Genetics exam 1 ( Chap 5 book
problems) Questions and Answers
100% Pas
A woman has blood-type AM. She has a child with blood-type AB MN. Which of the
following blood types could not be that of the child's father? Explain your reasoning.
George:O N
Tom: AB MN
Bill: B MN
Claude: A NN
Henry: AB M - ✔✔The child's blood type has a B allele and an N allele that could not
have come from the mother and must have come from the father. Therefore, the child's
father must have a B and an N. George, Claude, and Henry are eliminated as possible
fathers because they lack either a B or an N.
Turkeys have black, bronze, or black-bronze plumage. Examine the results of the
following crosses:
Parents
Cross 1: black + bronze = All black
Cross 2: black + black = 3/4 black, 1/4 bronze
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,Cross 3: black-bronze + black-bronze = all black-bronze
Cross 4: black + bronze = 1/2 black, 1/4 bronze, 1/4 black-bronze
Cross 5: bronze + black-bronze = 1/2 bronze, 1/2 black-bronze
Cross 6: bronze + bronze = 3/4 bronze, 1/4 black-bronze
Do you think these differences in plumage arise from incomplete dominance between
two alleles at a single locus? If yes, support your conclusion by assigning symbols to
each alleleand providing genotypes for all turkeys in the crosses. If your answer is no,
provide an alternative explanation and assign genotypes to all turkeys in the crosses. -
✔✔The results of Cross 2 tell us that black is dominant to bronze. Similarly, the results
of Cross 6 tell us that bronze is dominant to black-bronze. We can use BL for black, BR
for bronze, and b for black-bronze.
In the pearl-millet plant, color is determined by three alleles at a single locus: Rp1 (red),
Rp2 (purple), and rp (green). Red is dominant over purple and green, and purple is
dominant over green (Rp1>Rp2>rp). Give the expected phenotypes and ratios of
offspring produced by the following crosses:
a. Rp1/Rp2 × Rp1/rp
b.Rp1/rp × Rp2/rp
c. Rp1/Rp2 × Rp1/Rp2
d. Rp2/rp × rp/rp
e.rp/rp × Rp1/Rp2 - ✔✔a. We expect:
1⁄4 Rp1 /Rp1 (red),
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, 1⁄4 Rp1 /rp (red),
1⁄4 Rp2 /Rp1 (red),
1⁄4 Rp2 /rp (purple),
for overall phenotypic ratio of 3⁄4 red, 1⁄4 purple.
b. 1⁄2 red, 1⁄4 purple, 1⁄4 green.
c.3⁄4 red, 1⁄4 purple.
d.1⁄2 purple, 1⁄2 green.
e.1⁄2 Rp1/rp (red), 1⁄2 Rp2/rp (purple)
The LM and LN alleles at the MN blood group locus exhibit codominance. Give the
expected genotypes and phenotypes and their ratios in progeny resulting from the
following crosses.
a. L(M)L(M) × L(M)L(N)
b. LNLN × LNLN
c. LMLN × LMLN
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problems) Questions and Answers
100% Pas
A woman has blood-type AM. She has a child with blood-type AB MN. Which of the
following blood types could not be that of the child's father? Explain your reasoning.
George:O N
Tom: AB MN
Bill: B MN
Claude: A NN
Henry: AB M - ✔✔The child's blood type has a B allele and an N allele that could not
have come from the mother and must have come from the father. Therefore, the child's
father must have a B and an N. George, Claude, and Henry are eliminated as possible
fathers because they lack either a B or an N.
Turkeys have black, bronze, or black-bronze plumage. Examine the results of the
following crosses:
Parents
Cross 1: black + bronze = All black
Cross 2: black + black = 3/4 black, 1/4 bronze
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,Cross 3: black-bronze + black-bronze = all black-bronze
Cross 4: black + bronze = 1/2 black, 1/4 bronze, 1/4 black-bronze
Cross 5: bronze + black-bronze = 1/2 bronze, 1/2 black-bronze
Cross 6: bronze + bronze = 3/4 bronze, 1/4 black-bronze
Do you think these differences in plumage arise from incomplete dominance between
two alleles at a single locus? If yes, support your conclusion by assigning symbols to
each alleleand providing genotypes for all turkeys in the crosses. If your answer is no,
provide an alternative explanation and assign genotypes to all turkeys in the crosses. -
✔✔The results of Cross 2 tell us that black is dominant to bronze. Similarly, the results
of Cross 6 tell us that bronze is dominant to black-bronze. We can use BL for black, BR
for bronze, and b for black-bronze.
In the pearl-millet plant, color is determined by three alleles at a single locus: Rp1 (red),
Rp2 (purple), and rp (green). Red is dominant over purple and green, and purple is
dominant over green (Rp1>Rp2>rp). Give the expected phenotypes and ratios of
offspring produced by the following crosses:
a. Rp1/Rp2 × Rp1/rp
b.Rp1/rp × Rp2/rp
c. Rp1/Rp2 × Rp1/Rp2
d. Rp2/rp × rp/rp
e.rp/rp × Rp1/Rp2 - ✔✔a. We expect:
1⁄4 Rp1 /Rp1 (red),
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, 1⁄4 Rp1 /rp (red),
1⁄4 Rp2 /Rp1 (red),
1⁄4 Rp2 /rp (purple),
for overall phenotypic ratio of 3⁄4 red, 1⁄4 purple.
b. 1⁄2 red, 1⁄4 purple, 1⁄4 green.
c.3⁄4 red, 1⁄4 purple.
d.1⁄2 purple, 1⁄2 green.
e.1⁄2 Rp1/rp (red), 1⁄2 Rp2/rp (purple)
The LM and LN alleles at the MN blood group locus exhibit codominance. Give the
expected genotypes and phenotypes and their ratios in progeny resulting from the
following crosses.
a. L(M)L(M) × L(M)L(N)
b. LNLN × LNLN
c. LMLN × LMLN
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