In: Biology
A botanist who loves the smell of the plant Titan arum sets up the following cross. They mate a wild type pure-breeding Titan arum trisomic for chromosome 5 to a normal diploid Titan arum that is homozygous for a recessive mutation (s) that makes it smell even more aromatic all the time (the super smelly mutant). A trisomic F1 plant is then back-crossed to the super-smelly parent. From this cross:
A] What is the ratio of wild type Titan arum to those that are super smelly when you assume that s is situated on chromosome 5?
B] What is the ratio of wild type Titan arum to those that are super smelly when you assume that s is not situated on chromosome 5?
Parents- phenotype trisomy wild x super smelly
genotype- SSS ss
Gametes - SS , S s
cross-
SS | S | |
s | SSs - trisomy | Ss - normal diploid |
F2 cross,
Parents - phenotype trisomy x super smelly
genotype SSs ss
Gametes SS, Ss, S, s s
cross,
SS | Ss | S | s | |
s | SSs | Sss | Ss | ss |
a) super smelly condition , will be expressed only in ss condition.
if the s is present on chromosome number 5,
then, obtained progenies will be - SSs- trisomy , Sss- super smelly(as s is in 5th chromsome, ss will be get expressed), Ss- normal diploid condition, ss - super smelly.
number of wild type arum in the progeny = 1
number of super smelly arum in the progeny = 2
total number of progeny obtained = 4
ratio of wild type arum to super smelly = 1/2 = 0.5
b) if the s is not present on chromosome number 5 , then ss in he trisomy condition won't be expressed,
then obtained progenies will be- SSs- trisomy, Sss- trisomy, Ss- normal diploid, ss- super smelly
number of wild type arum in the progeny = 2
number of super smelly arum in the progeny = 1
ratio of wild type arum to super smelly arum = 2/1 = 2