Info > Genetics > Punnett square Practice page

Hello. On this page is a set of "typical" genetics questions that are best answered using a punnett square. It would be handy for you to have a pencil & some paper to work out the problems, & then you can click to see an explained solution to each.
For those who would benefit from a step-by-step explaination of how to use a p-square, click the link below to my "Baby Steps through the Punnet Square" page.

As always, do your best!

P-squARe prActICE QueSTioN #1

Let's say that in seals, the gene for the length of the whiskers has two alleles. The dominant allele (W) codes long whiskers & the recessive allele (w) codes for short whiskers.

a) What percentage of offspring would be expected to have short whiskers from the cross of two long-whiskered seals, one that is homozygous dominant and one that is heterozygous?
b) If one parent seal is pure long-whiskered and the other is short-whiskered, what percent of offspring would have short whiskers?

sOLUTioN TO QueSTioN 1

(Work it out 1st !)

P-sqARE PraCTice qUesTiON #2

In purple people eaters, one-horn is dominant and no horns is recessive. Draw a Punnet Square showing the cross of a purple people eater that is hybrid for horns with a purple people eater that does not have horns. Summarize the genotypes & phenotypes of the possible offspring.

SOlutiON TO qUeStion 2

(Work it out 1st !)

p-sqUaRe pRAcTicE QUestiON #3

A green-leafed luboplant (I made this plant up) is crossed with a luboplant with yellow-striped leaves. The cross produces 185 green-leafed luboplants. Summarize the genotypes & phenotypes of the offspring that would be produced by crossing two of the green-leafed luboplants obtained from the initial parent plants.

sOlUTion tO queStIoN 3

(Work it out 1st !)

P-squARE PRacTice qUeStION #4

Mendel found that crossing wrinkle-seeded plants with pure round-seeded plants produced only round-seeded plants. What genotypic & phenotypic ratios can be expected from a cross of a wrinkle-seeded plant & a plant heterozygous for this trait (seed appearance)?

soLutION to QueStiON 4

(Work it out 1st !)

NOTES:
There are only so many possible crosses that you could be asked about. They are:

PARENT GENOTYPES	OFFSPRING PHENOTYPES
pure (homozygous) dominant x anything	100% of offspring with dominant trait
hybrid x homozygous recessive	50% dominant trait, 50% recessive trait
hybrid x hybrid	75% with dominant trait & 25% with recessive trait
homozygous recessive x homozygous recessive	100% recessive trait

Seem like too much to memorize? Maybe it is. But the thing is if you can use the Punnett Square, you can work out ANY problem & get reliable results, so memorizing that chart ISN"T necessary

Understanding the vocabulary in the questions is extremely important. I've noticed that once students get the square set up they do just fine, it's that interpretation of the words in the question that they find most challenging. So LEARN THE VOCAB (pure/homozygous, hybrid/heterozygous, genotype, phenotype, cross, etc.).
The questions on this page are about as basic as they come. None of them involved any "advanced" genetic concepts like incomplete dominance, codominance, sex-linkage, or multiple alleles, we will practice those on another page. OK?

Solutions

P-squARe prActICE QueSTioN #1 - SOLUTION

In seals, the gene for the length of the whiskers has two alleles. The dominant allele (W) codes long whiskers & the recessive allele (w) codes for short whiskers.

a) What percentage of offspring would be expected to have short whiskers from the cross of two long-whiskered seals, one that is homozygous dominant and one that is heterozygous? ANSWER: 0%.

I personally like to write down the info given in the question on my paper first. So I start by writing:
W = allele for long whiskers
w = allele for short whiskers
A homozygous dominant seal would be "WW" (homozygous dominant = 2 CAPITAL letters).
A heterozygous seal would be "Ww" (heterozygous = 1 CAPITAL & 1 lowercase).
The cross is in the question therefore: WW x Ww.

The P-Square would look like this:

The possible gametes from the homozygous parent seal are on the left in front of the rows, & the possible gametes from the heterozygous parent are above the columns. We fill in the boxes by copying "one letter from the left, one letter from the top".

Analyzing our results, we find that 50% of our offspring (2 of 4 boxes) are "WW", and 50% (2 of 4 boxes) are "Ww". In terms of phenotype (what they would look like) 100% would have long whiskers (because all of the offspring have at least one "W", which codes for long whiskers).

So the answer to question 1a is: 0% would have short whiskers. The only way to have short whiskers is to be "ww", and that combo is not possible from the parents in this cross.

b) If one parent seal is pure long-whiskered and the other is short-whiskered, what percent of offspring would have short whiskers? ANSWER: 0%.

Again, I suggest starting by defining symbols like so:
W = allele for long whiskers
w = allele for short whiskers
"Pure" is the same as homozygous, so "pure long-whiskered" would be "WW".
If you're a seal, the only way to have short whiskers is to have the homozygous recessive genotype, in other words be "ww".
So our cross is: WW x ww.

The trusty p-square would be:

The alleles from the long-whiskered parent (WW) are out in front of the rows (at the left), & the alleles of the short-whiskered parent are above the columns. By the way, we could switch that around & it would not change our answer at all. What I'm saying is: it doesn't matter where you put the parents (top or side).
Anyway, all our offspring (4 of 4 boxes) have the same genotype: "Ww" & would all end up with long whiskers. To summarize the offspring:
genotype = 100% heterozygous (Ww)
phenotype = 100% long-whiskered.

So our answer to Question 1b is also: 0% would be short-whiskered.

TIP:
In any cross involving at least one parent that is homozygous dominant (2 CAPITAL letters),
100% of the offspring will have the dominant trait in their phenotype.
This is illustrated by Questions 1a & 1b.

BACK TO QUESTION 1

P-sqARE PraCTice qUesTiON #2 - SOLUTION

ANSWER:

Genotypes of Offspring	Phenotype(s) of Offspring
50% hybrid (Hh) 50% homozygous recessive (hh)	50% one-horn 50% no horns

No specific letter is given in the question to use as an abbreviation, so it's UP TO YOU! Being a real rebel, I'll use this:
H = dominant allele for one horn
h = recessive allele for no (zero) horns

A purple people eater that is "hybrid" has one of each letters (the definition of hybrid), so that parent is "Hh". A purple people eater without horns has the recessive phenotype and the only way to have a recessive phenotype is to have a homozygous recessive genotype, which is 2 lowercase letters, "hh".
So our cross for this question is: Hh x hh.

The p-square should be:

Alright, there we have it. The alleles carried in the sex cells of the purple people eaters are split up & placed "outside" the p-square. The alleles from the one-horn eater are on the left, and the alleles of the eater without horns are above each column. Copy one letter from the left & one from the top to fill-in the boxes. The combinations inside the boxses are the possible genotypes (with respect to horns) of purple people eater offspring from these two parent purple people eaters.
Analyzing the data is simple count how many of each genotype & phenotype are found in each of the four boxes. So, here we have 2 of 4 boxes "Hh" (50% hybrid, one horn), and 2 of 4 boxes "hh" (homozygous recessive, no horns).
Is you confidence soaring?

BACK TO QUESTION 2

p-sqUaRe pRAcTicE QUestiON #3 - SOLUTION

A green-leafed luboplant (I made this plant up) is crossed with a luboplant with yellow-striped leaves. The cross produces 185 green-leafed luboplants. Summarize the genotypes & phenotype of the offspring that would be produced by crossing two of the green-leafed luboplants obtained from the initial parent plants.

ANSWER:

Genotypes of the F2 Offspring	Phenotype(s) of F2 Offspring
25% homozygous dominant (GG) 50% hybrid (Gg) 25% homozygous recessive (gg)	75% green-leafed 25% yellow-striped leaves

OK, first let's jot down some letters & what they stand for. Since the parent luboplants have different leaf colors and 100% of the offspring resemble only one parent (i.e. they are all green), green is the dominant trait. It makes sense then to use:
G = dominant allele for green leaves
g = recesssive allele for yellow-striped leaves

TIP: This is important to recognize
When two parents have opposite traits, and all their offspring look like only one of the parents, the trait the offspring have is the DOMINANT TRAIT

The 185 "F1" offspring are all hybrids. How do I know? Lots of practice. The yellow-striped parent MUST BE "gg". The 185 offspring had to have inherited a "g" from that parent plant because that parent plant has no "G's" to pass on. Since the 185 offspring are ALL green, they must have a dominant allele for green ("G"), so their entire genotype is "Gg".
Don't believe me? That first cross must have been GG x gg, & its p-square would look like this:

Notice that 100% are hybrid (Gg) and 100% would look green. IF that green parent had "Gg" for a genotype, then we would get half of the offspring with a homozygous recessive genotype (gg), which would give us 50% yellow-striped luboplants. THIS IS NOT WHAT HAPPENED. The questions clearly states that all fo the 185 plants are green, pretty good evidence that green-leafed parent luboplant is "GG" & not "Gg".
The offspring of this cross, by the way, are refferred to as the "first filial" or "F1" generation.

Now, our question has to do with crossing two memebers of this F1 generation. That cross would be: Gg x Gg.
The punnett square showing this cross of two hybrids is:

Genotypes of the F2 Offspring	Phenotype(s) of F2 Offspring
1 of 4 boxes (25%) homozygous dominant (GG) 2 of 4 boxes (50%) hybrid (Gg) 1 of 4 boxes (25%) homozygous recessive (gg)	3 of 4 boxes (75%) green-leafed 1 of 4 boxes (25%) yellow-striped leaves

BACK TO QUESTION 3

P-squARE PRacTice qUeStION #4 - SOLUTION

ANSWER: 50% HYBRID ROUND-SEEDED, & 50% HOMOZYGOUS RECESSIVE WRINKLE-SEEDED

The first thing to figure out is which trait is dominant & which is recessive. We get this from the 1st sentence. If a wrinkled x round cross produces all round, then round is dominant & wrinkled is recessive.
Define our symbols:
R = dominant allele for round seeds
r = recessive allele for wrinkled seeds

Our wrinkle-seeded parent MUST be "rr", because the only way for a recessive trait to show up is if the genotype is homozygous recessive, which is 2 lowercase letters (rr). Our parent that is "heterozygous for this trait" is "Rr", because heterozygous = hybrid= 1 CAPITAL & 1 lowercase.
So our cross for this problem is: rr x Rr.

Again, you may have your "r's" on top & the "R" & "r" on the left, the combos inside the p-square will end up the same. No problem. Remember, "one from the left & one from the top" when you are filling in the boxes. Of the offspring in this cross, 2 of 4 (50%) are hybrid (Rr) and would have round seeds, and 2 of 4 (50%) are homozygous recessive (rr) and would have wrinkled seeds.
Good work.

BACK TO QUESTION 4