Question

either blue shels and long antenna, or green shells and short antenna. Crossing these lines yields progeny with blue shells and long antenna. Crossing F, progeny with beetles that have green shell and short antenna yield the following progeny: 44. You are doing ab work with a new species of beetle. You have isolated lines that breed true for Blue shell, long antenna 82 Green shell, short antenna78 Blue shell,short antenna 37 Green shell, long antenna 43 Total 240 You want to determine whether the genes for shell color and antenna size are linked. a. What null hypothesis should you test? b. What are the observed and expected values for the parental classes? at are the observed and expected values for the recombinant classes? d. Calculate the chi-squared value. What p value do you see? t or reject the null hypothesis? Why or w the er why not? Are the genes linked e. Do you accept

Answer 1

a) Null hypothesis: Genes are present in different chromosomes

alternative hypothesis: Genes are linked

b) let the alleles be B for Blue shell and b for green shell and L for long antenna and l for short antenna and B is dominant over b and L is dominant over l

then /

BbLl * BbLl

	BL	Bl	bL	bl
BL	BBLL (blue, long)	BBLl (blue, long)	BbLL (blue, long)	BbLl (blue, long)
Bl	BBLl (blue, long)	BBll (blue, short)	BbLl (blue, long)	Bbll (blue, short)
bL	BbLL (blue, long)	BbLl (blue, long)	bbLL (green, tall)	bbLl (green, tall)
bl	BbLl (blue, long)	Bbll (blue, short)	bbLl (green, tall)	bbll (green, short)

so here Blue shell and long antenna, and green shell and short antenna are parental phenotypes

so expect Blue, long be  9/16 and observed is 82/240

for green, short, expected is 1/16 and observed is 78/240

c) recombinant classes are a Blue shell and short antenna

and green shell and long antenna

so for Blue, short expected is 3/16 observed is 37

for green, long expected is 3/16 and observed is 43

d) let`s take the ratio of observed

Blue, long: Blue short;Green long:Green short=82/37:37/37:43/37:78/37=2.216:1:1.16:2.108

phenotype	Observed (O)	Expected (E)	(O-E)^2/E
Blue, long	2.216	9	5.11
Blue, short	1	3	1.33
green, long	1.16	3	1.128
green, short	2.108	1	1.227

chi-square value= 5.11+1.33+1.128+1.227=8.947

For df=3 ( df,degrees of freedom= number of phenotypes-1=4-1=3)

and P<0.05, the value from the table is 7.815

e) value we got is greater than the value from the table so we reject null hypothesis so these genes are linked in a chromosome.