How do you solve the phenotype and genotype of a punnet square?

Answer 1

Well it depends on how many of each that you have. Let's start simple. Let's say your mom has brown eyes which is a dominant gene meaning it has a capital B and your dad has blue eyes which is a recessive gene which means it has a lower case b. So mom can either be Bb or BB in order to have brown eyes, but dad is definitely bb because he has blue eyes. So let's just say mom is BB. So for this punnet square you would only need four squares. On the top two you put a B and a B on each square which stands for mom's gene. On the side of the square you put b and a b on each of the two squares which stand for dad's genes. Now all you do is write the new gene in each box. For the top left hand box you would write bB since that b is what dad gives and B is what mom gives. For the top right hand box you write bB again since that is what mom and dad is giving off. Then you do the bottom two squares the same way. Now let's say we want to add height to this. Dad is tall so he is T. Mom is short so she is t. You use the punnet squares again in the exact same way. Place mom's genes on the top and dad are on the bottom. You can also figure out probabilities of genotypes and phenotypes by doing fractions. Someone who is BB or bb is homozygous (means same) for that gene. Someone who is Bb is heterozygous (means opposite). Anyone who is homozygous dominant (BB) that mates with a homozygous recessive (bb) person will result in a punnet square with ¼ Bb, ¼ Bb, ¼ Bb, and ¼ Bb meaning that all of their children will have brown eyes. Anyone who is homozygous dominant (BB) that mates with a heterozygous (Bb) will have ¼ BB, ¼ Bb, ¼ Bb, and ¼ bb. So ¾ of the children will be brown eyed and ¼ blue eyed. If you have heterozygous (Bb) mate with a homozygous (bb) you will have ½ Bb and ½ bb. Using fractions would be easier if you can't get the punnet squares correct.

Below is an example of how to do more than one trait.

Ex. In peas, tall (D) is dominant to dwarf (d) and yellow cotyledons (G) is dominant to green (g). If a tall, heterozygous pea plant with green cotyledons is crossed with a dwarf pea plant heterozygous for yellow cotyledons, what will be the phenotypic results in the progeny?

Ddgg x ddGg

Dg dg x dG dg

dG

dg

Dg

DdGg

Ddgg

dg

ddGg

ddgg

½ D ½ d 1g ½ Dg

Probability of tall ½

Probability of short ½

Yellow ½

Green ½

Tall and yellow (1/2 x ½)= ¼

Tall and green (1/2 x ½)= ¼

Dwarf and yellow (1/2 x ½)= ¼

Dwarf and green (1/2 x ½)= ¼