Question

(4 points) In Arabidopsis thaliana, two genes, HOT2 and STO, are involved in heat and salt tolerance, respectively. You want to make a plant that is resistant to both stresses. As a first step in the generation of this plant, you cross a true breeding plant with a mutation in HOT2 that causes increased heat tolerance with a true breeding plant that has a mutation in STO that causes increased salt tolerance. This cross produces Fi double heterozygotes that do not have either high heat or salt tolerance. These two genes are both 35 centimorgans apart on chromosome 1. If h and t are the mutant alleles and H and I are the wild-type alleles, what fraction of gametes produced by the Fi double heterozygote would you expect to have each of the following genotypes? Show your work. (3 points) A. 토 초 B. If the F, individuals undergo self-fertilization, what proportion of the F2 progeny would exhibit both high heat and salt tolerance? (1 point)

Answer 1

A. y trom the question it is T are wild a type alleles alloles. evident that & hand t the are it and mufant → If the genes are 35 cm abort, then recombination frequency ore penge pares fortion of recombinanto is 35.1, This proportion of parental genotypes is 651 ng dibybrid cross to obtar F X HH tt Cheat BTT resistant) (salt tolerant) Нh T- ouble beterezy gole with no beat on salt tolerance Gamefes produced by th Tt HThе не are 57 as follows : OXFORD Recombinant Parental Teacher's Signature ......

Page No. Date. Thus 35 % fraction of ore 35 each recombinant garmele will be 200 HT - 35 - I on 0.175 200 ho kt - 35 200 - I on 0.175 40 each parental gamele will be The fraction of 65.1 ore 65 HE = 65 = 200 OR 0.325 LO 200 T = 65 - 13 or 0.325 40 200 3. Self- Ferefilization of f individuals ie thit thTt x HITE HTT and I bhat (Frp progeny with both heat tolerance and salt tolerance) ceffer performing a construction (This result is arrived of Pennett Square) Teacher's Cinatura