DOC PREVIEW
Berkeley STATISTICS 246 - Mapping mouse traits

This preview shows page 1-2-3 out of 10 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 10 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 10 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 10 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 10 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

1How many genes?How many genes?Mapping mouse traits, cont.Mapping mouse traits, cont.Lecture 2B, Statistics 246January 22, 20042Let’s estimate the recombination fractionr between D12Mit51 and D12Mit132129326136Total2823 50B6594610H36 01026ATotal BHA13251 2-locus genotypes at D12Mit51 and D12Mit132.129 offspring from H×H, where A×B→H.3Estimation of r First note that we can’t simply count recombinants. Why?Because recombination can occur in the paternal or thematernal meiosis, or both, and all we see are the genotypes ofthe offspring. In most cases, the parental origin of therecombination can be inferred, but not in every case. Denoting the two markers by 1 and 2, the NOD alleles by a, and B6 alleles by b, then the parental haplotypes are a1a2 onone chromosome, and b1b2 on the other. Each parent passeson a1a2 with probability(1-r)/2, and similarly for b1b2 , while theypass on each of the recombinant haplotypes a1b2 and b1a2with probability r/2. In practice, recombinations have slightly different frequencies inmale and female meioses, but we ignore this refinement.4Probabilities of parentally transmittedhaplotype combinations (×4) Haplotype combinations resulting from crossing doublyheterozygous parents, each a1/b1 at locus 1 and a2/b2 at locus2. This table is for coupling: the parental haplotypes are a1a2and b1b2, i.e. the mother and father are both a1a2/b1b2. Here P and M denote the Paternally and Maternally transmittedhaplotypes, respectively.(1-r)2r(1-r)r(1-r)(1-r)2b1b2r(1-r) r2 r2r(1-r)b1a2r(1-r) r2 r2r(1-r)a1b2(1-r)2r(1-r)r(1-r)(1-r)2a1a2b1b2b1a2a1b2a1a2P M5From the Punnett square to the tableof 2-locus genotype probabilities Terms in the Punnett square table can be summed to build up atable of probabilities for the 9 different 2-locus genotypeprobabilities. For example, we observe A (=a1/a1 ) at locus 1 and H (=a2 /b2)at locus 2, if and only if the transmitted male and femalehaplotypes are the pairs a1a2 &a1b2 or a1b2 &a1a2 , and thisoccurs with a combined probability of 2r(1-r)/4. The other terms are built up similarly, the most complex casebeing the 2-locus genotype HH, where 4 different terms need tobe considered, corresponding to the fact that a doubleheterozygote can result from 4 different combinations ofparental or recombinant haplotypes.6Probabilities of 2-locus genotypes (×4)(1-r)22r(1-r)r2B2r(1-r)2[r2+(1-r)2]2r(1-r)Hr22r(1-r)(1-r)2ABHAL1 L2Looking at this table, we see that recombinations(or not) can be inferred, apart from the parent, in allbut the HH case. We can almost count recombinants.7Estimation of r, cont. Using the table of probabilities we can write down a loglikelihood function for any set of 2-locus frequencies. Label the cells of the table 1,…,9, and denote thecorresponding probabilities by p1(r) …,.p9 (r), and thefrequencies by n1, …, n9. Then the log-likelihood for theresulting multinomial model is log L = ∑i ni log pi (r). The parameter r is then estimated by maximizing thisfunction, and an approximate standard error orconfidence interval obtained using the Fisher informationor the asymptotic chi-square approximation.8A frill: the M-step of an EM-algorithm The function log L(r) can be maximized in a number ofways, but in general there is no closed formexpression for the maximum likelihood estimate r^. Ifwe were able to decompose the count n5 of HHs intothe n5P that are pairs of parental haplotypes, and n5Rthat are pairs of recombinant haplotypes, withfrequencies (1-r)2 and r2, resp, the recombinanthaplotypes can then be counted directly and the MLEis = 2(n3 + n7 + n5R)+ n2 + n4 + n6 + n8)/2n.€ ˆ r € ˆ r9The E-step In general we don’t know n5R but can estimate it using thefollowing formula: In practice, we need a value of r to begin with. Next we usethe above estimate, then get the next , and then iterate. Exercise: Prove the above formula, and that the iteration isan instance of the EM-algorithm.€ E (n5R| n5) =r2(1− r)2+ r2n5€ ˆ r102-locus genotype frequencies for D12Mit132 and D13Mit6131307130Total 32 621 5B 61172915H 38 72110ATotal BHA132 | 6 Exercise: Estimate r for these two loci. Is it different from


View Full Document

Berkeley STATISTICS 246 - Mapping mouse traits

Documents in this Course
Meiosis

Meiosis

46 pages

Meiosis

Meiosis

47 pages

Load more
Download Mapping mouse traits
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Mapping mouse traits and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Mapping mouse traits 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?