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Critters to Grow
Meiosis and Mendelian Genetics
MATERIALS
Aluminum cans, tin cans, newspaper, plastic bottles, glass bottles/jars, etc. Something to hold them
together with: string, tape, or glue.
Blue and pink construction paper, scissors, pencil, 2 plastic baggies
Overview:
You are making 2 parent critters. Both parents will be heterozygous for all traits. The
parents will undergo meiosis to produce sperm and eggs. One egg will be fertilized
with one sperm to produce a zygote. The zygote will grow to become a baby critter.
Day 1: PLANNING
In your notebook: write notes for each step of the process. Be sure to number
each step to go with the steps on this protocol.
1. Pick materials that you can use for your critter: a body might be a 2-L bottle, a
soda can, an old ball, etc.
2. List the physical characteristics (traits) of your critter. [A trait might be “eye
color”, “skin type”, “amount of feathers”.] You must have at least 5 traits.
3. For each trait, pick two phenotypes. [Eye color might be yellow or pink.]
4. Randomly choose which trait will be the dominant phenotype and which will be
the recessive phenotype. [Suggestion: Toss a coin.]
5. Create a genotype symbol for each phenotype. The dominant phenotype will use
a capital letter, and the recessive will use the same letter, just lower-case.
Geneticists use letters be used to represent alleles.
•
•
A capital letter = Dominant trait, a lowercase letter = a recessive trait.
The same letter is used to indicate both alleles.
Examples
= Flower color: P= purple, p= white
= Seed color: Y= yellow, y = green
= Seed shape: W = wrinkled, w = round
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Yellow or pink eyes would be “Y” for yellow eyes; “y” for pink eyes. No two
traits can have the same letter, so if there are two traits that will have the same
letter, choose the second or third letter of the dominant trait.
6. Make a list of the traits with the dominant and recessive phenotypes and their
genotype symbols. This is an example of the chart you will make.
Trait
Eye Color
Dominant
phenotype
Yellow
Recessive
phenotype
Pink
Dominant
genotype
Y
Recessive
genotype
y
7. Use pink construction paper to make mom's chromosomes. Cut strips of equal
size (approx. 1 cm X 8 cm).
8. Label your chromosomes with the appropriate allele, using the genotype
symbols you have on your chart.
9. Label 2 more strips, each with the letter “X”. These are the sex genes.
10.If you have been assigned a sex-linked trait (such as color-blindness), label one
of the “X” alleles as “XC”
11.If you have been assigned to include blood type as a trait, label 3 strips with
these symbols: “IA”, “IB”, “i”. Blood type is an example of a multiple allele
trait that is also co-dominant (“IA” and “IB” are dominant over the recessive “i”,
but “IA” and “IB” are co-dominant). Turn the strips over so that you cannot see
the genotype symbols and mix them up. Randomly choose 2 of the strips. This
is your parent critter blood type. IA and i will be blood type “A”; IB and i will be
blood type “B”; IA and IB will be blood type “AB”.
12.If your team has been assigned a polygenic trait, choose 8 more strips and label
with the letter for the polygenic trait. (Polygenic traits are those like skin color
or height – there are an infinite number of possibilities.) Each of the strips will
have the same letter, 4 of them will be uppercase, and 4 will be lower-case.
13.Take all of the pink strips (mom critter’s chromosomes) and place them in a
plastic baggie. The baggie represents the nucleus of a diploid somatic (body)
cell. A somatic cell is any cell in the body other than gametes (sex cells).
14.Repeat steps #7, #8 using blue construction paper for the dad critter.
15.For step #9, label 2 blue strips, one with the letter “X”, and the other with the
letter “Y”. These are the sex genes for the dad critter.
16.Do not complete step #10 for the dad critter.
17.Complete steps #11, #12, and #13 for the dad critter using blue paper.
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18.Decide on a sexual dimorphism for your critter. Choose a genotype symbol and
place it on the “Y” chromosome for the dad critter to exhibit the trait.
Which phenotype will BOTH mom and dad critter be for ALL of their traits?
Circle the answer: dominant
or recessive?
19. Sketch both mom and dad critter. The polygenic trait will be right in the
middle. ie: If you choose “skin color” to be polygenic (ex: from very light
green to very dark green), your critter will be medium green.
20.Have your teacher sign your team’s planning sheet.
Day 2: PARENT CONSTRUCTION
Construct the parent critters.
Remember that each of the parents will look the same except for sexual
dimorphism.
Day 3: MEIOSIS
1. On four sheets of paper, draw a circle on each one so that the circle will take up
all the space on the paper. Label each paper “Mom Critter’s egg cell”.
2. Take each strip of paper (the chromosomes) and duplicate the genotype symbol
that it has on it. [If the strip of paper has a capital letter “Y” on it, you will write
another capital letter “Y” on it.] This is the “S” Synthesis portion of the cell
cycle when replication of genetic material takes place.
3. Cut each chromosome (strip of paper) so that a genotype symbol appears on
each half.
4. In meiosis, the homologous chromosomes must be lined up next to each other.
[The two dominant (upper-case “Y” pink strips) must be lined up next to the
recessive (lower-case “y”) alleles.]
5. Turn the four strips of one trait over so that you cannot see the genotype letters
and “stir” them. Randomly choose 1 of the strips and place it in the first “cell”
on the “Mom critter’s egg cells” paper so that you can see the genotype letter.
Choose another of the strips and place it in the second cell; the third strip in the
3rd cell, and the last strip in the 4th cell.
6. Repeat step #3 for all the traits.
7. Glue the strips onto the paper so that all the genotype letters can be seen. These
are haploid gametes (sex cells).
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8. Draw one circle on each of 4 sheets of paper so that the circle will take up all the
space on the paper. Draw a flagellum on each of these circles. Label each paper
“Dad Critter’s sperm cell”.
9. Repeat steps #2 through #5 for the dad critter traits.
FERTILIZATION
1. Turn Mom Critter’s egg cells upside down and mix them up. Randomly choose
one egg cell.
2. Turn Dad Critter’s sperm cells upside down and mix them up. Randomly
choose one sperm cell.
3. FERTILIZE the egg and sperm. Unite the egg and sperm cells by pairing up the
chromosomes so their alleles/genotypes match. Make a chart like the one below.
Baby critter trait
Genotype
Phenotype
(Blue "A" or "a" matches to either pink "a" or "A".) When an egg and a sperm
fertilize, a single cell called a zygote forms which grows into an embryo, then a
fetus, and then a new baby. CONGRATULATIONS! You are now the proud
expectant parents of a new baby Critter.
4. You went in to have an amniocentesis done, where doctors draw amniotic fluid
to obtain some of the baby's cells. They then can predict the baby's
chromosomes and genotype. Record the new baby's gender in your charts.
5. Record the new baby's phenotype. Sketch the new baby (Baby's first photo!)
6. Have your teacher sign for approval.
Day 4: THE BIRTH!
7. Construct the new critter baby.
8. CONGRATULATIONS! YOU ARE NOW THE PROUD PARENTS OF A
HEALTHY WEE CRITTER! SHOW HIM/HER OFF! BE PROUD.
9. Now calculate the percent probability that it took for the two parents to produce
that baby. Construct a Punnett square for each of your traits.
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Notes for the teacher:
Steps 10, 11, 12, 17 and 18 of the planning can be omitted. Those concepts are
not in the Next Generation Sunshine State Standards.
Punnett squares are not included in the standards.
Encourage students to make “critters”; not any recognizable animal, since “X”
and “Y” chromosomes do not determine gender the same way in all creatures.
Be sure to emphasize the process of meiosis, including independent assortment
and crossing over; explain how this process leads to genetic variation.
Also emphasize that meiosis is a reduction division of the genetic material
resulting in haploid gametes (or spores).
Some Suggested Questions:
Offspring that result from crosses between parents with different traits
a.
are true-breeding.
c.
make up the parental generation.
b.
make up the F2 generation.
d.
are called hybrids.
Answer: D
If a critter has a recessive allele for green eyes, it will produce
a.
green eyes if it also has a dominant allele for yellow eyes.
b.
both green and yellow eyes if it also has a dominant allele for yellow eyes.
c.
green eyes if it does not also have a dominant allele for yellow eyes.
d.
yellow eyes if it does not also have a dominant allele for green eyes.
Answer: C
Organisms that have two identical alleles for a particular trait are said to be
a.
hybrid.
b.
homozygous. c.
heterozygous.
Answer: B
d.
dominant.
The number of chromosomes in a gamete is represented by the symbol
a.
Z.
b.
X.
c.
N.
Answer: C
d.
Y.
If an organism’s diploid number is 12, its haploid number is
a.
12.
b.
6.
c.
24.
Answer: B
d.
3.
Gametes have
a.
homologous chromosomes.
b.
twice the number of chromosomes found in body cells.
c.
two sets of chromosomes.
d.
one allele for each gene.
Answer: D
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Gametes are produced by the process of
a.
mitosis.
b.
meiosis.
Answer: B
c.
crossing-over.
What is shown in the figure above?
a.
independent assortment
b.
anaphase I of meiosis
Answer: C
c.
d.
crossing-over
replication
d.
Unlike mitosis, meiosis results in the formation of
a.
diploid cells.
c.
b.
haploid cells.
d.
Answer: B
2N daughter cells.
body cells.
Unlike mitosis, meiosis results in the formation of
a.
two genetically identical cells.
c.
b.
four genetically different cells.
d.
Answer: B
four genetically identical cells.
two genetically different cells.
Which of the following assort independently?
a.
chromosomes
b.
genes on the same chromosome
Answer: A
multiple alleles
codominant alleles
c.
d.
replication.
In humans, a male has
a.
one X chromosome only.
b.
two X chromosomes.
c.
one X chromosome and one Y chromosome.
d.
two Y chromosomes.
Answer: C
Human females produce egg cells that have
a.
one X chromosome.
b.
two X chromosomes.
Answer: A
c.
d.
one X or one Y chromosome.
one X and one Y chromosome.
If a man with blood type A and a woman with blood type B produce an offspring, what might be the
offspring’s blood type?
a.
AB or O
c.
A, B, AB, or O
b.
A, B, or O
d.
AB only
Answer: C
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Most sex-linked genes are located on
a.
the autosomes.
b.
the X chromosome only.
c.
the Y chromosome only.
d.
both the X chromosome and the Y chromosome.
Answer: B
Colorblindness is more common in males than in females because
a.
fathers pass the allele for colorblindness to their sons only.
b.
the allele for colorblindness is located on the Y chromosome.
c.
the allele for colorblindness is recessive and located on the X chromosome.
d.
males who are colorblind have two copies of the allele for colorblindness.
Answer: C
What is the difference between genotype and phenotype?
a.
Genotype is the general term for chromosomes; phenotype is the specific term for chromosomes.
b.
Genotype is the information in the genetic material; phenotype is the expression of the information the way it looks.
c.
Genotype is the same as the gametes; phenotype is another name for the zygote.
d.
Genotype is the expression of the information; phenotype is the information in the genetic material.
Answer: B
A rooster with gray feathers is mated with a hen of the same phenotype. Among their offspring, 15
chicks are gray, 6 are black, and 8 are white. What is the simplest explanation for the inheritance of
these colors in chickens? What offspring would you predict from the mating of a gray rooster and a
black hen?
Answer: P = Gray x Gray; F1 = 15 Gray, 6 black, 8 white. Say that BX = black, WX = white, and WB =
gray. Then a WB x WB should result in a ratio of 1:2:1 for white (WW), gray (WB), and black (BB),
respectively. The ratio observed is 1.1:2.0:0.8. Therefore I would conclude that the simplest
explanation is the semidominance (i.e., incomplete dominance) where gray is the heterozygote (one
loci, two alleles). The phenotypes of the offspring which would arise from a mating between a gray
rooster and a black hen would have to been half black and half gray since we know that the black hen
has to be a homozygote (otherwise she would be grey).
A black guinea pig crossed with an albino guinea pig produced 12 black offspring. When the albino
was crossed with a second black one, 7 blacks and 5 albinos were obtained. What is the best
explanation for this genetic situation? Write genotypes for the parents, gametes, and offspring.
Answer: The first black guinea pig was homozygous for fur color, the second was a heterozygote.
Called the black allele B and the albino allele O, the parental genotypes are BB, OO, and BO,
respectively. The first guinea pig produced only B gametes, the second only O gametes, and the third
both gamete types in a 1:1 ratio. The offspring from the first mating all had the genotype BO. The
offspring from the second mating were nearly 1:1 BO and OO
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Color patterns in a species of duck is determined by one gene with three alleles. Alleles H and I are
semidominant (i.e., incomplete dominance), and allele i is recessive to both. How many phenotypes
are possible in a flock of ducks that contains all the possible combinations of these three alleles.
Answer: Possible genotypes are HH, HI, Hi, II, Ii, and ii. Note that Hi has the same phenotype
as HH and Ii has the same phenotype as II. Therefore there are six minus two possible
phenotypes. Note that the genetics of this system are identical to the genetics of the ABO
blood group.
A man has six fingers on each hand and six toes on each foot. His wife and their daughter have the
normal number of digits. Extra digits is a dominant trait. What fraction of this couple's children would
be expected to have extra digits?
Answer: One-half. Why? The wife and daughter we know are homozygous recessive from
their phenotype. If this man can father a child which does not have the dominant allele, then
the man must be a heterozygote. Therefore, half of his children will carry the dominant allele,
so half will express the dominant phenotype.
In cats with the Manx trait, the M (dominant) allele causes a short or absent tail, whereas the m allele
confers a normal, long tail. Cats of genotype MM die as embryos. If two Manx cats mate, what is the
probability that each living kitten has a long tail?
Answer: You would argue that short tails is the trait exhibited by the heterozygote, and long and
short tails by the two homozygotes. That is, you would argue that inheritance of tail length in cats is
controlled by a one loci, two allele system which displays semidominance (i.e., incomplete
dominance). If a double dose of the allele T results in long tails and a double dose of the allele N
results in no tails, then short tails would be associated with the genotype TN.
In 1986, National Geographic magazine conducted a survey of its readers' abilities to detect odors.
About 7% of Caucasians in the United States could not smell the odor of musk. If both parents could
not smell musk, then none of their children were able to smell it. On the other hand, two parents
who could smell musk generally had children who could smell it, too, but a few of the children in
those families were unable to smell it. If a single pair of alleles governs this trait, is the ability to smell
musk best explained as an example of dominant or recessive inheritance?
Answer: This is probably an example of recessive inheritance.