Trihybrid cross Assuming Independent Assortment

There are two possible solutions

• Classical Solution
• Simplify and Solve

Classical Solution

In Drosophila the following wild type alleles are used for analysis v (vermilion eyes), cv (cross-veinless) and ct (cut or snipped wing edges). Flies doubly recessive and homozygous (v+v+,cvcv,ctct) are crossed with flies singly recessive and homozygous (vv, cv+cv+, ct+ct+). Female progeny that are vv+, cvcv+, ctct+ are then test crossed with the tripley recessive males (vv, cvcv, ctct)

P1=v+v+,cvcv,ctct x vv, cv+cv+, ct+ct+

F1 =Wild Type - V+V, cv+cv, ct+ct x vv, cvcv, ctct (gametes v, cv, ct)

F1 gamete types			Obs.	Description
v	cv+	ct+	580	Parental
v+	cv	ct	592	Parental
v	cv	ct+	45	Crossover Region I
v+	cv+	ct	40	Crossover Region I
v	cv	ct	89	Crossover Region II
v+	cv+	ct+	94	Crossover Region II
v	cv+	ct	3	Double Crossover
v+	cv	ct+	5	Double Crossover
			1448

Finding the center gene by comparing parental to double crossover.

Map data
v<---13.2--->ct<--6.4-->cv

Can we calculate the Interference?
   The double cross overs are always the smallest in number. In this problem that is 5+3.   The expected frequency of double crossovers can be calculated by multiplying the   frequency of a crossover between v and ct by the frequency of a crossover between ct  and cv. 0.132 x 0.064 = 0.0084

The expected number of double crossovers is equal to the expected frequence of  double crossovers times the total number of individuals . 0.0084 x 1448 = 12

   Interference (I) = 1 -([# of dble c/o]/[exp # of dble c/o]) = 1- (8/12) = 0.33 }

Text iGenetics by Peter J. Russell

This web site is provided for instruction in Botany and Zoology 342

by Kenneth G. Wilson,
Professor of Botany
Miami University
wilsonkg@muohio.edu