# Chapter 1 End Problems

## Problem 3

Aa X Aa produces 6 offspring, so the number of trials is 6.
Expect aa 25% of the time
and expect not aa 75% of the time.

Let’s try a brute force approach. Populate a vector of length 1000 with 250 AA, 500 Aa and 250 aa. Sample with replacement 1000 times 6 geneotype. Calculate the fraction with aa present 2 or fewer times.

## Problem 4

Jack, queen, king are the face cards so Jack is likely 1/3 of the time a face card is drawn.

There are 4 Jacks so Jack of Hearts is 1 of 4.

The probability that a randomly drawn face card (of which there are 12) is the jack of hearts is 1 of 12.

Figure out potentially relevant probabilities then answer the question.

P{FaceCard}: probability of a Face card = 12/52 = 3/13

P{Jack}: probability of a Jack = 4/52 = 1/13.

P{Hearts}: probability of a hearts = 13/52 = 1/4.

P{Jack | FaceCard}: probability of a Jack given it’s a face card = 4/12 = 1/3.

P{Jack | Hearts}: probability of a Jack given it’s a heart = 1/13.

P{Heart | Jack}: probability of a heart given a Jack = 1/4

P{FaceCard | Jack}: probability of a face card given it’s a Jack = 1

What is the probability that a randomly drawn face card is the jack of hearts?

P{JackHearts | FaceCard} = P{ Heart | Jack } X P{Jack | FaceCard}
= 1/4 x 1/3 = 1/12

Perform a trivial Bayesian analysis

Bayes Rule: P{A | B} = P{B | A}*P{A}/P{B}

What is the probability of a Jack given a face card?

There are 12 Face cards and 4 are Jacks so 4/12 or 1/3

Rewrite the Bayesian rule as: P{Jack|FaceCard} = P{FaceCard|Jack}*P{Jack}/P{FaceCard}

P{Jack|FaceCard} = (1*1/13)/3/13 = 1/3

## Problem 6

r = 40/100 = 0.4

standard error = sqrt(( 0.4*( 1-0.4 ))/100 = 0.049

## Problem 12

This R solution requires the package “seqinr”

For a plotting example using seqinr, see

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# Chapter 1 Box Problems

## Box B

A 9:3:3:1 test ratio is a total of 16 possibilities

Ab/aB with r= 0.2 will produce nonrecombinants 80% of the time and recombinants 20% of the time.

 Genotype Proportion Ab 4 aB 4 AB 1 ab 1

## Box C b

$$\frac{dN}{[N(1-\frac{N}{K})]}=rdt$$

## Box C c

Given that:

$$\int \frac{1}{x(1+ax)} dx = ln \frac{x}{1+ax}$$

then:

$$\int_{N_0}^{N_t} \frac{1}{N(1- \frac{1}{K}N)} dN = ln \int_{t=0}^{t}r dt$$

Solve to get:

$$ln \frac{N_t}{1- \frac{1}{K} N_t} - ln \frac{N_0}{1- \frac{1}{K}N_0} = rt$$

ln(x) - ln(y) = ln(x/y) = ln(x*1/y)

$$ln \frac{N_t}{1- \frac{1}{K} N_t} . ln \frac{1- \frac{1}{K}N_0}{N_0} = rt$$

Inverse natural log of both sides:

$$\frac{N_t}{N_0} . \frac{1- \frac{1}{K}N_0}{1-\frac{1}{K}N_t} = e^{rt}$$

$$\frac{N_t - \frac{N_tN_0}{K}}{N_0 - \frac{N_tN_0}{K}} = e^{rt}$$

Multiply the left side by K/K

$$\frac{N_tK - N_tN_0}{N_0K - N_tN_0} = e^{rt}$$

$$\frac{N_t(K-N_0)}{N_0(K-N_t)} = e^{rt}$$

$$N_0e^{rt}(K-N_t) = N_t(K-N_0)$$

$$\frac{N_0e^{rt}}{K-N_0} = \frac{N_t}{K-N_t}$$

## Box D preamble

With statistical software such as R, the manipulations described in Box D are unnecessary except to educate you as to what is going on behind the scenes. Here is one method to perform the manipulations as described in the box:

## Box D a & b

Obtaining quantities 8 and 9 (slope and y intercept) for problems a and b is now trivial using the built in R function lm

# Box E a

$$0 = r_1N_1[ 1 - \frac{N_1 + \alpha_{21} N_2}{K_1} ]$$ $$0 = r_1N_1 - r_1N_1[\frac{N_1 + \alpha_{21} N_2}{K_1} ]$$ $$0 = r_1N_1 - \frac{r_1N_1N_1}{K_1} - \frac{r_1N_1 \alpha_{21} N_2}{K_1}$$ Subtract $r_1N_1$ from both sides and multiply by -1 $$r_1N_1= \frac{r_1N_1N_1}{K_1} + \frac{r_1N_1 \alpha_{21} N_2}{K_1}$$ Multiply by $\frac{K_1}{r_1N_1}$ $$K_1= N_1 + \alpha_{21} N_2$$ $$N_1= K_1 - \alpha_{21} N_2$$ Substitute for $N_1$ in $$0 = r_2N_2[ 1 - \frac{N_2 + \alpha_{12} N_1}{K_2} ]$$ $$0 = r_2N_2[ 1 - \frac{N_2 + \alpha_{12} (K_1 - \alpha_{21} N_2) }{K_2} ]$$ $$0 = r_2N_2 - r_2N_2[\frac{N_2 + \alpha_{12}K_1 - \alpha_{12}\alpha_{21} N_2 }{K_2} ]$$ $$0 = r_2N_2 + \frac{-r_2N_2N_2 - r_2N_2\alpha_{12}K_1 + r_2N_2\alpha_{12}\alpha_{21} N_2 }{K_2}$$ $$-r_2N_2K_2= -r_2N_2N_2 - r_2N_2\alpha_{12}K_1 + r_2N_2\alpha_{12}\alpha_{21} N_2$$ Divide by $-r_2N_2$
$$K_2= N_2 +\alpha_{12}K_1 - \alpha_{12}\alpha_{21} N_2$$ $$K_2 -\alpha_{12}K_1= N_2 - \alpha_{12}\alpha_{21} N_2$$ $$K_2 -\alpha_{12}K_1= N_2(1 - \alpha_{12}\alpha_{21})$$ $$\frac{K_2 -\alpha_{12}K_1}{1 - \alpha_{12}\alpha_{21}}= N_2$$

## Box E b

$r_1$ and $r_2$ affect the rate of achieving equilibrium, not the final population count.

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# Troop 272

## 1973

Back Mr. McKinney Mr. Davis Mark Lovejoy Jim Russo Neil Duval
Third Paul Desrocher Paul McKiney David Trask Ed Hinton Dennis Lavoie ?
Second Bruce Badeau Mark Lavoie Ken Gifford John Cloutier Bill Courtemanche
Front Jim Terriault ? Joe Zulich Peter LaPan ? Mark Cote

## 1976

Ed Hinton Ron Tetreault George Karageorge Mr. J. Bruce Davis Peter LaPan Joe Zulich Bill Courtemanche
Mark Lovejoy ? Ken Soares ? Steve Benson Tim Tetreault Jeff Greene
Tim Paiva ? Jim Stys Dennis Lavoie ? Ron Russo Rick Horne Mike Horne ? Roland Lavoie
? Mark Lavoie ?

## 1977

Back Bob Cormier Ron Tetreault Peter LaPan George Karageorge Bill Courtemanche
Middle Mike Horne Jim Stys Rick Horne Ron Russo Tom Lavoie Tim Paiva Dennis Lavoie
Front Alan Clarke Billy Lemire Roland Lavoie Jeff Greene Steve Benson Tim Tetreault Ken Soares

## 1978

Back Mark Lovejoy Paul Zeeley Tim Paiva George Karageorge Peter LaPan Bill Courtemanche Ed Hinton Mr. J. Bruce Davis
Front Mike Zeeley Scott Turcotte Jim Rice Tim Tetreault Steve Benson Ron Russo Tom Lavoie
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