Two genes interact to produce various phenotypic ratios among
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Chapter 4 Solutions
Pearson eText Genetic Analysis: An Integrated Approach -- Instant Access (Pearson+)
- In rice, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of rice plants (i.e. the stamen) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile rice plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male-sterile lines. Give the result(s) of the cross and explain the phenotype of the offspring.arrow_forwardIn corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of the corn plants (i.e the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male sterile lines Using the cardboard chips, simulate the crosses indicated below. Give the genotypes and phenotypes of the offsprings in each cross, and properly label the nucleus and the cytoplasm of each individual in the cross Legend male sterile cytoplasm Male fertile cytoplasm FF nucleus Ff nucleus ff nucleus A. Male sterile female x FF male Explain the phenotype of the offspring B. Male sterile female x Ff male Explain the phenotype of the offspringarrow_forwardIn c. elegans, genetics model organism, movement problems (unc) and small body size (sma) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (unc+ and sma+). A worm homozygous for movement problems and small body is crossed with a worm homozygous for the wild-type traits. The F1 have normal movement and normal body size. The F1 are then crossed with worms that have movement problems and small body size in a testcross. The progeny of this testcross is: Normal movement, normal body size 210 Movement problems, normal body size 9 Normal movement, small body size 11 Movement problems, small body size 193 a)From the test cross results, can you tell if the two genes are on the same chromosome or not? Explain your reasoning. b)What phenotypic proportions would be expected if the genes for round eyes and white body were located on different chromosomes? (please explain hot to get to these conclusions)arrow_forward
- In the model plant Arabidopsis thaliana, the following alleles were used in a cross: T = presence of trichomes t = absence of trichomes D = tall plants d = dwarf plants W = waxy cuticle w = nonwaxy A = presence of purple anthocyanin pigment a = absence (white) The T/t and D/d loci are linked 26 m.u. apart on chromosome 1, whereas the W/w and A/a loci are linked 8 m.u. apart on chromosome 2. A pure-breeding double-homozygous recessive trichomeless nonwaxy plant is crossed with another pure-breeding double-homozygous recessive dwarf white plant. a. What will be the appearance of the F1? b. Sketch the chromosomes 1 and 2 of the parents and the F1, showing the arrangement of the alleles.c. If the F1 is testcrossed, what proportion of the progeny will have all four recessive phenotypes?arrow_forwardIn a wild-type fungus, protein E (encoded by the haplosufficient gene E) normally dimerizes to catalyzes a biochemical reaction necessary for the production of a dark pigment. Ed represents a mutant, dominant negative allele of gene E. What is the predicted phenotype of a fungus cell of genotype E*/Ed, and why? O wild type (normal production of the dark pigment), as E is haplosufficient mutant (no pigment production), as no dimers will form in the heterozygous mutant (no pigment production), as the mutant allele Eg is dominant O wild type (normal production of the dark pigment), as dimers of wild-type and mutant protein E will be formed in the heterozygousarrow_forwardConsider the following crosses in Drosophila. The two traits being investigated involve eye color and the presence or absence of wing crossveins. The outcomes of four crosses are shown below.›arrow_forward
- In corn snakes, the wild-type color is brown. Oneautosomal recessive mutation causes the snake to beorange, and another causes the snake to be black. Anorange snake was crossed to a black one, and the F1offspring were all brown. Assume that all relevantgenes are unlinked.a. Indicate what phenotypes and ratios you wouldexpect in the F2 generation of this cross if there isone pigment pathway, with orange and black beingdifferent intermediates on the way to brown.b. Indicate what phenotypes and ratios you wouldexpect in the F2 generation if orange pigment is aproduct of one pathway, black pigment is the product of another pathway, and brown is the effect ofmixing the two pigments in the skin of the snakearrow_forwardAn ebony strain of flies was discovered to be sensitive to carbon dioxide. Crossing a female sensitive strain with male resistant strain gave all sensitive offspring. The offspring of an F1 female crossed with a resistant male were all sensitive. Using the following key to your illustrations using shapes, make a reciprocal cross up to the F2 generation. Put your illustrations in the space provided below. Label the phenotypes of all individuals in the reciprocal cross. Adjust spacing, if necessary. Make sure that the complete cross(es) can fit the same page. Big blue circle - male cytoplasm Big pink circle - female cytoplasm Small half-blue-half-pink circle - F1 nucleus Small blue circle - male nucleus Small pink circle - female nucleusarrow_forwardAn ebony strain of flies was discovered to be sensitive to carbon dioxide. Crossing a female sensitive strain with male resistant strain gave all sensitive offspring. The offspring of an F1 female crossed with a resistant male were all sensitive. Using the following key to your illustrations using shapes, make a reciprocal cross up to the F2 generation. Put your illustrations in the space provided below. Label the phenotypes of all individuals in the reciprocal cross. Adjust spacing, if necessary. Make sure that the complete cross(es) can fit the same page. Big blue circle - male cytoplasm Big pink circle - female cytoplasm Small half-blue-half-pink circle - F1 nucleus Small blue circle - male nucleus Small pink circle - female nucleus give a diagram pleasearrow_forward
- Two plants with white flowers, each from true-breeding strains, were crossed. All the F1 plants had red flowers. When these F1 plants were intercrossed, they produced an F2 consisting of 177 plants with red flowers and 142 with white flowers. (a) Propose an explanation for the inheritance of flower color in this plant species. (b) Propose a biochemical pathway for flower pigmentation and indicate which genes control which steps in this pathway.arrow_forwardIn corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of corn plants (i.e. the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male-fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male-sterile lines. Using the following color-coded circles, simulate the crosses indicated below. Put the illustrations of crosses in the spaces provided. Be sure to include in the labels the genotypes and phenotypes of the offspring in each cross. Big light green circle - male-sterile cytoplasm Big orange circle - male-fertile cytoplasm Small orange circle - FF nucleus Small half-light green-half-orange circle - Ff nucleus Small light-green circle - ff nucleusarrow_forwardConsider the first category of test-cross offspring shown in figure 8.2 (+b, LS). Consider also that the parents of the heterozygous female flies in the test cross had the following genotypes: bb, SS, and +, LL. A. What would be the physical phenotype of these flies? B. If PC was conducted with the DNA of one of these flies using the primers for the molecular marker, what would be the appearance of the bands on an electrophoresis gel with the PC products? C. If the gene for black body and the locus for the molecular marker (L long or S short) were unlinked, what proportion of the test-cross progeny would be black flies that are heterozygous for the molecular marker? What proportion would be flies with normal body color, which are homozygous for one form of the molecular marker? D. If the gene for black body and the locus for the molecular marker were linked, how would the proportion of flies be different?arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning