2. Your son is interested in obtaining a spotted floppy-eared rabbit for
entry into the fair. You have a male FfSs and a female ffss, and female rabbits
often produce about eight babies per litter. Figure out the ratio of poffspring
for each phenotype you can expect from crossing these rabbits, so you can
decide if it is likely that your son can enter the fair with a spotted floppy-eared
rabbit.
-
A. Ratio of the offspring with:
-
Monocolored fur and straight ears: One-fourth, or 25%
Monocolored fur and floppy ears: One-fourth, or 25%
Spotted fur and straight ears: One-fourth, or 25%
Spotted fur and floppy ears: One-fourth, or 25%
Each of the parent rabbits will pass on one of its alleles for spots and one
of its alleles for ear shape. The female rabbit can only pass on one combination
of these alleles: fs. The male, however, can pass on any one of four combinations:
FS, Fs, fS, and fs. This sets up a 4x4 Punnett square.
There will be four types of offspring, all equally common: FfSs,
Ffss, ffSs, and ffss.
Remember that mono-colored fur (F) is dominant over spotted fur (f), and straight
ears (S) is dominant over floppy (s). This means that you'd get 25% straight-eared
solid-colored rabbits (FfSs), 25% floppy-eared monocolored rabbits
(Ffss), 25% straight-eared spotted rabbits (ffSs), and 25%
floppy-eared spotted rabbits (ffss).
B. A male rabbit is mated to a whole flock of spotted floppy-eared rabbits. The number
of individuals and their phenotypes produced were: 30 monocolored straight-eared
rabbits; 27 monocolored floppy-eared rabbits; 29 spotted straight-eared rabbits;
and 31 spotted floppy-eared rabbits. Using what you know about the genotype of
the mothers, decide what the genotype of the father is.
-
The females are all ffss. For the male to father all these kinds of
baby rabbits, he must be FfSs -- this is the same cross as in 2A above.
You don't get exactly 25% of each phenotype -- it's fairly close, but
unless you have really huge numbers of offspring, there are usually noticeable
deviations from Mendel's predictions, just because of chance.
3. The ruffed chicken has an under-chin wattle. A smooth wattle (S) is
dominant over a wrinkled wattle (s). A red wattle (HR) is
incompletely dominant with a white wattle (HW) so that an individual
with HRHW will have a pink wattle.
-
A. A chicken with a wrinkled red wattle is mated to a homozygous chicken
with a smooth white wattle. What is the genotype and phenotype of the chicks?
-
Genotype of the parents: ssHRHR
x SSHWHW
Genotype of the chicks: SsHRHW
Phenotype of the chicks: smooth pink wattles
4. In turkeys, bronze body color (B) is dominant over red (b). Normal feathers
(N) are dominant over hairy feathers (n). A bronze male turkey is mated to a
bronze female, and some of the poults (baby turkeys) produced by this cross
are red. What are the genotypes of the male and female parents and the red
poults?
-
Ignore the hairy vs. normal feathers thing for a moment. In this cross, some of the
baby turkeys are red. Since bronze is dominant to red, this means that the red poults
must have the genotype bb. This in turn means that the red poults must
have received a b allele from Mom and a b allele from Dad. But we
know that Mom and Dad Turkey both have bronze feathers, so they must have either the
BB or the Bb genotype. The only way that both parents can be bronze,
but both can pass the red allele on to some of their babies, is for both parents to
have the Bb genotype.
5. Some of the poults from the cross described in #1 above are bronze.
What proportion of the poults would you expect to have the BB genotype?
What proportion would have the Bb genotype? What proportion would
have the bb genotype?
-
Do the Punnett square for a cross between a Bb parent and another Bb
parent -- you'll get 1/4, or 25% of your offspring, having the BB genotype;
1/4, or 25%, with the bb genotype; and 1/2, or 50%, with the Bb genotype.
6. If you were to cross a Bbnn turkey with a bbNn turkey,
what proportion of the offspring will have red, normal feathers? What
proportion will have bronze hairy feathers? red, hairy feathers? bronze,
normal feathers?
-
The Bbnn turkey passes on one of each pair of alleles to each of his kids.
He (let's say it's a he, although it's not important here) can only pass on the n
allele for hairy feathers, but he also passes on either the B or b
allele. So he can pass on either the combination Bn or the combination
bn. The other parent, by the same reasoning, can pass on either bN or
bn. So your Punnett square should look like this:
Bn bn
----------------------------
| | |
| BbNn | bbNn |
bN | | |
|--------------+-------------|
| | |
bn | Bbnn | bbnn |
| | |
|----------------------------|
You'll end up with one-quarter of the offspring with bronze, normal feathers;
one-quarter with red, normal feathers; one-quarter with bronze, hairy feathers;
and one-quarter with red, hairy feathers.
7. A breeder of hamsters crosses a golden hamster with a black hamster.
All the baby hamsters are golden. Which trait is dominant and which is
recessive? What are the genotypes of the parents? What are the genotypes
of the babies?
-
Golden fur is dominant to black fur. Let's write the golden fur
allele as G and the black fur allele as g. Thus the golden
parent must be GG and the black parent must be gg, which
means that all the offspring must have the genotype Gg.
8. The breeder takes one of the babies, raises it to maturity, and then
mates it to a black hamster. Use a Punnett square to predict the results
of this crossbreeding. Predict the genotype(s) and phenotype(s) of
the offspring, and the expected ratios.
-
The baby hamster has the genotype Gg, and the black hamster it's
mated with must have the genotype gg. So the Punnett square looks
like this:
G g
----------------------------
| | |
| Gg | gg |
g | | |
|--------------+-------------|
| | |
g | Gg | gg |
| | |
----------------------------
One-half of the offspring will have the Gg genotype and will have
golden fur. One-half of the offspring will have the gg genotype
and will have black fur.
9. In hamsters, long fur is recessive and short fur is dominant. Suppose
the breeder takes a hamster that is heterozygous for both golden fur
and short fur, and mates it to a hamster with long black fur. Use a
Punnett square to predict the results of this crossbreeding. Predict
the genotype(s) and phenotype(s) of the offspring, and the expected ratios.
-
OK, let's represent the allele for short fur as S and the allele for
long fur as s. The hamster with long black fur must have the genotype
ggss (any other genotype and he'd have golden and/or short hair).
The other parent I told you is heterozygous, which makes her have the genotype
GgSs. Now: The longhaired black hamster can only pass on the allele
combination gs. The heterozygous parent, however, can pass on any one of
four possible combinations: GS, Gs, gS, and gs.
That sets up the Punnett square, which should look more or less like this:
GS Gs gS gs
---------------------------------------------------------
| | | | |
gs | GgSs | Ggss | ggSs | ggss |
| | | | |
|--------------+-------------+--------------+-------------|
| | | | |
gs | GgSs | Ggss | ggSs | ggss |
| | | | |
|--------------+-------------+--------------+-------------|
| | | | |
gs | GgSs | Ggss | ggSs | ggss |
| | | | |
|--------------+-------------+--------------+-------------|
| | | | |
gs | GgSs | Ggss | ggSs | ggss |
| | | | |
---------------------------------------------------------
So that's. . . one-quarter of the offspring will have the GgSs genotype
and have short gold fur; one-quarter will have Ggss and have long gold
fur; one-quarter will have ggSs and have short black fur; and one-quarter
will have ggss and have long black fur.
10. Humans have four possible blood types (A, B, AB, and O) and these blood types are
controlled by three alleles (IA, IB, i). The
IA and IB alleles are codominant (they share
expression, thus we have an AB blood type), but they are both dominant over the
i allele.
-
A. List the possible genotypes from the phenotypes:
-
Blood Type A: IAIA,
IAi
Blood Type B: IBIB,
IBi
Blood Type AB: IAIB
Blood Type O: ii
B. A court case has been filed by a mother with type O blood who has a son
with type O blood. There are two fathers being accused; one has AB blood and
the other A blood. Which one of the men could be the father of the child?
-
The son, with type O blood, must have the genotype
ii. He must have received one i from his mother -- which
makes sense, because she has type O blood and thus the the genotype
ii as well -- and one i from his
father. The accused man with type AB blood cannot be the father, because his
genotype must be IAIB -- he does not have an
i allele to pass on to any of his children. The man with type A blood
is presumably the father, since he could have the IAi
genotype, with an i allele to pass on. (He could also have
IAIA blood, but that would mean that he's
not the father, and we're assuming that one of the suspects is the father.)
C. In another court case there are three possible fathers. The mother has type B
blood, and the child has type O blood. One suspect father has type AB blood,
another has type A blood (both his parents were AB), and the third father has type
A blood (one parent had AB and one parent had A blood). Construct a pedigree and
decide who the father is:
-
Again, the child has type O blood and so has the ii genotype. The mother
has type B blood, but she must have given an i allele to her child, so
her genotype must be IBi. The father must have contributed
the other i allele. . .
- Father 1: Blood type AB. He must have the IAIB
genotype. This means that he does not have an i allele to pass on to
any of his children. He cannot be the father.
- Father 2: Blood type A. He could have either the
IAIA or the IAi genotype. Which is it?
We know that both his parents had blood type AB. That means that both his parents
had the IAIB genotype. For Father 2 to have type
A blood, he would have to have received an IA allele from each
parent. This makes him IAIA. He cannot be the father.
- Father 3: Blood type A. Again, he could have either the
IAIA or the IAi genotype. If
he has IAIA blood, he's off the hook -- but we
are assuming that one of the three suspects is the father, and the other two
cannot be the father. So is there any way for Father 3 to have the
IAi genotype? Yes: One of his parents had type AB blood and
so had the IAIB genotype. The other had type A blood.
If that parent had the IAi genotype, then Father 3 could
have received IA from the parent with AB blood, and i
from the parent with A blood. That would mean that Father 3 would have an
i allele to pass on to the kid. Father 3 is a daddy!
11. Red-green color blindness, in humans, is a sex-linked trait controlled by
alleles on the X chromosome. Normal color vision (X+) is
dominant to colorblindness (Xc).
[NOTE: You could write this as XC for
the normal allele and Xc for the recessive colorblind
allele. Unfortunately, capital C and lowercase c
are hard to tell apart as superscripts, especially in type. I will use
X+ -- "X-plus" -- for the normal allele.]
-
A. If a colorblind man marries a woman with normal vision and they have
a colorblind son, what are the genotypes of the individuals?
-
Remember that men are XY and women are XX. A colorblind man
must be XcY, and his colorblind son must also be
XcY. Men pass on either an X chromosome or
a Y chromosome to their children -- and when a man has a son, he
passes on his Y chromosome; that's what makes his kid a son and not
a daughter. So the son did not get his colorblindness from Dad. This
means that Mom, who is female and therefore XX, must have the
X+Xc genotype, since we know she's not
colorblind.
B. If the mother and father were to have more children, what proportion
of the girls would be colorblind? Why?
-
We know that the man is XcY and the woman is
X+Xc. You can do the Punnett square, but in
brief, the answer is that half the girls would be colorblind. Why? A colorblind
woman must have the XcXc genotype. Each
girl gets one X chromosome from Dad and one from Mom. Each girl
automatically gets one Xc allele from Dad (because the
only other option is that she gets the Y chromosome from Dad, which
would make her a boy and not a girl). There's a fifty-fifty chance
that each girl will get Mom's X+ allele, in which case
she'll have the X+Xc genotype and normal vision;
and a fifty-fifty chance that she'll receive the Xc allele
from Mom, and end up with the XcXc genotype --
and be colorblind.
12. Budgies (Melanopsittacus undulatus) are attractive birds that come in
many colors. Budgies may produce either or both of two types of pigment in their
feathers: a blue pigment and a yellow pigment. Whether or not each pigment is
produced is governed by one gene with two alleles. Blue pigment (B) is
dominant to no blue pigment (b), and yellow pigment (Y) is dominant
to no yellow pigment (y). A budgie that produces neither blue nor yellow
pigment in its feathers looks white. A budgie that produces both blue and yellow
pigment in its feathers looks green.
- A. What is/are the possible genotype(s) of a yellow budgie?
- Either bbYY or bbYy. Just YY or
Yy isn't correct; you have to take both genes into account. (A YY
budgie could be either yellow or green, depending on whether or not it also carries
B alleles.)
- B. What is the phenotype of a budgie with the genotype BbYy?
- It's green.
- C. Suppose you cross a homozygous blue budgie with a homozygous white budgie.
Predict the results of this cross, both genotype and phenotype.
- Both budgies are homozygous -- meaning that the two
genes in each pair are both of the same allele. A blue budgie also can't produce
any yellow pigment -- if it did, it wouldn't be blue, but green. So the blue parent
must have the genotype BByy. The white parent, which produces no blue
and no yellow, must be bbyy. The blue parent can only pass on the allele
combination By to its offspring, the white parent can only pass on the
combination by, and the result is that all the bouncing baby budgies will
have the genotype Bbyy, and all will have blue feathers.
- D. Now suppose you cross one of the offspring from the mating in part C above
with a budgie whose genotype is BbYy. Predict the results of your cross.
What are the expected genotypes and phenotypes of the offspring from this cross,
and in what ratios?
- OK, you're crossing a Bbyy budgie with a BbYy
budgie. The Bbyy budgie may pass on either of two combinations of alleles to its
offspring: either By or by. However, the BbYy budgie may pass on
any one of four allele combinations to each of its babies: BY, By,
bY, or by. So the Punnett square looks like this:
By By by by
---------------------------------------------------------
| | | | |
BY | BBYy | BBYy | BbYy | BbYy |
| green | green | green | green |
|--------------+-------------+--------------+-------------|
| | | | |
By | BByy | BByy | Bbyy | Bbyy |
| blue | blue | blue | blue |
|--------------+-------------+--------------+-------------|
| | | | |
bY | BbYy | BbYy | bbYy | bbYy |
| green | green | yellow | yellow |
|--------------+-------------+--------------+-------------|
| | | | |
by | Bbyy | Bbyy | bbyy | bbyy |
| blue | blue | white | white |
---------------------------------------------------------
In other words, out of all the offspring from the cross, three-eights of the
babies would be green, three-eighths would be blue, one-eighth would be yellow,
and one-eighth would be white.