Single Gene Inheritance

Topics: Allele, Phenylketonuria, Meiosis Pages: 62 (12353 words) Published: September 29, 2013
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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 (as in corn plants) that A encodes an allele that produces starch in pollen and allele a does not. Iodine solution stains starch black. How would you demonstrate Mendel’s first law directly with such a system? Answer: One would use this iodine dye to color the starch producing corn pollen. Since pollen is a plant gametophyte generation (haploid) it will be produced by meiosis. Mendel’s first law predicts segregation of alleles into gametes, therefore we would expect 1 : 1 ratio of starch producing (A) versus non-starch producing (a) pollen grains, from a heterozygous (A/a) parent/male flower. It would be easy to color the pollen and count the observed ratio.

7.

In the text figure on page 43, assume the left-hand individual is selfed. What pattern of radioactive bands would you see in a Southern analysis of the progeny?
Answer: If an individual is selfed, the restriction fragments should be identical to the parents fragments. In this case, a heterozygous parent to the left had three bands (two from a mutant allele “a” and one from dominant allele “A”).

8.

Considering Figure 2-15, if you had a homozygous double mutant m3/m3 m5/m5, would you expect it to be mutant in phenotype? (Note: This line would have two mutant sites in the same coding sequence.)

Answer: Yes, this double mutant m3/m3 and m5/m5 would be a null mutation, because m3 mutation changes the exon sequence.

9.

In which of the stages of the Drosophila life cycle (represented in the box on page 52) does meiosis take place?
Answer: Meiosis happens in adult ovaries and testes, therefore before fertilization. After fertilization, fruit flies would lay their eggs (with now diploid embryos). That would be Stage 1 on the figure.

8 Chapter Two
10.

If you assume Figure 2-17 also applies to mice and you irradiate male sperm with X rays (known to inactivate genes), what phenotype would you look for in progeny in order to find cases of individuals with an inactivated SRY gene? Answer: If we inactivate the SRY...
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