A researcher studied five independently assorting genes in a plant. Each gene ha
ID: 204043 • Letter: A
Question
A researcher studied five independently assorting genes in a plant. Each gene has a dominant and a recessive allele: R black stem, r red stem; D tall play, d dwarf plant; C full pods, c constricted pods; O round fruit, o oval fruit; H hairless leaves, h hairy leaves. The three questions below refer to the following cross:
RR Dd cc Oo Hh X Rr DD cc Oo Hh
(a) How many different genotypes are possible among the progeny of this cross? (Please show what you do to find this)
(b) What is the probability of obtaining the Rr Dd cc Oo hh genotype in the progeny?
(c) Of the 64 progeny, how many are expected to be black, tall, constricted, oval, and hairy?
Appreciate the help:)
Explanation / Answer
Answer a.
No of different types of genotypes possible in the below cross:
Cross given is selfing
No of alleles in the plant = 8
That is allelles - R,D,d,c,O,o,H,h
Formula for no of different genotypes in the selfing cross = n(n+1)/2 where n = no of alleles
Here the no. of alleles is eight. Hence no of differenrt genotypes from cross= 8(8+1)/2= 8*9/2= 36
Answer b.
Probability of getting RrDdccOohh
The gametes involved in the cross are:
Total no of progenies formed =8*8= 64
No of RrDdccOohh =2 (do crossing in punnet square method)
Hence probability of RrDdccOohh = 2/64 = 1/32
Answer c.
No of progenies that are black tall constricted oval and hairy
The gametes involved in the cross are:
Do the crossing with punnet square. By douing the punnet square, we will get no of progeny with below phenotype.
No of progeny that are tall, black,constricted, oval and hairy = 14
Note : Thereis no shortcuts available other than punnet square crossing
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