18 offspring of Phenotype B.

Total number of offspring =37 + 18 = 55

Percentage of Phenotype A offspring = 37 / 55 = 67.27 %, which is very close to 75%.

Percentage of Phenotype B offspring = 18/55 = 32.72 %, which is close to 25%

Thus the ratio of Phenotype A : Phenotype B offspring is approximately 3 : 1.

This type of ratio is produced only when two heterozygous individuals are crossed.

For the above question, we cross 2 Xx individuals. The offspring genotype can be predicted using the Punnett square as below :

	X	x
X	XX (Phenotype A)	Xx (Phenotype A)
x	Xx (Phenotype A)	xx (Phenotype B)

Thus we observe a 3 :1 or 3/4 : 1/4 ratio of Phenotype A and Phenotype B.

If this is the case, then the mode of inheritance is Autosomal Dominant pattern. In this type of inheritance, the trait is determined by the presence of a dominant allele and is linked to the dominant allele.

Chi-Square Test :

Total Number of offspring = 55

According to the above question, the observed offspring phenotype numbers are : 37 offspring of Phenotype A and 18 of Phenotype B.

The expected number of offspring according to the above cross conducted in the Punnett Square will be :

Phenotype B = (1/4) x 55 = 13.75

Phenotype A = (3/4) x 55 = 41.25

χ2 = ∑ (Observed - Expected)2 / Expected

= (18 - 13.75)2 /13.75 + (37 - 41.25)2 /41.25

= (4.25)2 /13.75 + ( - 4.25)2 /41.25

= 18.0625 / 13.75 + 18.0625 / 41.25

= 1.313 + 0.4378

= 1.75

Degree of Freedom (dF) = Number of phenotypes - 1 = 2 - 1 = 1

At p = 0.05 level of significance and dF = 1, the χ2 value is 3.84.

The calculated χ2 value is far lesser than the χ2 value in the table . Thus the difference between the observed and expected number of phenotype is not significant. This means that the assumed mode of inheritance can be accepted satisfactorily.

PS - Punnett Square is a square diagram which is used to predict the genotype of the offspring using the genotype of the parents.

Homozygous means the individual contains alleles of only one type.Heterozygous means the individual contains alleles of both types (dominant and recessive).