Now let’s consider diploid organisms:
The genotype of the zygote will depend on which alleles are carried in t▯he gametes.
in gamete A a
A A/A A /a
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
(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
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.
– + – +
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)