Answer:
The correct answer is - 1) rough
2) [(9+14)/1786]* 100 = 1.2877 percentage
Step-by-step explanation:
In this case there is mutated genes are :
mutated : ebony (e), rough (r) and brevis (b)
wild type: e+, r+ and b+
Given: wild type 169 (e+ r+ b+)
ebony 619 (e b+ r+)....... parent
rough, brevis 621 (e+ b r) .... parent
ebony, rough, brevis - 165 (e r b)
rough - 93 (e+ r b+)
ebony, brevis - 97 (e r+ b)
ebony, rough - 9 (e r b+) ....... double cross
brevis - 14 (e+ r+ b) ..... double cross
solution:
In this question, there is the three-point cross is performed, in such case, if one is mutated then the other two are normal. For instance, here ebony is mutated then the rest two are norma - e r+ b+.
We know the most frequent genotypes in such case, are parental and least frequent are double crossover genotypes. We can get the correct order of genes by the double-cross of parental genotypes and if we get double cross over then the gene order is correct.
(e b+ r+) -----------> (e r b+)
(e+ b r) ------------> (e+ r+ b)
double cross double crossover
between
parental gentype
Thus, the correct order is e r b that means rough is between ebony and brevis.
The percentage of double cross over =
(number of first double-cross + second double-cross over/1787)*100
[(9+14)/1786]* 100 = 1.2877 percentage