What is the probability that each of the following pairs of parents will produce the indicated offspring? (assume independent assortment of all gene pairs.)
(a) aabbcc × aabbcc-> aabbcc
(b) aabbcc × aabbcc-> aabbcc
(c) aabbcc × aabbcc-> aabbcc
(d) aabbcc × aabbcc-> aabbcc

a. The probability for a cross of AABBCC × aabbcc to produce AaBbCc is: 1.

b. The probability for a cross of AABbCc × AaBbCc to produce AAbbCC is:

c. The probability for a cross of AaBbCc × AaBbCc to produce AaBbCc is:

d. The probability for a cross of aaBbCC × AABbcc to produce AaBbCc is:

Recall:

Thus, the images below shows the various outcome of each cross.

a. Figure A shows the cross between:

AABBCC × aabbcc

The genotype of all offspring produced are only AaBbCc.

Therefore, the probability for a cross of AABBCC × aabbcc to produce AaBbCc is: 1.

b. Figure B shows the cross between:

AABbCc × AaBbCc

From the cross, there are only 2 AAbbCC out of 64 offspring produced.

Therefore, the probability for a cross of AABbCc × AaBbCc to produce AAbbCC is:

c. Figure C shows the cross between:

AaBbCc × AaBbCc

From the cross, there are only 8 AaBbCc out of 64 offspring produced.

Therefore, the probability for a cross of AaBbCc × AaBbCc to produce AaBbCc is:

d. Figure D shows the cross between:

aaBbCC × AABbcc

From the cross, there are only 32 AaBbCc out of 64 offspring produced.

Therefore, the probability for a cross of aaBbCC × AABbcc to produce AaBbCc is:

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one centimetre is equal to 0.01 metres

answer: that would be cheating

explanation: