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Now let’s consider diploid organisms:
The genotype of the zygote will depend on which alleles are carried in the gametes.
When heterozygotes mate their offspring will have different phenotypes: If AAAAA is domi- nant to aaaaa, the two possible phenotypes will be the phenotype of aaaaa/aaaaa or the phenotype of AAAAA/AAAAA and AAAAA/aaaaa.
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 either A/AA/AA/AA/AA/A or A/aA/aA/aA/aA/a. A method to control this type of variation is to start with populations 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 homozygous.
A / A A / a
a / A a / a
A a
A
a
sperm
egg
Allele in gamete
Zygote
True Breeding:True Breeding:True Breeding:True Breeding:True Breeding: homozygous for all genes
Say we have a true breeding line of shibire flies these flies are paralyzed and have geno- type shishishishishi–– – ––/shishishishishi – ––––.
First, we can test to see whether the shibire allele is dominant or recessive.
shishishishishi – ––––/shishishishishi – ––––^ x (wild type) shishishishishi +++++/shishishishishi +++++
all are shishishishishi – ––––/shishishishishi +++++
(The offspring from a cross of two true breeding lines is known as the F 1 or first filial generation). The F 1 flies appear like wild type therefore shishishishishi––– – –^ is recessive (not expressed in heterozygote)
Say we have isolated a new paralyzed mutant that we call parparparparpar.
We start with a true breeding parparparparpar–––––^ strain 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 parparpar–parpar––––^ is the same as shishishishishi–––––^ we can do a complementation test since both mutations are recessive. For this test, we cross a true breeding parparparparpar–––––^ strain to a true breeding shishishi–shishi––––^ strain.
parparparparpar – ––––/parparparparpar – ––––^ xxxxx shishishishishi – ––––/shishishishishi – ––––
F 1 (these flies must inherit both shishishishishi – ––––^ and parparparparpar – ––––)
Possible outcome Complementation? Explanation Inferred genotype
F 1 not paralyzed
F 1 paralyzed
Let’s look more carefully at gene segregation in a cross between F 1 flies.
shishishishishi – ––––/shishishishishi +++++^ xxxxx shishishishishi – ––––/shishishishishi +++++
What is the probability of a paralyzed fly in the next (F 2 ) generation?
parpar parparpar–––––^ genotype can supply function missing in (^) shishishishishi––––– and vice versa
parparparparpar–––––^ has lost function needed to restore (^) shishishishishi–––––
shishishishishi – ––––^ and parparparparpar – –––– complement
shishishi shishi^ – ––––^ and^ parparparparpar^ – –––– do not complement
parparparparpar – ––––/parparparparpar +++++ shishishishishi +++++/shishishishishi––– – –
shishishishishi – ––––/shishishishishi––– – –
How many genes contribute to the differences between the two kinds of plants?
Let’s designate the genes that differ as AAAAA, BBBBB, CCCCC, DDDDD ...............
For each gene there are two alleles: the allele present in Teosinte and the allele present in Maize.
For the AAAAA gene we will designate these alleles AAAAA (^) TTTTT and AAAAA (^) MMMMM respectively. For the BBBBB gene there will be alleles BBBBB (^) TTTTT and BBBBB (^) MMMMM and so on for all the genes that differ.
Let’s follow the AAAAA gene through the cross between Maize and Teosinte
AAAAA (^) TTTTT/AAAAATT (^) TTT x AAAAA (^) MMMMM/AAAAA (^) MMMMM F 1 : AAAAATT (^) TTT/AAAAA (^) MMMMM
Because the F 1 don’t look like either parent, let’s assume that the alleles are codominant.
CodominantCodominantCodominantCodominantCodominant: heterozygote different than either homozygote.
Incomplete dominance:Incomplete dominance:Incomplete dominance:Incomplete dominance:Incomplete dominance: heterozygote expresses the traits of both homozygous parents.
(Alternatively, the genes that differ could have a mixture of dominant and recessive alleles)
F 2 : AAAAA (^) TTTTT/AAAAA (^) TTTTT AAAAA (^) TTTTT/AAAAA (^) MMMMM AAAAA (^) MMMMM/AAAAAMM (^) MMM 1 : 2 : 1
1/4 will look like Teosinte.
For two genes that differ: AAAAA (^) TTTTT/AAAAA (^) TTTTT BBBBB (^) TTTTT/BBBBB (^) TTTTT
1/4 x 1/4 = 1/16 will look like Teosinte.
Similarly, for three genes the probability will be 1/64. For four genes it will be 1/256, and for five genes it will be 1/1024.
Since ~1/500 look like Teosinte the conclusion is that 4–5 genes differ between wild corn (Teosinte) and domestic corn (Maize). Using modern methods, it has been confirmed that there are about five significantly different alleles and several of these have been located using mapping methods.
