Review Genetics
Concept Check 14.1
C O N C E P T C H E C K 14.1
1. Pea plants heterozygous for flower position and stem length (AaTt) are allowed to selfpollinate, and 400 of the resulting seeds are planted.
Draw a Punnett square for this cross. How many offspring would be predicted to have terminal flowers and be dwarf? (See Table 14.1.)
1. According to the law of independent assortment, 25 plants (1⁄16 of the offspring) are predicted to be aatt, or recessive for both characters. The actual result is likely to differ slightly from this value.
A–13
2. List all gametes that could be made by a pea plant heterozygous for seed color, seed shape, and pod shape (YyRrIi; see Table 14.1). How large a
Punnett square would you need to draw to predict the offspring of a self-pollination of this “trihybrid”?
2. The plant could make eight different gametes (YRI, YRi, YrI, Yri, yRI, yRi, yrI, and yri). To fit all the possible gametes in a self-pollination, a Punnett square would need 8 rows and 8 columns. It would have spaces for the 64 possible unions of gametes in the offspring.
3. In some pea plant crosses, the plants are self-pollinated. Refer back to Concept 13.1
(pp. 248–249) and explain whether self-pollination is considered asexual or sexual reproduction.
3. Self-pollination is sexual reproduction because meiosis is involved in forming gametes, which unite during fertilization.
As a result, the offspring in self-pollination are genetically different from the parent. (As mentioned in the footnote on p. 263, we have simplified the explanation in referring to the single pea plant as a parent. Technically, the gametophytes in the flower are the two “parents.”)
Concept Check 14.2
1. For any gene with a dominant allele A and recessive allele a, what proportions of the offspring from an
AA Aa cross are expected to be homozygous dominant, homozygous recessive, and heterozygous?
1. 1⁄2 homozygous dominant (AA), 0 homozygous recessive (aa), and 1⁄2 heterozygous
C O N C E P T C H E C K 14.1
1. Pea plants heterozygous for flower position and stem length (AaTt) are allowed to selfpollinate, and 400 of the resulting seeds are planted.
Draw a Punnett square for this cross. How many offspring would be predicted to have terminal flowers and be dwarf? (See Table 14.1.)
1. According to the law of independent assortment, 25 plants (1⁄16 of the offspring) are predicted to be aatt, or recessive for both characters. The actual result is likely to differ slightly from this value.
A–13
2. List all gametes that could be made by a pea plant heterozygous for seed color, seed shape, and pod shape (YyRrIi; see Table 14.1). How large a
Punnett square would you need to draw to predict the offspring of a self-pollination of this “trihybrid”?
2. The plant could make eight different gametes (YRI, YRi, YrI, Yri, yRI, yRi, yrI, and yri). To fit all the possible gametes in a self-pollination, a Punnett square would need 8 rows and 8 columns. It would have spaces for the 64 possible unions of gametes in the offspring.
3. In some pea plant crosses, the plants are self-pollinated. Refer back to Concept 13.1
(pp. 248–249) and explain whether self-pollination is considered asexual or sexual reproduction.
3. Self-pollination is sexual reproduction because meiosis is involved in forming gametes, which unite during fertilization.
As a result, the offspring in self-pollination are genetically different from the parent. (As mentioned in the footnote on p. 263, we have simplified the explanation in referring to the single pea plant as a parent. Technically, the gametophytes in the flower are the two “parents.”)
Concept Check 14.2
1. For any gene with a dominant allele A and recessive allele a, what proportions of the offspring from an
AA Aa cross are expected to be homozygous dominant, homozygous recessive, and heterozygous?
1. 1⁄2 homozygous dominant (AA), 0 homozygous recessive (aa), and 1⁄2 heterozygous