Helpful tips for crosses
Supplemental Instruction
Iowa State University / Leader: / Kalyca
Course: / BIOL 313
Instructor: / Vollbrecht
Date:

Below is the strategy I like to use when doing multi-trait crosses. This is just another way to approach the problem. There are other ways that work just as well such as the branching method.

Steps to approaching a dihybrid or multi-hybrid cross

  1. Separate the traits of interest into two different Punnett squares.
  2. Within each of the individual Punnett squares calculate the proportion of each type of genotype (i.e. AA, aa, or Aa).
  3. Read and understand what the problem is asking. But normally this is where you will multiply the individual probabilities or proportions of the separate genotypes together to get one complete genotype.

Ex: Crossing parents with AaBb and aaBb genotypes

  1. The two traits of interest are the A trait and the B trait
  2. So I will have two separate Punnett squares, one looking just at the A trait and the other looking at just the B trait. Then I will fill in the squares like I would with any monohybrid cross.

A trait:

a / a
A / Aa / Aa
a / aa / aa

Proportion of offspring with certain genotype:

AA = 0

Aa = ½

aa = ½

B trait

B / b
B / BB / Bb
b / Bb / bb

Proportion of offspring with certain genotype:

BB = 1/4

Bb= 1/2

bb = 1/4

  1. If the problem asked for the probability that first offspring will be aaBB then I take the proportion in the A trait that was aa genotype and multiply it with the proportion in the B trait that were BB. This can be thought of as an “and” problem where you have to multiple the events together because to get an individual with the genotype aaBB you have to have both the A trait AND the B trait coming together to give a complete genotype.

Probability of offspring having aaBB genotype= (probability of being aa) * (probability of being BB)

Probability of offspring having aaBB genotype= (1/2) * (1/4)

Probability of offspring having aaBB genotype= 1/8

This means that the first offspring has a 1/8 chance of having the genotype aaBB. This also means of the total offspring 1/8 of them should have this genotype.

Use this same procedure when doing larger crosses that include more traits. For example you could get parents AAbbCcDd and AaBBccDD. In this case you would have four traits of interest and so four “mini” or individual Punnett squares. The same basic steps mentioned above still apply, the only difference is you will have more individual Punnett squares and will be multiply more things together.

This way may work for some people but it may not work for others. Find what works best for you when working these problems! Hope this helped!