Answer:
(a) AABBCC × aabbcc->AaBbCc   → 1
(b) AABbCc × AaBbCc->AAbbCC  → 1/32
(c) AaBbCc × AaBbCc->AaBbCc   → 1/8
(d) aaBbCC × AABbcc->AaBbCc   → 1/2
Explanation:
In such cases we calculate the probability of each allelic pair separately.
(a) AABBCC × aabbcc -> AaBbCc Â
Aa  Aa  Aa  Aa      Bb  Bb  Bb  Bb     Cc  Cc  Cc  Cc
↓                  ↓                 ↓ Â
4/4 = 1 Â Â Â Â Â Â Â Â Â Â Â Â Â 4/4 = 1 Â Â Â Â Â Â Â Â Â Â Â 4/4 = 1 Â Â
In case of gene A, out of the 4 probable allelic combinations, 4 will be Aa so the probability is 4/4 = 1.
In case of gene B, out of the 4 probable allelic combinations, 4 will be Bb so the probability is 4/4 = 1.
In case of gene C, out of the 4 probable allelic combinations, 4 will be Cc so the probability is 4/4 = 1.
So, the total probability of getting AaBbCc  = 1 x 1 x 1 = 1.        Â
(b) AABbCc × AaBbCc -> AA bb CC
AA  AA  Aa  Aa      BB  Bb  Bb  bb     CC  Cc  Cc  cc
↓                             ↓      ↓ Â
2/4 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1/4 Â Â Â Â Â 1/4
In case of gene A, out of the 4 probable allelic combinations, 2 will be AA so the probability is 2/4.
In case of gene B, out of the 4 probable allelic combinations, 1 will be bb so the probability is 1/4.
In case of gene C, out of the 4 probable allelic combinations, 1 will be CC so the probability is 1/4.
So, the total probability of getting AA bb CC = 2/4 x 1/4 x 1/4 = 1/32.
(c) AaBbCc × AaBbCc -> AaBbCc
AA  Aa  Aa  aa      BB  Bb  Bb  bb     CC  Cc  Cc  cc
    ↓                  ↓                ↓ Â
    2/4                2/4              2/4
In case of gene A, out of the 4 probable allelic combinations, 2 will be Aa so the probability is 2/4.
In case of gene B, out of the 4 probable allelic combinations, 2 will be Bb so the probability is 2/4.
In case of gene C, out of the 4 probable allelic combinations, 2 will be Cc so the probability is 2/4.
So, the total probability of getting AaBbCc = 2/4 x 2/4 x 2/4 = 1/8.
(d) aaBbCC × AABbcc->AaBbCc
Aa  Aa  Aa  Aa        BB  Bb  Bb  bb      Cc  Cc  Cc  Cc
   ↓                     ↓                 ↓ Â
4/4 = 1 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2/4 Â Â Â Â Â Â Â Â Â Â Â Â Â 4/4 = 1 Â Â
In case of gene A, out of the 4 probable allelic combinations, 4 will be Aa so the probability is 4/4 = 1.
In case of gene B, out of the 4 probable allelic combinations, 2 will be Bb so the probability is 2/4.
In case of gene C, out of the 4 probable allelic combinations, 4 will be Cc so the probability is 4/4 = 1.
So, the total probability of getting AaBbCc = 1 x 2/4 x 1 = 1/2.
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: [tex]\frac{1}{32}[/tex]
c. The probability for a cross of AaBbCc × AaBbCc to produce AaBbCc is: [tex]\frac{1}{8}[/tex]
d. The probability for a cross of aaBbCC × AABbcc to produce AaBbCc is: [tex]\frac{1}{2}[/tex]
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: [tex]\frac{1}{32}[/tex]
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: [tex]\frac{1}{8}[/tex]
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: [tex]\frac{1}{2}[/tex]
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