2008problemset8key

2008problemset8key - Problem Set#8 KEY BICD 100 W’ 08 1 A...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem Set #8 KEY BICD 100 W’ 08 1). A horse breeder wants to develop a breed of small horses for children to ride. From a heterogenous population of 100 horses with a mean height of 60 inches, a subpopulation of horses with a mean height of 45 inches was chosen and interbred. The progeny of these matings had a mean height of 55 inches. a. What is the narrow sense heritability (h N 2 ) of height in this population of horses? f8e5 y = 60, y p = 45, y o = 55 h N 2 = y o – f8e5f8e5 y /y p- f8e5 y = 5/15 = 0.33 b. From the progeny of the first round of selective matings, a subpopulation of horses is again chosen with a mean height of 45 inches and interbred. What do you expect to be the mean height of the progeny of this second round of matings? f8e5 y = 55, y p = 45, h N 2 = 0.33 y o = f8e5 y + h N 2 (y p – f8e5 y ) = 55 + 0.33 (45 – 55) = 51.7 2.) A study was carried out to determine the heritability of physical endurance training potential. Thirty pairs of male monozygotic (identical) twins raised in different families were recruited for the study. All 60 individuals participated in the same 10 week training regimen. Before and after the training period, each participant was tested to determine how long he could perform repetitive elbow flexion while holding a 15 kg barbell, yielding an “endurance training” score. The scores for 9 pairs of twins were as follows: twin 1 twin 2 70 65 72 76 71 75 64 70 66 64 70 68 74 78 70 74 73 69 a. What type of heritability can be calculated from this data and what degree of heritability is indicated by this data? Narrow sense heritability can be estimated -> h N 2 = r obs /r exp where r obs = cov(x,y)/(s x )(s y ) and cov(x,y) = Σ (x – f8e5 x)( y – f8e5 y)/ n-1 where n is the number of x,y pairs = 8.75 r obs = cov(x,y)/(s x )(s y ) = 8.75/(3.20)(4.97) = 0.55 Since r exp = 1 (identical twins), h N 2 = 0.55 b. In 1998, a group of scientists reported in Nature a correlation between inheritance of “I” vs. “D” 1 alleles for the ACE gene and physical endurance. ACE stands for angiotensin-converting enzyme – this enzyme degrades a class of vasodilator proteins and also converts angiotensin I into a vasoconstrictor. Therefore, by promoting the constriction of blood vessels, ACE has an impact on blood flow, which in turn impacts physical endurance (more blood flow, better endurance)....
View Full Document

{[ snackBarMessage ]}

Page1 / 4

2008problemset8key - Problem Set#8 KEY BICD 100 W’ 08 1 A...

This preview shows document pages 1 - 3. Sign up to view the full document.

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