Name Fall 2018 The plant Columbia. The populations are dimorphic fo blotches, an
ID: 150788 • Letter: N
Question
Name Fall 2018 The plant Columbia. The populations are dimorphic fo blotches, and others don't. Near Nanaimo, one plant in nature had blotched leaves. This plant which had not yet flowered, was dug up and taken to a laboratory, Seeds were collected and grown into progeny. One randomly selected (but typical) leaf from each of the progeny is shown in the figure below t blue-eyed Mary grows on Vancouver Island and on the lower mainland of British r purple blotches on the leaves where it was allowed to selfExplanation / Answer
Hi, in the given case some plants have bolches and others don't. The plant which the geneticist has taken to lab has purple bolches. It was allowed to self and resulting progeny characterised. There are in total leaf samples from 107 progenies. Out of which 28 have no bolches. Rest 79 have purple bolches.
Let us say the genotype is caused by one dominant allele P which makes the leave purple bolched. The recessive form of this Gene p gives no purple bolches when present in homozygous form .
So purple leaf can be from two genotypes : PP or Pp
No purple is from : pp genotype.
The plant has given rise to no purple leafed progeny along with purple leafed. So it has both the alleles P and p. Hence the genotype of mother plant is Pp. When this plant was selfed,, Pp x Pp we got followonf combination.
PP, Pp, Pp and pp : the first 3 are purple and last one no purple. These occur in ratio 79:28 = 3:1
It appears like a monohybrid cross.
Let us test it by chi square test.
The chi square formula is (observed -expected )^2 / expected^2
The observed purple = 79 : no purple= 28
The expected is according to monohybrid ratio 3:1 ; 107/4 = 26. So expected no purple = 27 The remainig 107-27=80 are purple.
Chi square = (79-80)^2 /80^2 + (28-27)^2 / 27^2
= 0.000156 + 0.00137
=0.00152
Here the degree of freedom is 1. Because no. Of character is purple and no purple. Look at the chisquare table at degree of freedom 1 and probability of 0.05.
The value from table is 0.004. this is larger than the obtained value of 0.00152. this means the probability to get such numbers is smaller than 0.05. So we can say that there is no difference in the obtained value and expected value. The mode of inheritance in the given plant leaves is monohybrid.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.