Punnett Square Solutions

P-Square Practice Question 1:

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:

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.

P-Square 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.

ANSWER:

Genotypes of Offspring: 50% hybrid (Hh) and 50% homozygous recessive (hh)

Phenotype(s) of Offspring: 50% one-horn and 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?

P-Square Practice Question 3:

A green-leafed stinkbutt is crossed with a stinkbutt with yellow-striped leaves. The cross produces 185 green-leafed stinkbutts. Summarize the genotypes & phenotypes of the offspring that would be produced by crossing two of the green-leafed stinkbutts obtained from the initial parent plants.

ANSWER:

Genotypes of the F2 Offspring: 25% homozygous dominant (GG), 50% hybrid (Gg), 25% homozygous recessive (gg)

Phenotype(s) of F2 Offspring: 75% green-leafed and 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

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:

Summary of results:

Genotypes of the 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)

Phenotype(s) of F2 Offspring 3 of 4 boxes (75%) green-leafed, 1 of 4 boxes (25%) yellow-striped leaves

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?

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.

The p-square you drew should look something like this:

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.