Question

Ruby Eyes, Curved Wings, Males	124
Curved Wings, Males	127
Wild Type, Females	825
Wild Type, Males	412
Ruby Eyes, Males	418
Curved Wings, Females	261

Conduct a Chi-Square Test. Please state the mode of inheritance for both the curved wing allele and the ruby eyes allele. Also, would we need to do the test separately for males and separately for females? Thanks :)

Edit: I'm not sure what genetic ratios to use to get the observed values as I do not know what mode of inheritance (ex. autosomal dominant, autosomal recessive, X-linked dominant, X-linked recessive, Y-linked) to assign to the curved wing allele and the ruby eyes allele. I think the ruby eyes allele is X-linked recessive. In class, they mentioned that if a gene is X-linked we have to conduct the chi-squared differently cause the ratio would be different according to males and females. Hence, I was confused about the mode of inheritance and thus I did not know how to begin the chi-squared test. Sorry for the confusion and thank you!

Answer 1

тх"у Now, P- jat e ty ct Try to lxtyct Hypothesi's - Eye oodau im Droso phelta - in X-linked while wing fyße la autosomal amd ruby eyeing and aurred wing X-linked recessire to dominant wild type, recessive to wild type Gt). {curred wing- cc wild fuybe it +/40 Ruby eye x"x", Wild tubea x+x/x** xent yety of C (2) xt yet 1 Purnette square X9+ yre x++ yte xx*cc x + x + %+CC lytyrtt lyryter text ut 1x*x+ 4C རྔབྱས ༡ ༨ х*у се xty to xtycc y + xy + xy te x+y++ Expected ratio wild femalen (6) ! Curved wing (2) - female wild male €3) Curs red wing (1)! Ruby eye (2): maten Curred, ouby (1) Chi square test. Phenotube observed! () Expected (e) ole 825 0,19 Curred 26) 270.875 0.36 would re) 412 406.3125 Curved (ory 127 135. 4375 Ruby con 418 0.34 Curred, ouby (6) 124 135. 4375 0.97 2= 21671 [x² = 2.47 • Asi Calculated chicsquare Nature (x² = 2.47) is tower than tabulan ralue (11.07) for 5 degree of freedom the mull my bortheris. wild (4) S12625 0.08 0:53 406.3125 at poo.o5; so must accept