please check the work that is done

part check if hypotheis is rejected or failed to be rejected by the numbers given:

1)608.63

2)7.35

3)6.44

4)0.707

Can't reject null hypothesis | Reject null hypothesis p values - 0.99 0.90 0.50 0.10 0.05 0.01 0.001 trees of eedom 0.02 0.45 2.71 3.84 6.64 10.83 0.02 0.21 1.39 4.61 5.99 9.21 13.82 UAWNE 0.11 0.58 2.37 6.25 7.81 11.35 16.27 0.30 1.06 3.36 7.78 9.49 13.28 18.47 0.55 1.61 4.35 9.24 11.07 15.09 20.523 F1 offspring observed stst*t 95 sstt 100 s*stt 76 sst*t 72 Parentals : (95+ 100)= 195 1 10.5) ( 343) = 1715 Recombinants ( 76 + 72 )- 148, (0.5) (343 = 171.5 Total = 343 X2_ ( 195- 171.5) 2 + (148 - 171.5) 2 171.5 = 6.44 1715 df - 2-1 = 1 p-value cc 0.0Of wpothesis that these genes are Rest Re/4. Parents: RaBb x AABB wulinked and segregationg F1 offspring observed AABB 87 randonly . AaBb 86 AABb 95 AaBB 94 Parentals: ( 87 + 86) = 173, (0.5) ( 362 ) = 189 Recombinants ( 95 + 94) = 189, (0.5) ( 362) = 181 Total = 362 X 2 17 3 - 181 ) 2 + 189 - 181 /2 = 0.707 181 df = 2-1 = 1 p -value > >0.009 Must support hypothesis that thes genes are unlinked and segregating randomly.Homework 2_6 Test if the loci observed in the following crosses are unlinked. Be sure to indicate the parental types and recombinant types. 1. Parents: CcDd x codd F1 offspring observed CcDd 165 codd 168 Codd 63 ccDd 69 Parentals : (165 + 168) = 333 , (0.5) ( 465) = 232.5 Recombinants: ( 6 3 + 69) = 132, (0.5 ) 456) = 232.5 Total= 465 x2- (333 - 232.5) 2 + 132 - 232.5 ) 2- 608.63 2 32. 5 232.5 df = 2- 1=1 , p- value c