12. Meiosis

What is an outcome of genetic recombination?

a. The synapsing of homologs during prophase of meiosis I

b. The new combination of maternal and paternal chromosome segments that results when homologs cross over

c. The new combinations of chromosome segments that result when self-fertilization occurs

d. The combination of a haploid phase and a diploid phase in a life cycle

A wild-type fruit fly (heterozygous for gray body color and normal wings) is mated with a black fly with vestigial wings. The offspring have the following phenotypic distribution: wild-type, 778; black vestigial, 785; black normal, 158; gray vestigial, 162. What is the recombination frequency between these genes for body color and wing size? Is this consistent with the results of the experiment in Figure 15.9?

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A planet is inhabited by creatures that reproduce with the same hereditary patterns seen in humans. Three phenotypic characters are height (T=tall,t=dwart), head appendages (A=antennae,a=no antennae), and nose morphology (S=upturned snout,s=downturned snout). Since the creatures are not 'intelligent,' Earth scientists are able to do some controlled breeding experiments using various heterozygotes in testcrosses. For tall heterozygotes with antennae, the offspring are tall antennae, 46; dwarf antennae, 7; dwarf no antennae, 42; tall no antennae, 5. For heterozygotes with antennae and an upturned snout, the offspring are antennae upturned snout, 47; antennae downturned snout, 2; no antennae downturned snout, 48; no antennae upturned snout, 3. Calculate the recombination frequencies for both experiments.

Nondisjunction that leads to problems in offspring can occur in:

a. Mitosis

b. Meiosis I only

c. Meiosis I and II

d. Mitosis, meiosis I, and meiosis II

Using the information from problem 4, scientists do a further testcross using a heterozygote for height and nose morphology. The offspring are tall upturned snout, 40; dwarf upturned snout, 9; dwarf downturned snout, 42; tall downturned snout, 9. Calculate the recombination frequency from these data, and then use your answer from problem 4 to determine the correct order of the three linked genes.

A fruit fly somatic cell contains 8 chromosomes. This means that ___________ different combinations of chromosomes are possible in its gametes.

a. 8

b. 16

c. 32

d. 64

A wild-type fruit fly (heterozygous for gray body color and red eyes) is mated with a black fruit fly with purple eyes. The offspring are wild-type, 721; black purple, 751; gray purple, 49; black red, 45. What is the recombination frequency between these genes for body color and eye color? Using information from problem 3, what fruit flies (genotypes and phenotypes) would you mate to determine the order of the body color, wing size, and eye color genes on the chromosome?

What is the physical basis for the independent assortment of alleles into offspring?

a. There are chromosome divisions during gamete production.

b. Homologous chromosome pairs are separated during gamete production.

c. Sperm and eggs are produced by different sexes.

d. Each gene codes for more than one protein.

e. The instruction manual for producing a human is incomplete.

Assume that genes A and B are on the same chromosome and are 50 map units apart. An animal heterozygous at both loci is crossed with one that is homozygous recessive at both loci. What percentage of the offspring will show recombinant phenotypes resulting from crossovers? Without knowing these genes are on the same chromosome, how would you interpret the results of this cross?

Two genes of a flower, one controlling blue (B) versus white (b) petals and the other controlling round (R) versus oval (r) stamens, are linked and are 10 map units apart. You cross a homozygous blue oval plant with a homozygous white round plant. The resulting F1 progeny are crossed with homozygous white oval plants, and 1,000 offspring plants are obtained. How many plants of each of the four phenotypes do you expect?

Triploid (3n) watermelons, which are seedless, are produced by crossing a tetraploid (4n) strain with a diploid (2n) plant. Explain why this mating produces a triploid individual.

You design Drosophila crosses to provide recombination data for gene a, which is located on the chromosome shown in Figure 15.12. Gene a has recombination frequencies of 14% with the vestigial wing locus and 26% with the brown eye locus. Approximately where is a located along the chromosome?

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Meiosis results in independent assortment of maternal and paternal chromosomes. If 2n=6 for a given organism, and there is no crossing over, what is the chance that a gamete produced by this diploid organism will receive only paternal chromosomes?

a. 0

b. 1/16

c. 1/8

d. 1/3

Why are individuals with an extra chromosome 21, which causes Down syndrome, more numerous than individuals with an extra chromosome 3 or chromosome 16?

a. There are probably more genes on chromosome 21 than on the others.

b. Chromosome 21 is a sex chromosome, and chromosomes 3 and 16 are not.

c. Down syndrome is not more common, just more serious.

d. Extra copies of the other chromosomes are probably fatal.