Homework Answers
A.We need to describe crosses for Mendelian trait- crab color
1. blue crab and red crab resulting in F1 (red crab).
2. F1 cross with blue crab to give 70 red and 22 blue crabs
B.Perform Chi square test to check prediction
A. When blue crabs were crossed with red crabs, all individuals of F1 were red. So red crab color is a dominant trait. Blue crab color was not observed in F1 generation, so this is a recessive allele.
Let us assign R as dominant allele for red crab color and r recessive allele for blue crab color.
Then genotype can be written as: RR (homozygous dominant) - red crab, rr (homozygous recessive) - blue crab and Rr (heterozygote)-blue crab
A1. Since one of the parent is blue, it must be having both recessive alleles (rr). Red crab can either be heterozygote (Rr) or homozygous recessive (rr).
If red crab is homozygous dominant (RR) it will produce only one type of gamete R, and all individuals of F1 will be heterozygous (blue color). As seen from Punnet square below
If red crab is heterozygous for trait (Rr) it will produce gametes R and r and all individuals of F1 will be either heterozygous (Rr) or recessive (rr). Thus F1 ratio would have been 1:1 for Blue crab (rr). As given in problem, all crabs from F1 are blue.
So, the Parents were homozygous for crab color: Red crab (RR) , Blue crab (rr), to generate Rr (Red heterozygote)
A2. Crossing of F1 (Rr) to blue crab (rr)is a test cross, between F1 and recessive parents.
Red crab is heterozygous for trait (Rr) it will produce gametes R and r. Thus individuals will be either heterozygous (Rr) or recessive (rr). Thus phenotypic ratio will be 1:1 for Blue crab (rr): Red crab (Rr). As seen from Punnet square bellow
B. To check this hypothesis with Chi square test:
Expected phenotype values are 0.5 for red and 0.5 for blue
Observed phenotype values are 70 red + 22 blue = 92 Total individuals
Proportion of red crab color = 70 / 92 =0.76
Proportion of blue crab color = 22 / 92 =0.24
Null hypothesis will be: There is no difference in the expected values based on prediction and observed values
p value of 0.05 corresponds to chi square of 3.841 for degree of freedom =1.
Our value 0.2704 is less than 3.841. so we will accept the null hypothesis, that observed results are expected and predicted by us.
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