Question

re ТВ Tb тв TTBB ТТвь TIBB TIeb е тТВЫ Тbb тво е тівв TIBb IIBB tb TIBD tBb при і митні

From this Punnette square, what proportion of the offspring have a phenotype that is DIFFERENT from the phenotype of either parent?
a. 3/16
b. 6/16
c. 7/16
d. 9/16
e. 1/16

Answer 1

Answer) C 7/16

in the above question, they have made a dihybrid cross

T is one character and B is the other character

T and B are dominants while t and b are recessives.

now as per the cross that is given it shows that ma dihybrid is done with TtBb X TtBb

which will show the following combinations which are shown above: TB Tb tB tb X TB Tb tB tb

hence, we have to find out the number offsprings which have a different phenotype than that of the parent.

considering the genotypic consideration of the parents given above, we know that the dominant pair expresses itself inspite of the recessive being present, hence the orgnisms in which recessives express itself will actually be the ones who will have a different phenotype than that of the parent

those are

TTbb

Ttbb

ttBB

ttBb

Ttbb

ttBb

ttbb

therefore they are 7/16 offsprings, hence the answer is 7/16 !

i hope i was able to help you with the answer ! :)