Probability for Genetics

Punnett Squares

It is often of interest to predict the probabilities of offspring inheriting certain traits. Punnett squares are a useful tool for predicting these probabilities.

We will examine it with the following example: a heterozygous, blue oddish (Bb) is crossed with a homozygous, green oddish (bb).

You start by determining what potential alleles the parents can pass on. If P1 has the genotype Bb, that means they can pass on a B or a b to their offspring. Likewise, the bb oddish can only pass on a b.

Set up a 2x2 square, with one parent's potential alleles written on the top of the square, and the other's written on the left of the square, as can be seen in the square to the left.

Then, you fill in the four empty squares.


The way that it works is that you look for where the parent alleles intersect. For example, in the top left corner, it would receive a B from the parent along the side, and a b from the parent on top, giving it Bb as its genotype. Continue this process for all four squares.

Click here to see the answer!

You can use this square to determine the ratio of genotypes in comparison to each other (genotypic ratio) or to determine the ratio of phenotypes in comparison to each other (phenotypic ratio).

In this example, here are our ratios:

Genotypic Ratio: 1 Bb : 1 bb

Phenotypic Ratio: 1 Blue : 1 Green


This means there is a 50% chance of the offspring being blue, and a 50% chance of it being green!

Multiple Traits

Things get more complicated when you look at multiple traits in a single cross.

The rules work the same way - you just need to make a bigger square. You need to start by determining possible gametes from the parent genotypes.

I recommend using the FOIL method to determine these. 

First - combine the first of both genes

Outside - combine the first and last allele

Inner - combine the middle alleles

Last - combine the second of both genes

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As a note, the offspring genotypes are written by gene. OoPp is correct, but OPop would not be.

Here, we have a phenotypic ratio of 

9 brown/no pattern : 3 orange/no pattern : 3 brown/patterned : 1 orange/pattern

This square has 16 possibilities for offspring as opposed to the 4 from a cross using a single locus. If we wanted to look at three genes, there would be 64 squares inside it. Fortunately, there are ways to skip this by doing some simple math.

A Simpler Way

In koopas, green-shelled (G) is dominant to red-shelled (g). Wingless (W) is dominant to having wings (w). Two green -shelled, wingless koopas are heterozygous at both loci and are crossed.

While we could make a 4x4 Punnett Square like we did in the example above, there is a much simpler way. It starts with making a separate square for each trait.

Shell Color Square

So, based on this square, we have the following probabilities:

3/4 Green

1/4 Red

Wing Presence Square

So, based on this square, we have the following probabilities:

3/4 Wingless

1/4 Wings

AND

What's the chance of having a koopa who is red and winged?

OR

What's the chance of having a koopa who is red or winged?

To find out the chance of __ and __, you need to multiple the probabilities together.

1/4 x 1/4 = 1/16

To find out the chance of __ or __, you need to add the probabilities together.

1/4 + 1/4 = 1/2

Let's Get Some Practice

In house elves:

floppy ears (E) are dominant to non-floppy ears (e)

large ears (L) are dominant to small ears (l)

capability to do magic (M) is dominant to not being able to do magic (m)

being bald (H) is dominant to having hair (h)


If two house elves that are heterozygous for all traits have kids, what is the chance that they have a baby with small, non-floppy ears, hair, and who is able to do magic?

Click here to see the answer!

We are looking for an offspring who is eellM_hh. If you do a Punnett Square for each trait, you will see the following probabilities for offspring phenotypes:

1/4 small ears

1/4 non-floppy ears

1/4 hair

3/4 magic


1/4 x 1/4 x 3/4 x 1/4 = 3/256