To review:
Rice is grown in flooded conditions. There are many varieties of the same that can be tolerant and intolerant. The true breeding-tolerant varieties are FR13A, BKNFR, and Kurk, while the true breeding-intolerant varieties are
1. |
Number of plants | ||
Tolerant (Alive) | Intolerant (Dead) | Total | |
On the basis of the data given for the various crosses, justify that it supports that tolerance is a dominant trait with the help of genotypes and phenotypes.
Introduction:
A homozygous allelic pair is one that has the same genes, both coding either for dominant genes or recessive genes. A heterozygous allelic pair is one that contains two genes coding for dominant genes and recessive genes. Two plants that differ in one trait only are crossed in the case of a monohybrid cross. They can be either homozygous or heterozygous. When a homozygous dominant trait is crossed with a recessive trait, or heterozygous dominant traits are self-crossed, it produces a
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Chapter 12 Solutions
Biology: The Dynamic Science (MindTap Course List)
- Half of the worlds population eats rice at least twice a day. Much of this rice is grown in flooded conditions, and different strains of rice are tolerant (survive) or intolerant (die) under these conditions. Rice breeders used genetic crosses to test whether tolerance to flooding is a dominant trait. Researchers used three true-breeding flood-tolerant strains, FR143, BKNFR, and Kurk, and two true-breeding flood- intolerant strains, IR42 and NB, in the crosses. Results were obtained from three sets of crosses and are reported in the Table below: Results of cross of F1 to tolerant parent: F1 plants were crossed with the tolerant parent of the cross. Number of Plants Progeny Analyzed from Intolerant Tolerant Cross Alive Dead Total 1. F2 results of cross: IR42 FR13A 187 77 264 IR42 BKNFR 192 73 265 NB Kurk 142 52 195 2. Results of cross of F1 to intolerant parent: (F1 of IR42 FR13A) IR42 14 17 31 (F1 of IR42 BKNFR) IR42 15 10 25 (F1 of NB Kurk) NB 21 35 56 3. Results of cross of F1 to tolerant parent: (F1 of IR42 FR13A) FR13A 31 0 31 (F1 of IR42 BKNFR) BKNFR 28 0 28 (F1 of NB Kurk) Kurk 40 0 40 Do the data support the hypothesis that the tolerance trait is dominant? Justify your conclusion by explaining the results from each of the three sets of crosses in terms of genotypes and phenotypic ratios. Source: T. Setter et al. 1997. Physiology and genetics of submergence tolerance in rice. Annals of Botany 79:6777.arrow_forwardA mountain region has a population of 5,000 mountain goats. You score these animals for the R locus and find that this locus has two alleles, R (dominant) and r (recessive). 3200 individuals are homozygous dominant, 1,600 are heterozygous, and 200 are homozygous recessive. A deadly virus infects all of the RR mountain goats in the population (above), killing all individuals of this genotype and leaving the population with only the rr and Rr mountain goats. a) Calculate the new allele frequencies for this population. Show your work. b) Calculate the new (observed) genotypic frequencies for this population. Show your work. c) Does this population still appear to be at H-W equilibrium? Why or why not? (You do not need to analyze this statistically).arrow_forwardMalaria is a leading cause of human illness and mortality worldwide, with 200 million people infected and 600,000 deaths each year. In the 1960s, the incidence of malaria was reduced owing to the use of insecticides that killed mosquitoes in the genus Anopheles, which transmit the disease from person to person. But today, mosquitoes are becoming resistant to insecticides--causing a resurgence in malaria. In this exercise, you will investigate whether alleles encoding resistance to insecticides have been transferred between closely related species of Anopheles. To find out whether transfers have occurred, you will analyze DNA results from two species of mosquitoes that transmit malaria (Anopheles gambiae and A. coluzzii) and from A. gambiae × A. coluzzii hybrids. Resistance to DDT and other insecticides in Anopheles is affected by a sodium channel gene, kdr. The allele r of this gene confers resistance, while the wild type (+/+) genotype is not resistant. Researchers sequenced the kdr…arrow_forward
- A large, genetically heterogeneous group of tomato plants was usedas the original breeding stock by two different breeders, namedMary and Hector. Each breeder was given 50 seeds and began aselective breeding strategy, much like that described in Figure24.11. The seeds were planted, and the breeders selected the 10plants with the highest mean tomato weights as the breeding stockfor the next generation. This process was repeated over the courseof 12 growing seasons, and the following data were obtained: A. Explain these results.B. Another tomato breeder, named Martin, got some seeds fromMary’s and Hector’s tomato strains (after 12 generations),grew the plants, and then crossed them to each other. The mean weight of the tomatoes in these hybrids was about 1.7 pounds.For a period of 5 years, Martin subjected these hybrids to thesame selective breeding strategy that Mary and Hector had followed,and he obtained the following results: Explain Martin’s data. Why was Martin able to obtain…arrow_forwardThe gene for pea pod colour has two alleles. The dominant allele, G, is associated with the dominant trait of green pea pod colour. The recessive allele, g, is associated with the recessive trait of yellow pea pod colour. For many years, there has been a large collection of pea plants with a mixture of yellow and green pea pod colours in a savannah. But there has been a recent infestation of wild rabbits with a preference for eating plants with yellow pea pods. Assume that at the start of the infestation, the alleles in the pea plant population were in Hardy-Weinberg equilibrium. (a) What does it mean for the alleles in the pea plant population to be in Hardy-Weinberg equilibrium? (b) What is expected to happen to the allele frequency of the g allele in the subsequent pea generations after the rabbits have been introduced? Explain your answer using terms from Darwinian evolution.arrow_forwardSnuffleups are small, ant-eating mammals that live deep in the jungles of Brazil. They possess a nose that varies between 3 inches long and 19 inches long. The exact length of the nose is determined by quantitative loci. If one crosses a 19-inch nosed snuffleup with one that has a 3-inch nose, one gets F1 offspring with 11-inch noses. When one crosses two F1 individuals, one gets an array of nose lengths from 3 inches to 19 inches long. However, among the F2, 3-inch noses occur only about 1 in 256 offspring and 19-inch noses occur about one in 256 offspring. How many ALLELES are involved in forming this trait?arrow_forward
- Snuffleups are small, ant-eating mammals that live deep in the jungles of Brazil. They possess a nose that varies between 3 inches long and 19 inches long. The exact length of the nose is determined by quantitative loci. If one crosses a 19-inch nosed snuffleup with one that has a 3-inch nose, one gets F1 offspring with 11-inch noses. When one crosses two F1 individuals, one gets an array of nose lengths from 3 inches to 19 inches long. However, among the F2, 3-inch noses occur only about 1 in 256 offspring and 19-inch noses occur about one in 256 offspring. What is the contribution per allele for this trait (in inches/allele)?arrow_forwardYou are studying an isolated population of irises on an island off the coast of Maine. Plants in this population exhibit three different phenotypes for stem hair. ShSh plants are hairy, SsSs plants are smooth, and ShSs plants have intermediate amounts of stem hair. After characterizing stem phenotypes for a sample of the population, you classify 450 plants as having hairy stems, 701 as having intermediate hairy stems, and 112 as having smooth stems. Is the population in Hardy-Weinberg Equilibrium with respect to stem hair? ___YES______NO # Justify your answer. Edit View Insert Format Tools Table Paragraph✓ BI U 12pt v 80 F3 $ 000 000 F4 % F5 <arrow_forwardWhat is the chi-square value of the pumpkin population? round up and use four decimal points After doing some crosses and genotyping some individuals, you determined that the number of individuals homozygous dominant for the largest pumpkin allele is 98, 80 are heterozygous for medium size, and 20 are homozygous recessive for small size.arrow_forwardTABLE 1 Data for Bean Simulation for F1 and F2 Generations: F1 Generation GENOTYPES TOTALS Observed Frequency of Observed frequency of Genotypes = proportion alleles out of 50 BLACK, BLACK (AA) A= 12 BLACK, WHITE (Aa) 19 a= WHITE, WHITE (aa) 19 F2 Generation BLACK, BLACK (AA) A= 14 BLACK, WHITE (Aa) 24 a= WHITE, WHITE (aa) 12arrow_forwardA scientist is studying a wild population of Japanese morning glories. It is easy to separate genotypes at a flower color locus by their phenotypes. Red individuals are homozygous for the R allele, yellow individuals are homozygous for the Y allele, and orange individuals are heterozygous. If the number of individuals with the following genotypes are: RR: 831 RY: 33 YY: 442 Is the population in Hardy Weinberg Equilibrium?arrow_forwardA population of horses are having a problem. They are not mating randomly, they are mating with those that have the same flying speed as their own: Fast (FF), Medium speed (FS), or Slow (SS). You arrive to this land and count the following horse of each flying velocity: Fast = 71 Medium speed = 90 Slow = 31 If we consider these results Generation 1, what are the expected number of Fast individuals for each genotype in Generation 3 if we have only inbreeding, that is, FF only reproduces with FF; FS only reproduces with FS; and SS only reproduces with SS? round down to the whole number.arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
- Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning