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

Ignoring crossover events, how many kinds of gametes can be produced by an organism with a diploid number of 8?

What is an outcome of genetic recombination?a. the synapsing of homologs during prophase of meiosis Ib. the new combination of maternal and paternal chromosome segments that results when homologs cross overc. the new combinations of chromosome segments that result when self-fertilization occursd. 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?

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.

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. 8b. 16c. 32d. 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.