tHe
pUnneTT SquaRE
prACTice
PagE
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?
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.
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.
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)?
NOTES: ·
There
are only so many possible crosses that you could be asked about. They
are:
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: |
P-sqARE PraCTice qUesTiON #2 - SOLUTION
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 |
Phenotype(s) of Offspring |
50%
hybrid (Hh) |
50%
one-horn |
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?
Note: As far as I know, purple people eaters do not exist. They are ficticious creatures. |
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) |
75%
green-leafed |
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, |
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 |
Phenotype(s) of F2 Offspring |
1
of 4 boxes (25%) homozygous dominant (GG) |
3
of 4 boxes (75%) green-leafed |
P-squARE PRacTice qUeStION #4 - SOLUTION
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.
Good
work.