In a natural population of outbreeding plants, the variance of the total numberof seeds per plant is 16. From the natural population, 20 plants are taken intothe laboratory and developed into separate true-breeding lines by self-fertilization-with selection for high, low, or medium number of seeds-for 10generations. The average variance in the tenth generation in each of the 20 setsis about equal and averages 5.8 across all the sets. Estimate the broad-senseheritability for seed number in this population. (4 pts)12pt vParagraphBIDIT² v✓B°

Broad-sense heritability formula:H2=VG /VP where:VG is the genetic varianceVP is the total…

Answered: In a natural population of outbreeding…

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter5: The Inheritance Of Complex Traits

Section: Chapter Questions

Problem 22QP: Height in humans is controlled by the additive action of genes and the action of environmental...

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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?
If there is no genetic variation within a population for a given trait, what is the heritability for the trait in the population?
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?
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.
Why are monozygotic twins who are reared apart so useful in the calculation of heritability?
In sweet pea plant, an allele for purple flowers (P) is dominant when paired with a recessive allele for red flowers (p). An allele for long pollen grains (L) is dominant when paired with a recessive allele for round pollen grain (l). Bateson and Punnett crossed a plant having purple flowers/long pollen grains with one having white/flowers/round pollen grains. All F1 offspring had purple flowers and long pollen grains. Among the F2 generation, the researchers observed the following phenotypes: 296 purple flowers/long pollen grains 19 purple flowers/round pollen grains 27 red flowers/long pollen grains 85 red flowers/round pollen grains What is the best explanation for these results?
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
More Crosses with Pea Plants: The Principle of Independent Assortment Determine the possible genotypes of the following parents by analyzing the phenotypes of their children. In this case, we will assume that brown eyes (B) is dominant to blue (b) and that right-handedness (R) is dominant to left-handedness (r). a. Parents: brown eyes, right-handed brown eyes, right-handed Offspring: 3/4 brown eyes, right-handed 1/4 blue eyes, right-handed b. Parents: brown eyes, right-handed blue eyes, right-handed Offspring: 6/16 blue eyes, right-handed 2/16 blue eyes, left-handed 6/16 brown eyes, right-handed 2/16 brown eyes, left-handed c. Parents: brown eyes, right-handed blue eyes, left-handed Offspring: 1/4 brown eyes, right-handed 1/4 brown eyes, left-handed 1/4 blue eyes, right-handed 1/4 blue eyes, left-handed
Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?
Sunflowers with flowers 10 cm in diameter are crossed with a plant that has 20-cm flowers. The F1 plants have flowers 15 cm in diameter. In the F2 generation, 4 flowers are 10 cm in diameter and 4 are 20 cm in diameter. Between these are 5 phenotypic classes with diameters intermediate to those at the extremes. a. Assuming that the alleles that contribute to flower diameter act additively, how many genes control flower size in this strain of sunflowers? b. How much does each additive allele contribute to flower diameter? c. What size flower makes up the largest phenotypic class?
Monozygotic (MZ) twins have a concordance value of 44% for a specific trait, whereas dizygotic twins have a concordance value of less than 5% for the same trait. What could explain why the value for MZ twins is significantly less than 100%?
A pedigree analysis was performed on the family of a man with schizophrenia. Based on the known concordance statistics, would his MZ twin be at high risk for the disease? Would the twins risk decrease if he were raised in an environment different from that of his schizophrenic brother?

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