Class Notes (810,833)
Canada (494,303)
Biology (2,100)
BIO207H5 (65)

Dominance and Breeding Types.pdf

4 Pages
Unlock Document

University of Toronto Mississauga
Steven M Short

Lecture 3 Lecture 3 Now let’s consider diploid organisms: The genotype of the zygote will depend on which alleles are carried in t▯he gametes. Allele sperm in gamete A a A A/A A /a egg Zygote a a/A a /a When heterozygotes mate their offspring will have different phenotApes: ▯If A is domi- nant toaa, the two possible phenotypes will be the phea aype of a/a or the phenotype of A A and A a. When we do breeding experiments it is important to know the genotypes of▯ the parents. But as you can see from the example above individuals with the dominant ▯trait could be eitherA/AA oA//aa. A method to control this type of variation is to start with populatio▯ns that we know to be homozygous. One way to do this is to keep inbreeding▯ individuals until all crosses among related individuals always produce identical offspring▯. This is known as a true-breeding population and all individuals can be assumed to be homo▯zygous. Trrue Brreedding:mozygous for all genes Say we have a true breeding line of shibire flies these flies are paraly▯zed and have geno- – – type shi sshi . First, we can test to see whether the shibire allele is dominant or rece▯ssive. shiisshii x (wild type) shii sshii ↓ all are shi /hiii shii shii (The offspring from a cross of two true breeding lines is known as the ▯F or fi1st filial generation). The F flies appear like wild type therefore sshiiis recessive (not expressed 1 in heterozygote) Say we have isolated a new paralyzed mutant that we call par. par – We start with a true breeding par parain that we mate to wild type. We find that the mutation is not expressed in the F 1 heterozygotes and therefore is recessive. To find out whether parpar the same as shi weshii do a complementation test since both mutations are recessive. For this test, we cross a true breeding par strpar to a – true breeding shshiirain. parr–/par– x shii/shii ↓ F1(these flies must inherit both shi shiipar )paar– Possible outcome Complementation? Explanation Inferred genotype shii and ppar– paar–genotype can supply parr ppar++ F1 not paralyzed complement function missing ishii and vice versa shii/sshii shiiand ppaar– parrhas lost function – – F 1paralyzed do not complement needed to restoreshii shhii shii Let’s look more carefully at gene segregation in a cross between F flies. 1 – + – + shii/shii x shii/shii What is the probability of a paralyzed fly in the next (F ) g2neration? Definition: p(a) =n a n = number of outcomes that satisfy conditionaa a N a aa N = total number of outcomes (of equal probability) Probability pro
More Less

Related notes for BIO207H5

Log In


Don't have an account?

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.