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uman population geneticists can increase the accuracy of their linkage analysis by using a probability technique, developed by Newton Morton, called the lod score method (Log Odds). The geneticist asks what the probability is of getting a particular pedigree assuming a particular recombination frequency ( ), as compared with getting the same pedigree assuming independent assortment ( 0.50). In other words, he or she calculates the ratio of the probability of genotypes in a family given a certain crossover frequency compared with the probability of those genotypes if the loci are unlinked. Logarithms are used for ease of calculation, and the parameter is called z, the lod score. Using this method, a researcher can try different crossover frequencies until the one giving the highest lod score is found. For example, take the pedigree in gure 6.28. The father in generation

We chose to use a variable called objCon to point to this object using this code: objCon = new DataOleDbOleDbConnection(objBuilderConnectionString) Next, we built our SQL Insert statement using the data passed into the BugData parameter, as shown here: strInsert = " Insert Into ReportedBugs " strInsert += " (Tester, AppName, Build, DateReported, Description) " strInsert += " Values( " + BugData + " ) " To issue the SQL command, we created a Command object from the ADO Command class and set the Connection, CommandType, and CommandText properties as needed: objComm = new DataOleDbOleDbCommand() objCommConnection = objCon objCommCommandType = DataCommandTypeText objCommCommandText = strInsert.

(0.25)8. This is the denominator of the equation. Thus, z z log (0.4375)7(0.0625) (0.25)8 1.099

To use it, just pass this plug-in the ID of an object whose state has already been saved, like this:

log [12.566]

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II can have one of two allelic arrangements: A NPS1/B nps1 or A/B NPS1/nps1. The former assumes linkage, whereas the latter does not. Our initial estimate of recombination, assuming linkage, was (1/8) 100, or 12.5 map units. We now need to calculate the ratio of two probabilities:

RestoreState('myobject')

and executes the query. If there is an error, using the Catch block provides an error message to help with troubleshooting. One common issue happens when the application does not find the database file. We have chosen to place the file on the root of the C:\ drive to make this less likely to happen; but when testing real applications, this is something to watch for.

probability of birth sequence assuming 12.5 map units log probability of birth sequence assuming independent assortment

The following example extends the previous plug-in to both create a div and then change it twice, the first time by modifying a few of its style properties and the second to change it back by calling RestoreState() In between, the JavaScript alert() function is called to give you a chance to view the screen before moving on:

Any lod score greater than zero favors linkage. A lod score less than zero suggests that has been underestimated. A lod of 3.0 or greater (103 or one thousand times more likely than independent assortment) is considered a strong likelihood of linkage. Thus, in our example, we have an indication of linkage with a recombination frequency of 0.125. Now we can calculate lod scores assuming other values of recombination, as table 1 does. You can see that the recombination frequency as calculated, 0.125 (12.5 map units), gives the highest lod score.

Newton E. Morton (1929 ).

<div id='mydiv'>Example Text</div> <script> windowonload = function() { S('mydiv')width S('mydiv')height S('mydiv')backgroundColor S('mydiv')color S('mydiv')position SaveState('mydiv') alert('Click OK to see some changes')

When the ExecuteNonQuery() method is called, the Insert statement is sent to Access. After it completes, we close the connection and redirect the user back to the default web page using ASP.NET s Response.Redirect() method. This example only allows the user to submit bug reports, but more features could be added later (such as the ability to edit and remove a bug report). To allow changes like these, you would use this same code, but modify the statements to delete and update the bug reports by executing those types of SQL statements.

(Courtesy of Dr. Newton E. Morton.)

= = = = =

Assuming 12.5 map units (or a probability of 0.125 of a crossover; 0.125), the probability of child III-1 is 0.4375. This child would be a nonrecombinant, so his probability of having the nail-patella syndrome and type A blood is half the probability of no crossover during meiosis, or (1 0.125)/2. We divide by two because there are two nonrecombinant types. This is the same probability for all children except III-5, whose probability of occurrence is 0.125/2 0.0625, since he is a recombinant. Thus, the numerator of the previous equation is (0.4375)7(0.0625). If the two loci are not linked, then any genotype has a probability of 1/4, or 0.25. Thus, the sequence of the eight children has the probability of

Recombination Frequency ( )

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