asked 98.4k views
1 vote
All 113 baby mice in an experiment are Rr, the parents are probably (do a Punnett square to be sure)

a

RR x RR

b

RR x rr

c

Rr x Rr

asked
User InnaM
by
8.1k points

2 Answers

1 vote

Answer:

B

Step-by-step explanation:

answered
User Johan Davidsson
by
7.3k points
3 votes

Answer:

The correct answer is option B.

Step-by-step explanation:

If am individual cross between two true breed mice the probability of offspring with heterozygous condition is 100 percent in such cross. True breed is an individual organism with homozygous for the particular character either dominant or recessive.

So, if mice with RR (purebred or homozygous dominant) bred with mice with rr (homozygous recessive), all the offspring will be heterozygous in genotype which is Rr.

Thus, the correct answer is : option B. (Punnet square is attached)

All 113 baby mice in an experiment are Rr, the parents are probably (do a Punnett-example-1
answered
User Gavin Mannion
by
8.6k points
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