Figure: 03-06 Caption: Computation of the combined probabilities of each F 2 phenotype for two independently inherited characters. The probability of each plant being yellow or green is independent of the probability of it being round or wrinkled.
Figure: 03-09 Caption: Generation of the F 2 trihybrid phenotypic ratio using the forked-line method.
Punnett Squares• A Punnett square is a grid that enables one to predict the outcome of simple genetic crosses• Proposed by the English geneticist, Reginald Punnett
Plant genes can have dominant and recessive allelesExample: The dominant allele R R produces … … a red flower r The recessive allele r produces … … a white flower
A genetic cross in plants: Rr x RrThere is only one way to make a homozygous plant withtwo dominant genes… R R RR… and only one way to make a homozygous plant withtwo recessive genes r r rr
In the genetic cross Rr x Rr there are two ways of making a heterozygous plant… R r RrOR r R Rr… so, heterozygous offspring are twice as likely as eitherhomozygous dominant or homozygous recessive offspring
We can show the likeliness or probability of differentgenotypes in different ways:1. As a percentage - Probability of RR is… 25% Probability of Rr is… 50% Probability of rr is… 25%2. As a fraction - Probability of RR is… ¼ Probability of Rr is… ½ Probability of rr is… ¼3. As a ratio - RR : Rr : rr = 1 : 2 : 1
There are three ways to make a plant with red flowers…… one produces a homozygous plant…R R RR … and two produce heterozygous plants… R r Rr r R Rr
…but, there is only one way to produce a plant with white flowers…r r rrSo, in the genetic cross, Rr x Rr… …red flowers are three times more likely than white flowers!!!
We can show the likeliness or probability of differentphenotypes in different ways:1. As a percentage - Probability of red is… 75% Probability of white is… 25%2. As a fraction - Probability of red is… ¾ Probability of white is… ¼3. As a ratio - red : white = 3:1
The Punnett SquareWe can use the Punnett Square to work out the probability of the different genotypes and phenotypes in a genetic cross Step 1: Write down the genotypes of the parents. x = Rr x Rr Step 2: Write down the genotypes of the gametes that each parent produces. pollen Egg cells = R+r = R+r
Step 3: Draw your Punnett Square Egg cellsStep 4: Write the genotype R rof the different gametesinto your Punnett Square R RR R r PollenStep 5: Write in thedifferent ways in which the r rR rrgametes can be combined……this shows you thepossible genotypes of the Genotypes ofoffspring offspring
Probability of Genotypes Egg cellsThere are four possible ways tocombine egg cells and pollen R rCount up how many timeseach genotype appears in R RR R r Pollenthe Punnet Square – thisgives you its probability: r rR rr Probability of RR is… ¼ = 25 % Probability of Rr is… 2 4 = ½ = 50 % The ratio of Probability of rr is… ¼ = 25 % RR : Rr : rr is… 1:2:1
Probability of Phenotypes Egg cellsLook at the genotypes of theoffspring in your Punett Square… R r…decide which phenotypeeach one should produce. R RR R r PollenCount up how many timeseach phenotype appears in r rR rrthe Punnet Square – thisgives you its probability:Probability of red is… ¾ = 75 % The ratio of red : white is…Probability of white is… ¼ = 25 % 3:1
Mendel’s Experiments• Mendel also performed dihybrid crosses – Crossing individual plants that differ in two traits• For example – Trait 1 = Seed texture (round vs. wrinkled) – Trait 2 = Seed color (yellow vs. green)• There are two possible patterns of inheritance for these traits
DATA FROM ONE OF MENDEL’S DIHYBRID CROSSESP Cross F1 F2 generation generationRound, All round, 315 round, yellowYellow seeds yellow seedsX wrinkled, 101 wrinkled, yellowgreen seeds seeds 108 round, green seeds 32 green, wrinkled seeds
Interpreting the Data• The F2 generation contains seeds with novel combinations not found in the parentals – Round and Green – Wrinkled and Yellow• These are nonparentals• Occurrence contradicts the linkage model
• Independent assortment is also revealed by a dihybrid test-cross – TtYy X ttyy• Thus, if the genes assort independently, the expected phenotypic ratio among the offspring is 1:1:1:1