2007 final

2007 final - Biology 202 BASIC GENETICS Final Examination...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Biology 202 BASIC GENETICS Final Examination 2007 VERSION # 1 - 5 - SHORT ANSWER QUESTIONS 1. In a laboratory population of Drosophila melanogaster, the narrow-sense heritability ( h 2 ) of abdominal bristle number is 0.52 when measured at 14 degrees Celsius. Suppose the mean bristle number in this population is 35.2. (a) (2 points) What would you predict the mean number of bristles to be in the offspring generation if a selection differential ( S ) of 2.0 is applied for increased bristle number, and the response to selection ( R ) is measured in offspring grown at 14 degrees Celsius (Note: Assume that the covariance of genotype and environment is zero when answering this question)? Please show your calculations. One uses the “Breeders Equation” to solve this problem; namely the response to selection ( R ) is a product of the selection differential ( S ) and the narrow-sense heritability ( h 2 ). Or in other words, R = h 2 S = 0.52 x 2.0 =1.04. ( Award 1 point for recall of correct formula and its correct application in the calculation of R). This response to selection ( R ) must now be added to the mean of the parental generation (parental mean = 35.2) in order to arrive at the predicted offspring mean of 36.24 (i.e., Predicted offspring mean = parental mean + R = 35.2 + 1.04) ( Award 1 point for correct use and application of the calculation of R in predicting the offspring bristle number) (b) (1 point) Answer the above question (in one sentence), but now assume that response to selection ( R ) is measured in offspring grown at 20 degrees Celsius instead of at 14 degrees. Correct answers: One cannot predict R under these conditions, as at the new temperature of 20 C, changes in phenotypic expression among genotypes are expected to be different from those observed at 14 C. or/ One cannot predict R under these conditions, as at the new temperature of 20 C, the likely existence of non-parallel norms of reaction for quantitative traits mean that the response may be different from that observed at 14 C. Some Incorrect answers: The response will be the same because the covariance of genotype and environment is zero at each temperature. (Note: Zero genotype-environment does not change the fact that genotypes may respond differently to a new environment; in other words, the genotype-environment covariance can be zero when measured in each environment (14 and 20 C), but this does not imply that the norm of reaction is flat across temperatures). The response will be the same because the heritability is already known (calculated) at 14C and should, therefore, not change ( This answer is incorrect, as we are interested in the response to selection. Even though heritability may be different when measured at 20 C this is not really the point. Rather, it is inability to know in advance the norm of reaction that is the problem that enters into predicting the response, not the heritability; i.e., the issue is not the heritability, but rather the way that genotypes
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 18

2007 final - Biology 202 BASIC GENETICS Final Examination...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online