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There are three degrees of freedom since there are four categories of combs. (4 1 3). The critical chisquare value with three degrees of freedom and probability of 0.05 7.815 (table 4.4). Since our calculated chi-square value (0.533) is less than this critical value, we cannot reject our hypothesis. In other words, our data are consistent with the hypothesis of a 9:3:3:1 phenotypic ratio, indicative of a two-locus genetic model with dominance at each locus.

At the next line of code, if no object has the ID TT_0 , it means this is the first time the plug-in has been called, so all the image objects are created and set to style positions of absolute (so that they can be moved about) In addition, their opacity is set to different levels so that the ones furthest away from the mouse cursor are the most faded, the images are loaded from the URL supplied in image, and the X and Y properties of each image are assigned starting values of 9999 to place them well off screen, as follows:

P R O B L E M S*

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Using the DataReader object, you can execute a loop to run through each row retrieved by the SQL query "Select * from customers where ContactName like 'M%' " This SQL query, by the way, will return all rows in the Customers table for which the ContactName column has a value that starts with the letter M Using the DebugWriteLine() method, the results are printed into the Immediate window, but you could use this data in many different ways by capturing it into a string variable The complete syntax for all of the properties and methods of the DataReader and Command objects can be found by exploring the ADONET type library in the Object Browser Although, to use these properly, you will have to spend considerable time studying more about them (see Appendix C for good sources to learn more).

if (!O('TT_0')) { for (var j = 0 ; j < 10 ; ++j) { var newimg = documentcreateElement('img') newimgsetAttribute('id', c + j) documentbodyappendChild(newimg) Position(newimg, ABS) Opacity(newimg, (j + 1) * 9) newimgsrc = image O(c + j)X = -9999 O(c + j)Y = -9999 } }

1. Assuming a 1:1 sex ratio, what is the probability that ve children produced by the same parents will consist of a. three daughters and two sons b. alternating sexes, starting with a son c. alternating sexes d. all daughters e. all the same sex f. at least four daughters g. a daughter as the eldest child and a son as the youngest (See also USE OF RULES) 2. Phenylthiocarbamide (PTC) tasting is dominant (T) to nontasting (t). If a taster woman with a nontaster father produces children with a taster man, and the man previously had a nontaster daughter, what would be the probability that a. their rst child would be a nontaster b. their rst child would be a nontaster girl c. if they had six children, they would have two nontaster sons, two nontaster daughters, and two taster sons d. their fourth child would be a taster daughter (See also USE OF RULES)

3. Albinism is recessive; assume for this problem that blue eyes are also recessive (albinos have blue eyes). What is the probability that two brown-eyed persons, heterozygous for both traits, would produce (remembering epistasis) a. ve albino children b. ve albino sons c. four blue-eyed daughters and a brown-eyed son d. two sons genotypically like their father and two daughters genotypically like their mother (See also USE OF RULES) 4. On the average, about one child in every ten thousand live births in the United States has phenylketonuria (PKU). What is the probability that a. the next child born in a Boston hospital will have PKU b. after that child with PKU is born, the next child born will have PKU c. two children born in a row will have PKU 5. In fruit ies, the diploid chromosome number is eight. a. What is the probability that a male gamete will contain only paternal centromeres or only maternal centromeres

With everything prepared, the final command in the setup section starts the repeating interrupts to the DoCurTrail() subfunction:

*Answers to selected exercises and problems are on page A-4.

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