Question

Answer all my questions please..

9. (Problem 25, part 1) Groodies are useful (but fictional) haploid organisms that are pure genetic tools. A wild-type groody has a fat body, a long tail, and flagella. Mutant lines another (although they are so shy that we do not know how) and produce recombinants. A wiad-type groody mates with a thin-bodied groody lacking both tal and risgelia. Burton S. Guttman.) The genotypes of the groodies in the group of 9 is, and the genotypes of the groodies in the group of 5 is (Use the following symbols: F is fat, f is thin; T is tail, t is tail-less; FI is with flagella, I is no flagella.) The genotypes of the groodies in the group of 9 isand the genotypes of the groodies in the group of 5 is

Answer 1

Few points:

1. The phenotype showed the highest number of the individual is the parental phenotype.
2. The phenotypes showed the lowest numbers of individuals are the double cross over progeny.

9. The genotype of the groodies in the group of 9 is fTfl, and the genotype of the groodies in the group of 5 is FtFl.

10. F T Fl

In the double cross over progenies only the middle changes with respect to the parental genotype. So, by comparing parental genotype (FTFl and ftfl) and double (FtFl and fTfl) cross over progeny we can conclude that the middle gene is T gene.

11. F (9.3 m.u) T (15.3 m.u) Fl

Distance b/w F and T = [(44+35+ 9+5) /(398+370+72+67+44+35+9+5)]*100 = [93/1000]*100 = 9.3 mu

Distance b/w T and Fl = [(67+72+ 9+5) /(398+370+72+67+44+35+9+5)]*100 = [153/1000]*100 = 15.3 mu

12. W (3.5 m.u) S (18.4 m.u) Y

In the double cross over progenies only the middle changes with respect to the parental genotype. So, by comparing parental genotype (wSy and WsY) and double (WSY and wsy) cross over progeny we can conclude that the middle gene is s gene.

Distance b/w S and Y = [(626+601+2+4) /(113+108+2+2708+626+601+4+2538)]*100 = [1233/6700]*100 = 18.4 mu

13. W S Y

14. 0.97

Expected DCO = 0.35*0.184 = 0.0644

Observed DOC = (4+2)/6700 = 0.0009

Coefficient of coincidence = C = Observed DCO / Expected DCO = 0.0009/0.0644 = 0.01

Interference = 1 − coefficient of coincidence (c.o.c.) = 1 – 0.01 = 0.99 ~ 0.97

15. This values can be used to calculate the distance between the two alleles of the gene, which in this case is equal to 8 m.u.

The distance between two genes = cross over frequency*100 = percentage of cross over frequency.

Kindly revert for any queries and concerns.