3) (4 pts) Assume that investigators crossed a strain of fies Garnescodh vere th

Question

3) (4 pts) Assume that investigators crossed a strain of fies Garnescodh vere then Lobe on the second chromosome with a strain homozygous fo obe females recessive mutations smooth abdomen and straw body. The F1 0es and the fll backcrossed with homozy gous smooth abdomen, straw body male phenotypes were observed females were then the following ales, an smooth abdomen, straw body 820 780 42 58 148 152 Lobe smooth abdomen, Lobe straw body smooth abdomen Lobe, straw body normal, normal, normal Lobe, smooth abdomen, straw body5 2000 TOTAL (a) (2 pts) Give the gene order and map units between these three loci. (b) (2 pts) What is the coefficient of coincidence? What is the interference value?

Explanation / Answer

smooth abdomen straw body normal eye= 820 (parental)

normal abdomen, normal body, lobe eye= =780 (parental)

smoooth abdomen, lobe eye, normal body= 42 (single cross over between smooth abdomen and lobe eye)

normal abdomen, straw body, normal eye= 58 (single cross over between smooth abdomen and lobe eye)

smooth abdomen, normal body, normal eye= 148 (single acroos over between lobe eye and straw body)

normal abdomen, straw body, lobe eye= 152  (single acroos over between lobe eye and straw body)

normal, normal, normal= 5 (double cross over)

lobe, smooth abdomen, straw body= 5 (double cross over)

total =2000

a) give the gene order and map units between these three loci

Answer: To determine gene order which gene is changing position with respect to genotype of parents and double cross over genotype. According to this lobe eye position is changed so lobe eye gene will be in middle

smooth abdomen---------normal eye---------straw body

normal abdomen---------lobe eye-------------normal body

Distance between smooth abdomen and normal eye= (42+58+5+5)2000* 100= 0.055% or 5.5 cM

Distance between normal eye and straw body= (148+152+5+5)/2000 *100= 0.155% or 15.5 cM

b) cofficient of coincidence (coc)= observed number of double recombinants/ expected number of double recombinants

observed number of recombinants= 5+5=10

expected number of double recombinants= 0.055 X 0.155 X 2000 (total progeny)= 17

coc= 10/17= 0.58 or 58%

interference= 1-coc= 1-0.58= 0.42 or 42%

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