A fruit fly somatic cell contains 8 chromosomes. This means that ___________ different combinations of chromosomes are possible in its gametes.a. 8b. 16c. 32d. 64

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?

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=noantennae), and nose morphology (S=upturnedsnout,s=downturnedsnout). 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. mitosisb. meiosis I onlyc. meiosis I and IId. 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 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?