BIOLOGY
12th Edition
ISBN: 9781260169614
Author: Raven
Publisher: RENT MCG
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 53, Problem 2DA
Summary Introduction
To explain: The role of behavioral genetics between two closely-related species of North American rodents named prairie and montane voles.
Introduction: Both genetics and environment play a major role in shaping human behavior. Nature and nurture are two components influencing the behavior of organisms, which means that the genes, the environment, upbringing, and life experiences together are responsible for determining an organism's behavior.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Hardy Weinberg Equilibrium
In 1975, in a population of 88 garter snakes on the coast of Oregon, 54 of them exhibit the recessive genotype are therefore not immune to the rough-skinned newt’s toxin.
What is the frequency for the recessive genotype (q^2)?
The typical tail size for elephants is 150 cm, with a standard deviation of 10 cm. The
typical tail size for mice is 7 cm, with a standard deviation of 3 cm. Do elephants or
mice exhibit greater variability in tail size?
Elephants, because CV = 15 > 2.3
Elephants, because SD = 10 cm > SD = 3 cm
%3D
Elephants, because = 150 cm > 7 cm
O Mice, because CV = 0.43 > CV = 0.07
%3D
%3D
Researchers divided 100 mice, with identical heredity composition that makes them susceptible to coronary artery disease, into two groups. Group A ate standard rodent chow and group B ate low-fat rodent chow for 120 days. what would you conclude if 45 out of 50 mice in group A show coronary artery disease and 45 out of 50 mice in group B show coronary artery disease?
Chapter 53 Solutions
BIOLOGY
Ch. 53.1 - Prob. 1LOCh. 53.1 - Prob. 2LOCh. 53.1 - Prob. 3LOCh. 53.2 - Prob. 1LOCh. 53.2 - Prob. 2LOCh. 53.3 - Prob. 1LOCh. 53.3 - Prob. 2LOCh. 53.3 - Prob. 3LOCh. 53.4 - Prob. 1LOCh. 53.4 - Prob. 2LO
Ch. 53.4 - Prob. 3LOCh. 53.5 - Prob. 1LOCh. 53.5 - Prob. 2LOCh. 53.5 - Prob. 3LOCh. 53.6 - Prob. 1LOCh. 53.7 - Define migration.Ch. 53.7 - Prob. 2LOCh. 53.7 - Prob. 3LOCh. 53.8 - Prob. 1LOCh. 53.8 - Prob. 2LOCh. 53.8 - Prob. 3LOCh. 53.9 - Prob. 1LOCh. 53.9 - Prob. 2LOCh. 53.10 - Prob. 1LOCh. 53.10 - Prob. 2LOCh. 53.11 - Prob. 1LOCh. 53.11 - Prob. 2LOCh. 53.11 - Prob. 3LOCh. 53.12 - Prob. 1LOCh. 53.12 - Prob. 2LOCh. 53.12 - Prob. 3LOCh. 53.13 - Prob. 1LOCh. 53.13 - Prob. 2LOCh. 53.13 - Prob. 3LOCh. 53 - Prob. 1DACh. 53 - Prob. 2DACh. 53 - Prob. 3DACh. 53 - Prob. 4DACh. 53 - Prob. 5DACh. 53 - Prob. 1IQCh. 53 - Prob. 2IQCh. 53 - Prob. 3IQCh. 53 - What factors might be responsible for the slight...Ch. 53 - Prob. 1UCh. 53 - Prob. 2UCh. 53 - The study of song development in sparrows showed...Ch. 53 - Prob. 4UCh. 53 - Prob. 5UCh. 53 - Prob. 6UCh. 53 - Prob. 7UCh. 53 - Prob. 8UCh. 53 - In the haplodiploidy system of sex determination,...Ch. 53 - Prob. 10UCh. 53 - Prob. 11UCh. 53 - Prob. 1ACh. 53 - Refer to figure 54.25. Six pairs of birds were...Ch. 53 - Prob. 3ACh. 53 - Prob. 1SCh. 53 - Behavioral genetics has made great advances from...Ch. 53 - If a female bird chooses to live in the territory...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- Why is the intermembral index (IMI) of a biped similar to that of a vertical clinger and leaper?arrow_forwardPeter Boag, studied the inheritance of beak depth in Galapagos finches by looking at the relationship between parent beak depth (in mm) and that of their offspring. He collected two sets of parent offspring data, once in 1976 and again in 1978. For both years he followed birds to determine which pairs belonged to which nests. Most of the parents had been previously captured so their beak depths were known. He then captured the offspring when they fledged and measured their beak depth. He calculated the "midparent" beak depth (the average beak depth of the two parents) and then compared that to beak depth of their offspring. Here is what he found in 1976: Assume that in 1978, all the finches with beak depths less than 9.0 mm died due to prolonged drought. Use your knowledge of how quantitative traits respond to selection and the following relationship, R = h2S, to determine the following: Selection differential. Predicted response…arrow_forwardYou observe some bark beetles with different feeding behaviors. You find that the behavioral phenotypes you observed are closely associated with the following genotypes. genotype phenotype number observed BB feeds under bark 60 Bb feeds on top of bark 24 bb feeds on leaves 16 a) What are the allele frequencies and the expected number (out of 100 beetles) of genotypes? b) Calculate the Chi-Squared value (c2). Then look up the p-value only using this Chi-squared calculator: https://www.mathsisfun.com/data//chi-square-calculator.html Clear all the values in the box at upper right, enter the Chi Square value that you obtain FROM YOUR CALCULATIONS in the Chi-Square box, and enter 1 for the degrees of freedom. Is the population in H-W equilibrium? c) If you answered “No” to b): Provide at least three biologically valid reasons why the observed beetle population may not be in equilibrium. If you answered “Yes” to b): Why do you think this…arrow_forward
- Some people consider stress and crowding studies of laboratory animals highly applicable in understanding human behavior. Other people question the cross-species transferability of these results. What considerations would be important in interpreting these experiments?arrow_forwardYou observe some bark beetles with different feeding behaviors. You find that the behavioral phenotypes you observed are closely associated with the following genotypes. genotype phenotype number observed BB feeds under bark 60 Bb feeds on top of bark 24 bb feeds on leaves 16 a) What are the allele frequencies and the expected number (out of 100 beetles) of genotypes? b) Calculate the Chi-Squared value (c2). Then look up the p-value using this Chi-squared calculator: https://www.mathsisfun.com/data//chi-square-calculator.html Enter “1” for Degrees of Freedom. (There are 3 genotype categories, but only 1 DF because they are not fully independent—calculated from just 2 allele frequencies). The p-value is the probability that you would see the deviance you observed by chance, assuming the null hypothesis is true). Report both values. Is the population in H-W equilibrium? c) If you answered “No” to b): Provide at least three biologically…arrow_forwardGive typing answer with explanation and conclusion What is a phenotype and how is it determined? What is phenotypic plasticity? Over what time scale does phenotypic plasticity act? Why would animals want to have plastic phenotypes?arrow_forward
- I've observed varying egg mimic sizes on female fins: large, small, or none. I found that this trait follows Mendelian genetics with a single locus having two alleles: BB (large mimic), Bb (small mimic), and bb (no mimic). During a recent collection trip, we gathered 211 males with large egg mimics, 156 with small mimics, and 28 with no mimics. Is the locus in Hardy-Weinberg equilibrium, and what's the expected frequency of the small egg mimic phenotype in this population? 1.Yes it is in HW equilibrium, expected = 0.07 2.Yes it is in HW equilibrium, expected = 0.39 3.No it is not in HW equilibrium, expected = 0.53 4.No it is not in HW equilibrium, expected = 0.20 5.No it is not in HW equilibrium, expected = 1.0arrow_forwardExplain why tortoiseshell cats are almost always female and why they have a patchy distribution of orange and black fur.arrow_forwardConsider the data that are summarised in the figure i. The data in part (a) are consistent with the hypothesis that alarm calling only in female black-tailed prairie dogs evolved in accordance with kin selection. ii. On the basis of the data in part (a), male black-tailed prairie dogs without kin in their home coteries almost are as likely (i.e., within 10%) to produce alarm calls as are female black-tailed prairie dogs with kin in their home coteries.iii. On the basis of the data in the figure, males and females produce alarm calls more often when the kin that they live with are offspring than when the kin that they live with are parents and siblings.iv. On the basis of the data in part (b), alarm calls in black-tailed prairie dogs constitute merely parental care (i.e., rather than altruism).v. On the basis of the data in part (b), mean alarm calling percentage is greater for male black-tailed prairie dogs than for female black-tailed prairie dogs. Question 2 options:…arrow_forward
- In snouters, mucus color to catch prey shows incomplete dominance. Red mucus color is dominant to yellow mucus color. What kind of offspring could result from snouter parent 1 being homozygous dominant and snouter parent 2 having orange mucus?arrow_forwardA study was conducted on a population of snowshoe hares in Algonquin Provincial Park, Ontario, Canada. 110 one month-old hares were removed from their nests, micro-transmitters were inserted under their skin, and they were quickly returned to their homes. Scientists monitored their activities on a monthly basis, and during the breeding season checked nests and counted the babies produced for four years. Below is a summary of the data for only the female hares that were tagged (60 individuals). For simplicity, the monthly data has been summed into yearly totals. Survival: 60 were alive at the start the experiment (year = 0), 20 at the start on the next year (year 1), 8 at year 2, 1 at year 3, and 0 at year 4. Fecundity: during year 0, no babies were produced; year 1- females produced an average of 4 babies each; year 2 = 8 babies each; and year 3 = 8 babies each. A. Answer the following questions using the information above. 1. Create a life table for the snowshoe hare.arrow_forwardIn an assessment of learning in Drosophila, flies were trained to avoid certain olfactory cues. In one population, a mean of 8.5 trials was required. A subgroup of this parental population that was trained most quickly (mean = 6.0) was interbred, and their progeny were examined. These flies demonstrated a mean training value of 7.5. Calculate realized heritability for olfactory learning in Drosophila.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning
Biology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage Learning
Human Heredity: Principles and Issues (MindTap Co...
Biology
ISBN:9781305251052
Author:Michael Cummings
Publisher:Cengage Learning
Biology (MindTap Course List)
Biology
ISBN:9781337392938
Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg
Publisher:Cengage Learning