Single Gene Inheritance
2
Single Gene Inheritance
WORKING WITH THE FIGURES
(The first 14 questions require inspection of text figures.)
1.
In the left-hand part of Figure 2-4, the red arrows show selfing as pollination within single flowers of one F1 plant. Would the same F2 results be produced by cross-pollinating two different F1 plants?
Answer: No, the results would be different. While self pollination produces 3 : 1 ratio of yellow versus gene phenotype, cross pollination would result in 1 : 1 ratio, in the F2. This is because F1 yellow are heterozygous, while green are homozygous genotypes.
2.
In the right-hand part of Figure 2-4, in the plant showing an 11 : 11 ratio, do you think it would be possible to find a pod with all yellow peas? All green?
Explain.
Answer: Yes, it is possible to find a pod with only yellow peas or heterozygous for the seed color gene, if all the flowers had dominant allele in a given fruit/pod. This could be also one example of rare changes at a physiological level. 3.
In Table 2-1, state the recessive phenotype in each of the seven cases.
Answer: wrinkled seeds; green seeds; white petals; pinched pods; yellow pods; terminal flowers; short stems
4.
Considering Figure 2-8, is the sequence “pairing → replication → segregation
→ segregation” a good shorthand description of meiosis?
Answer: No, it should say either: “pairing, recombination, segregation, segregation” or: “replication, pairing, segregation, segregation.”
Chapter Two 7
5.
Point to all cases of bivalents, dyads, and tetrads in Figure 2-11.
Answer: Replicate sister chromosomes or dyads are at any chromatid after the replication (S phase). A pair of synapsed dyads is called a bivalent and it would represent two dyads together (sister chromatids on the right), while the four chromatids that make up a bivalent are called a tetrad and they would be the entire square (with same or different alleles on the bivalents).
6.
In Figure 2-12, assume
Single Gene Inheritance
WORKING WITH THE FIGURES
(The first 14 questions require inspection of text figures.)
1.
In the left-hand part of Figure 2-4, the red arrows show selfing as pollination within single flowers of one F1 plant. Would the same F2 results be produced by cross-pollinating two different F1 plants?
Answer: No, the results would be different. While self pollination produces 3 : 1 ratio of yellow versus gene phenotype, cross pollination would result in 1 : 1 ratio, in the F2. This is because F1 yellow are heterozygous, while green are homozygous genotypes.
2.
In the right-hand part of Figure 2-4, in the plant showing an 11 : 11 ratio, do you think it would be possible to find a pod with all yellow peas? All green?
Explain.
Answer: Yes, it is possible to find a pod with only yellow peas or heterozygous for the seed color gene, if all the flowers had dominant allele in a given fruit/pod. This could be also one example of rare changes at a physiological level. 3.
In Table 2-1, state the recessive phenotype in each of the seven cases.
Answer: wrinkled seeds; green seeds; white petals; pinched pods; yellow pods; terminal flowers; short stems
4.
Considering Figure 2-8, is the sequence “pairing → replication → segregation
→ segregation” a good shorthand description of meiosis?
Answer: No, it should say either: “pairing, recombination, segregation, segregation” or: “replication, pairing, segregation, segregation.”
Chapter Two 7
5.
Point to all cases of bivalents, dyads, and tetrads in Figure 2-11.
Answer: Replicate sister chromosomes or dyads are at any chromatid after the replication (S phase). A pair of synapsed dyads is called a bivalent and it would represent two dyads together (sister chromatids on the right), while the four chromatids that make up a bivalent are called a tetrad and they would be the entire square (with same or different alleles on the bivalents).
6.
In Figure 2-12, assume