The $\chi^2$ value means nothing on its own-it is used to find the probability that, assuming the hypothesis is true, the observed data set could have resulted from random fluctuations. A low probability suggests that the observed data are not consistent with the hypothesis, and thus the hypothesis should be rejected. A standard cutoff point used by biologists is a probability of 0.05(5%). If the probability corresponding to the $\chi^2$ value is $0.05$ or less, the differences between observed and expected values are considered statistically significant, and the hypothesis (that the genes are unlinked) should be rejected. If the probability is above $0.05$, the results are not statistically significant; the observed data are consistent with the hypothesis. To find the probability, locate your $\chi^2$ value in the $\chi^2$ Distribution Table in Appendix $F$. The "degrees of freedom" (df) of your data set is the number of categories (here, 4 phenotypes) minus 1 , so $\mathrm{df}=3$. Determine which values on the $d f=3$ line of the table your calculated $\chi^2$ value lies between.

The STRIP study (Special Turku Coronary Risk Factor Intervention Project for Children) is a longitudinal randomized prevention trial where repeated dietary counseling aiming at reducing intake of saturated fat took place from infancy to age 7. Subjects who had complete data on components of the metabolic syndrome (MetS) (waist circumference, blood pressure, triglycerides, glucose, and HDL cholesterol) at ages 15-20 were included in the study. There was an intervention group who received dietary counseling and a control group who received only standard clinical care.

One of the goals of the study was to look at the correlation among components of the MetS. The correlation coefficient between waist circumference and HDL-cholesterol ( $\mathrm{HDL}-\mathrm{C})$ was $-0.24$ among the 243 subjects in the intervention arm.

The mean $\pm$ sd for waist circumference was $72.8 \pm 7.8 \mathrm{~cm}$.

The mean $\pm$ sd for $\mathrm{HDL}-\mathrm{C}$ was $20.7 \pm 4.1 \mathrm{mg} / \mathrm{dl}$.

Determine the slope and intercept of the least squares regression line of HDL-C on waist circumference, and write out the equation of the least squares regression line. Hint: Use the relationship between the regression and correlation coefficient to estimate the regression coefficient of HDL-C on waist circumference.

Muons are the only particles that can demonstrate time dilation and length contraction.

Inspect the graph of the function to determine whether it is concave up, concave down, or neither, on the given interval. (If necessary, review Section $2.2$).

The identity function, $f(x)=x$, on $(-\infty, \infty)$

In this exercise, use vertical and horizontal representative rectangles to set up integrals for finding the area of the region bounded by the graphs of the equations. Find the area of the region by evaluating the easier of the two integrals.

$$y=1-\frac{x}{2}, y=x-2, y=1$$

Select the correct answer for the following question:
The perception of vision occurs in the
a. optic nerve.
b. occipital lobe of the cerebrum.
c. retina.
d. frontal lobe of the cerebrum.

Write an explicit rule and a recursive rule using the sequence. 12, 26, 40, 54, 68

Integrate $f(x, y)=x+y$ a. over the region $0 \leq x^2+y^2 \leq 1, \quad x \geq 0, y \geq 0$; b. over the region $1 \leq x^2+y^2 \leq 4, \quad x \geq 0, y \geq 0$.

A student whose hobby was fishing pulled a very unusual carp out of Cayuga Lake: It had no scales on its body. She decided to investigate whether this strange nude phenotype had a genetic basis. She therefore obtained some inbred carp that were pure-breeding for the wild-type scale phenotype (body covered with scales in a regular pattern) and crossed them with her nude fish. To her surprise, the $\mathrm{F}_{1}$ progeny consisted of wild-type fish and fish with a single linear row of scales on each side in a 1 : 1 : 1 ratio. a. Can a single gene with two alleles account for this result? Why or why not? b. To follow up on the first cross, the student allowed the linear fish from the $\mathrm{F}_{1}$ generation to mate with each other. The progeny of this cross consisted of fish with four phenotypes: linear, wild type, nude, and scattered (the latter had a few scales scattered irregularly on the body). The ratio of these phenotypes was 6 : 3 : 2 : 1, respectively. How many genes appear to be involved in determining these phenotypes? c. In parallel, the student allowed the phenotypically wild-type fish from the $\mathrm{F}_{1}$ generation to mate with each other and observed, among their progeny, wild-type and scattered carp in a ratio of 3 : 1. How many genes with how many alleles appear to determine the difference between wild-type and scattered carp? d. The student confirmed the conclusions of part c by crossing those scattered carp with her pure-breeding wild-type stock. Diagram the genotypes and phenotypes of the parental, $\mathrm{F}_{1},$ and $\mathrm{F}_{2}$ generations for this cross and indicate the ratios observed. e. The student attempted to generate a true-breeding nude stock of fish by inbreeding. However, she found that this was impossible. Every time she crossed two nude fish, she found nude and scattered fish in the progeny, in a 2 : 1 ratio. (The scattered fish from these crosses bred true.) Diagram the phenotypes and genotypes of this gene in a nude $\times$ nude cross and explain the altered Mendelian ratio. f. The student now felt she could explain all of her results. Diagram the genotypes in the linear $\times$ linear cross performed by the student (in part b). Show the genotypes of the four phenotypes observed among the progeny and explain the 6 : 3 : 2 : 1 ratio.

Find each of these differences mentally using equal additions. Write out the steps that you thought through.

b. $62-39$

Show that if $\alpha=a+b \sqrt{-5}$ and $\beta=c+d \sqrt{-5}$, where a, b, c. and d are integers. then $N(\alpha \beta)=N(\alpha) N(\beta).$

Prove each identity.

$$\tan \left( \frac { \pi } { 4 } + \alpha \right) = \frac { 1 + \tan \alpha } { 1 - \tan \alpha }$$

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.

$$2,2 i , 4 - \sqrt { 6 }$$

Use the Product Rule or the Quotient Rule to find the derivative of the function. f(x) = (5x²+8) (x²-4x-6)