bicd 100 - lecture 9

bicd 100 - lecture 9 - LECTURE - 9 1. Dihybrid crosses 2....

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: LECTURE - 9 1. Dihybrid crosses 2. Branch diagrams 3. Binomial Expansion Dihybrid Cross Rr R RR Rr r Rr rr Y y Y YY Yy y Yy yy Branch Diagrams Rr r Rr rr r Rr rr Y y y Yy yy y Yy yy Dihybrid Testcross INCOMPLETE DOMINANCE Probabilities and Binomial Expansion A A Aa a Aa ¼ albino a Aa aa ¾ normal pigmentation If the heterozygous patients (Aa) have 3 children, what is the probability that 1 child has albinism? There are 3 possible scenarios (events): Child 1 Child 2 Child 3 Albino Pigmented Pigmented Pigmented Albino Pigmented Pigmented Pigmented Albino Probability for event 1 (first child is albino AND second child is pigmented AND third child is pigmented) to occur is the product of the individual probabilities (muliplication rule) 1/4x3/4x3/4=9/64 : event 1 Similarly, for events 2 and 3, the probabilities are: 3/4x1/4x3/4=9/64: event 2 3/4x3/4x1/4=9/64: event 3 Only one of these scenarios will actually occur, event 1 OR event 2 OR event 3. Hence, the probability of having one child with albinism and two without (in any order) is determined by adding the probabilities of events 1, 2 and 3 (addition rule). 9/64 + 9/64 + 9/64 = 27/64 Binomial Expansion 1 1 1 1 1 1 2 3 4 5 1 1 3 6 10 1 4 10 1 5 1 Pascal's triangle is a geometric arrangement of the binomial coefficients in a triangle. When progeny of crosses segregate into two distinct classes, we can calculate the binomial probability of any particular combination in another way using the following formula: s individuals fall into the class with probability a t individuals fall into the class with probability b In the albinism example (two heterozygous parents with three children, one of whom has albinism), s (= 1) would be the number of children with albinism, with probability a (= ¼) t (= 2) would be the number of children with normal pigmentation with a probability b (= ¾) For our other albinism example: five children three of whom have albinism: ...
View Full Document

Ask a homework question - tutors are online