liDIe following crosses are made, with the results shown: (S) to a certain patho
ID: 148976 • Letter: L
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liDIe following crosses are made, with the results shown: (S) to a certain pathogenic fungus. th S 9 x R 6 all progeny S R 9 x S all progeny R What can you conclude about the location of the genetic determinants of R and S? 36. A presum ed dihybrid in Drosophila, B/b; Flf, is test- crossed with b/b; flf. (B- black body;b brown body; F = forked bristles,f= unforked bristles.) The results are black, forked 230brown, forked 240 black, unforked 210 brown, unforked 250 Use the 2 test to determine if these results fit the results expected from testcrossing the hypothesized dihybrid. 37. Are the following progeny numbers consistent with the results expected from selfing a plant presumed to be a dihybrid of two independently assorting genes, H/h; R/r? (H-hairy leaves; h smooth leaves; R round ovary; r- elongated ovary.) Explain your answer hairy, round 178 smooth, round56 hairy, elongated 62 smooth, elongated 24 38. A dark female moth is crossed with a dark male. All the male progeny are dark, but half the female progeny are light and the rest are dark. Propose an ex this pattern of inheritance. 39. In Neurospora, a mutant strain called stopper (stp) arose spontaneously. Stopper showed erratic "stop and start growth, compared with the uninterrupted growth of wild-type strains. In crosses, the following results wereExplanation / Answer
37. A dihybrid individual is heterozygous at two different loci. Therefore the dihybrid plant should be HhRr.
Selfing : HhRr x HhRr
Gametes : HR , Hr , hR , hr
Expected phenotypic ratio is : 9 hairy round : 3 hairy elongated: 3 smooth round : 1 smooth elongated
Total number of progeny observed = 178 + 56 + 62 + 24 = 320
We expect 9 out of 16 progeny to be hairy round ,3/16 to be hairy elongated, 3/16 to be smooth round and 1/16 to be smooth elongated.
We will perform chi-square analysis to see if the observed progeny is consistent with expected progeny. Before carrying out the chi-square analysis we have to put forward a null hypothesis as below.
Null hypothesis: There is no significant difference between the observed and expected progeny and any difference is due to sampling or experimental error
If p value calculated via chi-square analysis is greater than 0.5 (5%) we will accept our null hypothesis and if p value is less than 0.5 (5%) we will reject the null hypothesis.
Expected number of hairy round in the given progeny is 9/16 x 320 = 180
Expected number of hairy elongated in the given progeny is 3/16 x 320 = 60
Expected number of smooth round in the given progeny is 3/16 x 320 =60
Expected number of smooth elongated in the given progeny is 1/16 x 320 = 20
Degree of freedom is total number of phenotypic classes minus 1 . Here are four phenotypic classes : Hairy round , hairy elongated, smooth round, smooth elongated. Therefore degree of freedom is 3.
Look into the chi-square table for p values under the df =3 row.
Our chi-square value lie between 0.8 and 0.7. The average value comes out to be 0.75 i.e 75% . There is 75% probability that the observed progeny is following the mendelian 9;3:3:1 ratio typical of a dihybrid cross and any variation is due to chance alone. This p value of 75% is above the threshold value of 5% therefore we accept our null hypothesis that there is no difference between the observed and expected frequency of progeny and any difference observed is due to sampling or experimental error
HR Hr hR hr HR HHRR (Hairy, round) HHRr (Hairy, round) HhRR (Hairy, round) HhRr (Hairy, round) Hr HHRr (Hairy, round) HHrr ( Hairy, elongated) HhRr (Hairy, round) Hhrr (Hairy, elongated) hR HhRR (Hairy, round) HhRr (Hairy, round) hhRR (smooth, round) hhRr (smooth, round) hr HhRr (Hairy, round) Hhrr (Hairy, elongated) hhRr (smooth, round) hhrr (smooth, elongated)Related Questions
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