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Colored aleurone in the kernels of com is due to the dominant allele R. The recessive allele r, when homozygous, produces colorless aleurone. The plant color (not the kernel color) is controlled by another gene with two alleles, Y and y. The dominant Y allele results in green color, whereas the homozygous presence of the recessive y allele causes the plant to appear yellow. In a testcross between a plant of unknown genotype and
Explain how these results were obtained by determining the exact genotype and phenotype of the unknown plant, including the precise arrangement of the alleles on the homologs.
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- In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct+ is crossed to a sn+ct male. The F1 flies are interbred. The F2 males are distributed as follows: genotype number sn ct 15 sn ct+ 34 sn+ ct 33 sn+ct+ 18 What is the map distance between sn and ct?arrow_forwardIn com, colored aleurone (in the kernels) is due to the dominant allele R. The recessive allele r, when homozygous, produces colorless aleurone. The plant color (not the kernel color) is controlled by another gene with two alleles Y and y. The dominant Y allele results in green color, whereas the homozygous presence of the recessive y allele causes the plant to appear yellow. In a testcross between a plant of unknown genotype and phenotype and a plant that is homozygous recessive for both traits, the following progeny were obtained. Numbers of Offspring Phenotype Colored Aleurone & Green Plant 88 Colored Aleurone & Yellow Plant 12 Colorless Aleurone & Green Plant 8 Colorless Aleurone & Yellow Plant 92 Question: Determine the genotype and phenotype of the unknown plant, including the precise association of the two genes on the homologs (i.e. the map arrangement)arrow_forwardIn Drosophila, the sepia mutation (se, chromosome 3, position 26) results in dark brown eyes, while cinnabar (cn, chromosome 2, position 57.5) results in bright orange-red eyes. True breeding, wild type females are mated with true breeding males homozygous recessive for both traits. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1arrow_forward
- In a certain plant, the dominant form of gene B codes for blue fruit, while the recessive form results in pink fruit. The dominant form of another gene, E, inhibits the activity of the enzyme coded for by gene B, resulting in white fruit, while the recessive form is unable to inhibit this enzyme and results in colored (i.e., blue or pink) fruit. A doubly-homozygous dominant white-fruited plant is crossed with a pink-fruited plant. The F1 progeny were then self-crossed to generate the F2 generation. Determine the ratios of genotypes and phenotypes for each generation.arrow_forwardIn a species of roses, the alleles RP and RR code for pink roses and red roses, respectively. In this case, the wild-type phenotype is red roses in which the locus exhibits a paramutation. The presence of RP with RR leads to a paramutation of RR to RP", which results in pink roses. This paramutation is 100% penetrant. If a heterozygous rose is crossed with a homozygous red rose, what is the phenotypic ratio of the offspring? O % pink and % red none pink and all red O all pink and none red O % pink and redarrow_forwardIn roses, purple flower color is determined by the dominant P allele, while pphomozygotes are white. The presence of long stems is determined by the dominant S allele, while ss homozygotes have short stems. Both mutations are completely penetrant. A test cross was performed between a rose plant of unknown genotype with a white flowered, short stemmed rose plant (pp ss) and the following 200 progeny plants were obtained: 84 white flowers, long stems 16 purple flowers, long stems 82 purple flowers, short stems 18 white flowers, short stems Select two statements below that are TRUE. options: The P and S genes independently assort during meiosis. The map distance between P and S is 17 cM. The genotype of the progeny plants with purple flowers and short stems is PP ss. The map distance between P and S is 83 cM. The homologs in the plant with…arrow_forward
- Dominant allele R in corn results in purple kernel, while recessive allele r results in pearl white kernel in homozygous condition. The plant body colour is controlled by another gene with two alternative alleles, G for green colour and g for yellow colour. In a testcross between a corn plant with unknown genotype and a plant that is homozygous recessive for both genes, the following progeny were obtained: Phenotype Purple, green Purple, yellow Pearl white, green Pearl white, yellow Number 90 12 10 88 Explain how these results were obtained by determining the exact genotype of the unknown plant. Include the precise gene arrangement on the homologous chromosomes.arrow_forwardIn roses, purple flower color is determined by the dominant P allele, while pp homozygotes are white. The presence of long stems is determined by the dominant S allele, while ss homozygotes have short stems. Both mutations are completely penetrant. A test cross was performed between a rose plant of unknown genotype with a white flowered, short stemmed rose plant (pp ss) and the following 200 progeny plants were obtained: 84 white flowers, long stems 16 purple flowers, long stems 82 purple flowers, short stems 18 white flowers, short stems Select the statements below that are TRUE. Select 2 correct answer(s) The P and S genes independently assort during meiosis. The map distance between P and S is 17 CM. The genotype of the progeny plants with purple flowers and short stems is PP ss. The map distance between P and S is 83 CM. The homologs in the plant with unknown genotype are p S and Ps. The homologs in the plant with unknown genotype are PS and p s.arrow_forwardIn autotetraploid Chinese primrose (Primula sinensis L.), the gene controlling stigma color is very near the centromere of the chromosome carrying it. The allele G for green stigma is dominant to g for red stigmas. A homozygous green autotetraploid strain is crossed with a homozygous red autotetraploid strain. Each of the F1 GGgg plants would obtain 12 gametes which are 2GG, 8Gg, and 2g. How were these obtained?arrow_forward
- In Drosophila, the dominant Bar mutation (B, chromosome X, position 57) results in thin bar- shaped eyes, while the recessive singed (sn, chromosome X, position 21) results burnt looking bristles. True breeding, wild type females are mated with true breeding males with Bar eyes and singed bristles. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1 =1arrow_forwardIn corn, a colored aleurone is due to the presence of an R allele; r/r is colorless. Another gene controls the color of the plant, with g/g being yellow and G_being green. A plant of unknown genotype is test-crossed, and the following progeny plants were obtained. Colored green 89 Colored yellow 13 Colorless green 9 Colorless yellow 92 What is the recombination frequency between the R locus and the G locus? A. 45.6% B. 9.85% C. 91.15% D. 4.93% E. 6.4%arrow_forwardIn corn, a colored aleurone is due to the presence of an R allele; r/r is colorless. Another gene controls the color of the plant, with g/g being yellow and G_being green. A plant of unknown genotype is test-crossed, and the following progeny plants were obtained. Colored green 89 Colored yellow 13 Colorless green 9 Colorless yellow 92 What was the phenotype and genotype of the plant used for the test cross? A. Colorless green - rG/rg B. Colorless yellow - rg/rg C. Colored yellow - Rg/rg D. Colored green - RG/rgarrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning