You have the following data: Daily milk yield records of close relatives of a bull used to sire F1s: 24.4, 19.8, 23.0, 22.6, 20.2 kg, Daily milk yield records for cows: 18.1, 20.4, 17.5, 16.2, 25.2, 14.2 kg, Daily milk yield records for F1s: 31.0, 24.6, 25.4, 32.7, 26.3 kg. Calculate the heterosis (hybrid vigour) for this trait

You have the following data: Daily milk yield records of close relatives of a bull used to sire F1s: 24.4, 19.8, 23.0, 22.6, 20.2 kg, Daily milk yield records for cows: 18.1, 20.4, 17.5, 16.2, 25.2, 14.2 kg, Daily milk yield records for F1s: 31.0, 24.6, 25.4, 32.7, 26.3 kg. Calculate the heterosis (hybrid vigour) for this trait

Case Studies In Health Information Management

3rd Edition

ISBN:9781337676908

Author:SCHNERING

Publisher:SCHNERING

Chapter3: Informatics, Analytics, And Data use

Section: Chapter Questions

Problem 3.29.1C

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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.
If the genotype frequencies in a population are 0.60 AA, 0.20 Aa, and 0.20 aa, and if the requirements of the HardyWeinberg principle apply, the genotype frequencies in the offspring generation will be: a. 0.60 AA, 0.20 Aa, 0.20 aa. b. 0.36 AA, 0.60 Aa, 0.04 aa. c. 0.49 AA, 0.42 Aa, 0.09 aa. d. 0.70 AA, 0.00 Aa, 0.30 aa. e. 0.64 AA, 0.32 Aa, 0.04 aa.
Using the HardyWeinberg Law in Human Genetics In a given population, the frequencies of the four phenotypic classes of the ABO blood groups are found to be A = 0.33, B = 0.33, AB = 0.18, and i = 0.16. What is the frequency of the i allele?
Overall, the normal ratio is about 0.2 0.4 0.6 0.8
Use the following information to respond to the three questions posed below: (1) The proband (affected individual who led to the construction of the pedigree) exhibits the trait. (2) Neither her husband nor her only sibling, an older brother, exhibits the trait. (3) The proband has five children by her current husband. The oldest is a boy, followed by a girl, then another boy, and then identical twin girls. Only the second oldest fails to exhibit the trait. (4) Both parents of the proband show the trait. a. Construct a pedigree of the trait in this family. b. Determine how the trait is inherited (go step by step to examine each possible pattern of inheritance). c. Can you deduce the genotype of the probands husband for this trait?
Using the HardyWeinberg Law in Human Genetics Suppose you are monitoring the allelic and genotypic frequencies of the MN blood group locus (see Question 2 for a description of the MN blood group) in a small human population. You find that for 1-year-old children, the genotypic frequencies are MM = 0.25, MN = 0.5, and NN = 0.25, whereas the genotypic frequencies for adults are MM = 0.3, MN = 0.4, and NN = 0.3. a. Compute the M and N allele frequencies for 1-year-olds and adults. b. Are the allele frequencies in equilibrium in this population? c. Are the genotypic frequencies in equilibrium?
Why are monozygotic twins who are reared apart so useful in the calculation of heritability?
Imagine that you are performing a cross involving seed texture in garden pea plants. You cross true-breeding round and wrinkled parents to obtain F1 offspring. Which of the following experimental results in terms of numbers of plants are closest to what you expect in the F2 progeny? a. 8lOroundseeds b. 8lOwrinkledseeds c. 405:395 round seeds:wrinkled seeds d. 610:190 round seeds:wrinkled seeds
When a population has a gene with four alleles circulating, how many possible genotypes are there? a.3 b.6 c. 10 d. 16
Perfect pitch is the ability to identify a note when it is sounded. In a study of this behavior, perfect pitch was found to predominate in females (24 out of 35 in one group). In one group of seven families, individuals in each family had perfect pitch. In two of those families, the affected individuals included a parent and a child. In another group of three families, three or more members (up to five) of each had perfect pitch, and in all three families, two generations were involved. Given this information, what, if any, conclusions can you draw about whether this behavioral trait might be genetic? How would you test your conclusion? What further evidence would be needed to confirm your conclusion?
A walnut comb rooster is mated to three hens. Hen A, which has a walnut comb, has offspring in the ratio of 3 walnut to 1 rose. Hen B, which has a pea comb, has offspring in the ratio of 3 walnut to 3 pea to 1 rose to 1 single. Hen C, which has a walnut comb, has only walnut comb offspring. What are the genotypes of the rooster and the three hens?
Height in humans is controlled by the additive action of genes and the action of environmental factors. For the purposes of this problem, assume that height is controlled by four genesA, B, C, and Dand that there are no environmental effects. Assume further that additive alleles contribute two units of height and partially additive alleles contribute one unit of height. a. Given these assumptions, can two individuals of moderate height produce offspring that are much taller and shorter than either parent? If so, how can this happen? b. Can someone of minimum height and someone of intermediate height have children taller than the parent of intermediate height? Why or why not?