Mendel Quiz
1. Who was Gregor Mendel?
a) He was Charles Darwin’s British colleague
b) He was an Austrian monk
c) He was a botanist
d) He was an American environmentalist
Formative feedback
a) Darwin actually had no knowledge of Mendel’s work as Mendel’s work was thought to be
about hybridization instead of inheritance.
b) Mendel was born into a German family in the Heinzendorf bei Odrau region of Austrian
Empire and began his training as a priest in 1843. Maybe monks have a lot of time to think?
c) Although Mendel studied pea plants for his research, Mendel was actually majoring in
Physics.
d) Mendel was born in Austria and was an Austrian. Back then in the 1800s, do they even have
environmentalist?!
2. What was Gregor Mendel’s contribution to science?
a) His discoveries on finches initiated the theory of natural selection
b) His work on dinosaurs led to the theory of evolution
c) His work on pea plants inheritance initiated the study of modern genetics
d) His discovery on DNA molecule allowed for deeper understanding of the structure of genes
Formative feedback
a) Darwin was studying finches during his travels to the Galapagos Islands.
b) Mendel did not work on dinosaurs (although, quite cute to think about Mendel in his priest
robe, digging for bone evidence).
c) Mendel’s interest in studying inheritance stemmed from his professor, Johann Nestler. After
leaving the university, Mendel focused his pea plants study at the St. Thomas Abbey.
d) Watson and Crick discovered the structure of DNA molecule.
3) Mendel believed that the traits of the pea plants are determined by
a) inheritance of genes from both parents
b) inheritance of genes from the dominant parent
c) health of the parent plants at the time of pollination
d) types of pollinating insects available in the season
Formative feedback
a) Sexually reproducing organisms obtain their genes equally from both parents.
b) Sexually reproducing organisms obtain their genes equally from both parents, whether the
genes are dominant or recessive will determine the expression of the genes.
c) Health of the parent plants do not have an effect on which genes are passed on to the
offspring.
d) Genes are passed on regardless of external factors, such as pollinating insects.
4) The idea that different pairs of alleles are passed to offspring separately is Mendel’s principle
of :
a) unit inheritance
b) F1 generation law
c) segregation
d) independent assortment
Formative feedback
a) Genes and alleles are inherited, but inheritance does not explain how the different pairs of
alleles are passed on.
b) Alleles are passed on separately regardless of which generation it is.
c) Principle of segregation explains you get one allele from each parent.
d) Alleles of genes does not travel together, therefore are independent of each other.
Use the following diagram to answer question 5, 6 and 7.
Image obtained: http://naturalselection.0catch.com/Files/gregormendel.html
5) Assuming that both pea plant parents in the diagram above are homozygous, why would all
the f1 generation have yellow phenotypes?
a) because yellow phenotypes are dominant over green phenotypes
b) because both parents passed on the yellow alleles
c) because the f1 genotypes are homozygous
d) because the f1 genotypes are heterozygous
Formative feedback
a) Even though one of the parents is a recessive green pea plant, all the offspring are yellow;
therefore yellow is a dominant trait over green.
b) The green pea plant parent does not have the yellow allele, if it did, the phenotype would be
yellow.
c) The f1 genotypes cannot be homozygous if there is a green pea plant as one of the parents.
d) The f1 genotypes IS heterozygous, although the question asks for the reason for why the
yellow phenotype is expressed (shown) and not what genotype it is.
6) Predict the dominance of the colour phenotypes and the most likely genotypes of the f1
offspring possessing yellow colour
a) yellow offspring’s genotype is YY and yellow is the dominant gene.
b) yellow offspring’s genotype is Yy and yellow is the dominant gene.
c) yellow offspring’s genotype is YY and green is the dominant gene.
d) yellow offspring’s genotype is Yy and green is the dominant gene.
Formative feedback
a) Yellow offspring’s genotype cannot be YY due to the contribution of the “y” allele from the
green pea parent.
b) Yellow offspring’s genotype is Yy because the offspring gets one “Y” allele from the yellow
pea parent, and one “y” allele from the green pea parent. Yellow trait is indeed dominant as the
f1 generations are all yellow in colour.
c) Yellow offspring’s genotype cannot be YY due to the contribution of the “y” allele from the
green pea parent. Green is the recessive gene.
d) Yellow offspring’s genotype is Yy because the offspring gets one “Y” allele from the yellow
pea parent, and one “y” allele from the green pea parent. Green trait is not dominant as the f1
generations are all yellow in colour.
7) Assuming Y is the dominant allele, YY=yellow phenotype, Yy= yellow phenotype and
yy=green phenotype. Predict the most likely genotypes of the f2 offspring possessing yellow
colour and the green colour.
a) yellow offspring’s genotype is YY and green offspring’s genotype is YY.
b) yellow offspring’s genotype is Yy and green offspring’s genotype is YY.
c) yellow offspring’s genotype is YY and green offspring’s genotype is Yy.
d) yellow offspring’s genotype is Yy and green offspring’s genotype is yy.
Formative feedback
a) Yellow offspring’s genotype cannot be YY because f2 generation is heterozygous, possessing
one “Y” and one “y.” Green is the recessive trait; therefore the green pea plant’s genotype has to
be yy.
b) Yellow offspring’s genotype is Yy. Green is the recessive trait; if the green plant is “YY”, the
phenotype would be yellow.
c) because f2 generation is heterozygous, possessing one “Y” and one “y.” Green is the
recessive trait; if the green plant is “YY”, the phenotype would be yellow.
d) Yellow offspring’s genotype is Yy after obtaining one “Y” and one “y” from the yellow
heterozygous parents. Green offspring’s genotype is yy.
8) The idea that pair of alleles of either parent separate and only one allele from each parent is
passed on to the offspring is Mendel’s principle of :
a) unit inheritance
b) F1 generation law
c) segregation
d) independent assortment
Formative feedback
a) Genes and alleles are inherited, but inheritance does not explain how the each parent provides
one allele to the offspring.
b) Alleles are passed on separately regardless of which generation it is.
c) Principle of segregation explains you get one allele from each parent.
d) Independent assortment explains the alleles are not binded together, and are passed on
separately regardless of the genes.
9) Which of the following is not the reason Mendel selected pea plants to study?
a) Pea plants have a short life cycle
b) It is easier to breed pea plants
c) Pea plants could be cross-pollinated and self-pollinated
d) Pea plants produce pretty flowers
Formative feedback
a) Pea plants (and all plants in general) has a shorter life cycle, and will be easier to study the
generations compared to animals.
b) Plants are breed by putting the pollen (equivalent of plant sperm) of one plant, to the stigma
(equivalent of the female parts of the plant) of another plant. Each plant has both male and
female parts.
c) Cross-pollination refers to breeding with other plants, and self-pollination refers to breeding
with the same plant (basically having sex with your own self. Quite useful if you wanted to
focus on the particular genes the plant possesses).
d) Production of flowers is not an important factor relating to inheritance of the pea plant.
10) Which of the following was not a conclusion Mendel came up with at the conclusion of his
experiments on pea plants?
a) The inheritance of each trait is decided by “units” or “factors”
b) The number of offspring per plant affects the trait that is passed on in the previous generation
c) Traits can skipped generations
d) For each trait, an individual inherits one “units” or “factors” from each parent
Formative feedback
a) Mendel didn’t know what to call what he observed was being passed on through pea plants,
thus he called it “units” or “factors.” It was later coined alleles and genes.
b) The number of offspring per plant depends on the number of times it was bred and has nothing
to do with inheritance.
c) It is possible for an offspring to have their grandparents’ traits.
d) Since there are two parents, genes are obtained equally from each parent.
11. Genes are passed on
a) from one parent to the offspring
b) equally from both parents’ genes
c) from parent that is the same gender as the offspring
d) via the nucleus of the offspring’s cell
Formative feedback
a) Sexually reproducing organisms has two parents, and therefore obtain genes from both
parents.
b) Sexually reproducing organisms has two parents, and therefore obtain genes from both
parents.
c) Offspring obtain genes from each parent. If we only receive genes from the same gender
parents, all the daughters will look like their moms (gasp!) and all the sons will look like their
dads.
d) Genes are contained in the nucleus of the sperm and egg and not from the offspring’s cells (as
it hasn’t formed yet)
12. An allele is:
a) another word for a gene
b) a homozygous phenotype
c) a type of chromosome found in the cell
d) one of the possible forms of a gene
Formative feedback
a) Close, but not quite. For each gene, there are at least two variations (which you get from each
parent).
b) An allele is only one version of the gene. We don’t yet know if it is homozygous (same
alleles such as: YY, or yy) or heterozygous (different alleles such as: Yy).
c) An allele is a version of the gene found on the chromosome.
Image obtained from: http://www.lhasa-apso.org/articles/general/graphics/alleles.gif
d) Yes! Forms of the gene for eye colour could be: blue eyes, green eyes, brown eyes etc.
13. Genotype refers to
a) what genes the individual possess
b) the physical appearance of the individual
c) recessive alleles of the individual
d) inheritance of genes from the parents
Formative feedback
a) A tip to remembering this genetics vocabulary is if you see “geno-,” then most likely it is
related to genes.
b) The appearance of the individual is the phenotype.
c) We don’t yet know whether the genes are recessive or dominant yet. We need to see both
alleles to determine that.
d) Genes are inherited from parents, but the question asks for the definition of genotype.
14. Phenotype refers to
a) what genes the individual possess
b) the physical appearance of the individual
c) dominant alleles of the individual
d) homozygous genes from the parents
Formative feedback
a) The genes that are possessed by individuals are called genotype
b) Yes, genes determine what proteins you make; and thus what you look like.
c) We don’t yet know whether the genes are recessive or dominant yet. We need to see both
alleles to determine that.
d) We don’t yet know whether the genes are homozygous or heterozygous yet. We need to see
both alleles to determine that.
15. Which of the following is an example of homozygous dominant genotype?
a) height gene
b) short gene
c) TT
d) tt
Formative feedback
a) We don’t know if height is a dominant or recessive gene without looking at the alleles.
b) We don’t know if being short is a dominant or recessive gene without looking at the alleles.
c) Homozygous refers to same type. Therefore, having 2 of the same “TT” is correct. Dominant
refers to capital letters (meaning it is stronger than the lower case letters).
d) Homozygous refers to same type. Therefore, having 2 of the same “tt” is correct. Dominant
refers to capital letters. This is actually an example of homozygous recessive.
16. Which of the following is an example of homozygous recessive genotype?
a) height gene
b) tall gene
c) TT
d) tt
Formative feedback
a) We don’t know if height is a dominant or recessive gene without looking at the alleles.
b) We don’t know if being short is a dominant or recessive gene without looking at the alleles.
c) Homozygous refers to same type. Therefore, having 2 of the same “TT” is correct. Recessive
refers to lower case letters, thus the latter part is incorrect.
d) Homozygous refers to same type. Therefore, having 2 of the same “tt” is correct. Recessive
means you need 2 lower case versions of the alleles to show that trait (because it is not as strong
as the dominant alleles).
Answer Key
1) b
2) c
3) a
4) d
5) a
6) b
7) d
8) c
9) d
10) b
11) b
12) d
13) a
14) b
15) c
16) d
Punnet Square Quiz
1. A pair of mice parents that are black in colour produced some offspring that are white in
colour. Assuming black mice has “B” alleles, the genotypes of the parents are most probably
a) BB and bb
b) BB and Bb
c) Bb and Bb
d) bb and bb
Formative feedback
a) If both mice parents are black in colour, and “B” is the dominant allele because of its
capitalization, both mice will have at least one “B” on their alleles. The “bb” mouse is incorrect.
b) If both mice parents are black in colour, each of the mice parent will have at least one “B” on
their alleles. However, the white offspring must have “bb” to be white in colour because white is
the recessive trait. This meant that each parents has to donate “b” to the white offspring.
c) If both mice parents are black in colour, each of the mice parent will have at least one “B” on
their alleles. We know “B” is dominant because of its capitalization, thus to be white in colour,
you will need “bb.” This meant that each parents has to have one “B” to be black in colour, but
still have one “b” to donate to the white offspring.
d) If both mice parents are black in colour, and “B” is the dominant allele because of its
capitalization, both mice will have at least one “B” on their alleles. Two “bb” meant that neither
parents are black in colour.
2. Two heterozygous parents are crossed. What is the predicted ratio of the resulting offspring
with the dominant phenotype versus the recessive phenotype?
a) 1:1
b) 2:1
c) 3:1
d) 4:1
Formative feedback
a) Heterozygous parents have one dominant and one recessive allele. Drawing out the Punnet
square will help you with this question. Here is how to help you get started:
b) Heterozygous parents have one dominant and one recessive allele. Drawing out the Punnet
square will help you with this question. Here is how to help you get started:
c) Heterozygous parents can either donate one “A” or one “a.” The results indicate that in 3 out
of the 4 Punnet square boxes, at least one allele will be “A.” Since capital letters are dominant
over lower case letters, three of the four possible offspring will have the dominant phenotype.
d) Heterozygous parents have one dominant and one recessive allele. Drawing out the Punnet
square will help you with this question. Here is how to help you get started:
3. Most breeders of Dalmation dogs want dogs that are white in colour with small black spots.
One breeder bred two white Dalmation dogs with small black dots and the offspring produced
contained some offspring with solid white coats, some offspring with large black spots and the
majority of dogs with small black spots. What is the probable genotype of the Dalmatians with
small black spots?
a) homozygous dominant
b) homozygous recessive
c) heterozygous
d) sex-linked recessive
Formative feedback
a) Since the parents also possess small black dots, and the offspring that are produced varies in
spots, the trait small black dots cannot be dominant (if it was, then there would not be so many
phenotypic variation in offspring).
b) Since the parents also possess small black dots, the small black spots traits are not recessive.
c) Since there are 3 different variations in traits seen in the offspring, we know that this is not a
simple dominant recessive relationship. Another clue is that the majority of the offspring
possess small black dots (same as parents), therefore heterozygous proportion must be larger.
d) We do not have enough information to know if the traits are sex-linked, but if spots are sexlinked, you will see female dogs have more of one colouring than male dogs.
4. In guinea pigs, the phenotype for fur are rough hair (dominant) and straight hair (recessive). If
two heterozygous guinea pigs are crossed, the largest number of any one genotype of offspring
would be
a) homozygous straight hair
b) homozygous rough hair
c) heterozygous rough hair
d) intermediate between rough hair and straight hair
Formative feedback
a) Straight hair is recessive, and therefore offspring will only have 25% chance of having straight
hair.
b) Rough hair is dominant, and although the chances of having rough hair phenotype are high,
the question asks about genotype (genes).
c) 50% of the offspring will be heterozygous.
d) Rough hair and straight hair is actually a dominant and recessive gene. There is no in between
trait between rough and straight hair.
5. If an individual possesses two identical alleles for a certain trait. This individual’s gene is said
to be
a) homozygous
b) heterozygous
c) dominant
d) hybrid
Formative feedback
a) Homozygous (homo=same), so two identical alleles are homozygous.
b) Heterozygous (hetero=different), so heterozygous will have one different allele each.
c) Two identical alleles could be dominant or recessive. We cannot assume it is dominant.
d) Hybrid meant that there are different alleles on the same genes.
6. The physical appearance of a particular trait in an organism is referred to as a(n)
a) genotype
b) phenotype
c) allele
d) chromosome
Formative feedback
a) Genotype refers to the genes of the organism.
b) Phenotype is what you look like. (Yes, blame the big nose on that).
c) Allele is the variation of the gene.
d) Chromosome is the genetic information found in the nucleus of the cell.
7. The genetic disease called cystic fibrosis is a recessive gene. If both parents are heterozygous
for this trait, what is the probability that their offspring will be heterozygous?
a) 25%
b) 50%
c) 75%
d) 100%
Formative feedback
a) Heterozygous parents will have one dominant capital letter allele and one recessive lower case
allele. The chance of the offspring being heterozygous is higher than 25%.
b) Yes! 50% of the offspring will have one dominant capital allele and one recessive lower case
allele.
c) Heterozygous parents will have one dominant capital letter allele and one recessive lower case
allele. The chance of the offspring being heterozygous is lower than 75%.
d) Heterozygous parents will have one dominant capital letter allele and one recessive lower case
allele. There is a chance that some offspring is going to be homozygous.
8. In fruit flies, long wing is dominant over vestigial wing. If two flies that are homozygous
dominant and homozygous recessive for this trait are crossed, what is the probability that their
offspring will be heterozygous?
a) 25%
b) 50%
c) 75%
d) 100%
Formative feedback
a) The homozygous dominant parent will donate one dominant allele to the offspring. The
chances of the offspring being heterozygous are higher.
b) The homozygous dominant parent will donate one dominant allele to the offspring. The
chances of the offspring being heterozygous are higher.
c) The homozygous dominant parent will donate one dominant allele to the offspring. The
chances of the offspring being heterozygous are higher.
d) Yes! Surprising, this combination can only produce one type of trait for the wing gene. The
homozygous dominant parent can only donate one dominant allele to the offspring and the
homozygous recessive parent can only donate one recessive allele to the offspring.
9. In peas, tall is dominant over short. If a heterozygous plant is crossed with a homozygous
recessive plant, what is the probability that the offspring will be short?
a) 25%
b) 50%
c) 75%
d) 100%
Formative feedback
a) Since the homozygous recessive parent has two short alleles, the chances of having a short
offspring are higher than 25%.
b) Yes, 2/4 boxes in the Punnet square will have two lower case alleles.
c) The heterozygous parent will have a chance to donate one capital letter dominant allele 50%
of the time.
d) The heterozygous parent will have a chance to donate one capital letter dominant allele,
therefore at least a portion of the offspring will be tall.
10. Brown hair color is dominant to blond hair color. If two brown-haired parents (both are
heterozygous) have one blond-haired child, what is the probability that their second child will
have brown hair?
a) 25%
b) 50%
c) 75%
d) 100%
Formative feedback
a) Draw out the Punnet square by putting the heterozygous parents outside of the square.
Heterozygous parents will have one capital letter allele and one lower case letter allele. Brown
hair should be more dominant than 25%.
b) Draw out the Punnet square by putting the heterozygous parents outside of the square.
Heterozygous parents will have one capital letter allele and one lower case letter allele. Brown
hair should be more dominant than 50%.
c) Yes! 3 out of the 4 possible offspring will have at least one dominant allele, and will have
brown hair.
d) Draw out the Punnet square by putting the heterozygous parents outside of the square.
Heterozygous parents will have one capital letter allele and one lower case letter allele. If both
parents donate one lower case allele to the child, there is a chance the offspring will be blond.
11. In squash, the allele for white fruits (W) is dominant over the allele for yellow (w). If a
white-fruited plant is crossed with a yellow-fruited plant and 50% of the offspring are yellow,
what are the possible genotypes of the parents and the offspring?
a. parents ______________________________
b. offspring ____________________________
Formative feedback
a) The first hint is that white fruits are dominant over yellow fruits. We know one of the parents
is a white-fruited plant and one of the parents is a yellow-fruited plant. That tells you that for the
white-fruited parent, he/she will have at least one “W” and for the yellow fruited parent, he/she
will have two “ww.” Do the guess work in a Punnet square to find out what alleles the whitefruited parent will have to produce 50% (2/4) of yellow offspring.
b) After completing the Punnet square in question 11a, you should discover there are four
possible offspring. We already know the two yellow-fruited offspring’s genotype is “ww” (it has
to be to be the recessive trait). Now figure out what the other 50% of the white-fruited
offspring’s genotype is.
In humans, there are two alleles for the hairline trait. The dominant trait is “H” with a widow’s
peak and the recessive trait is “h” for non-widow’s peak.
Image obtained from http://1.bp.blogspot.com/_Wkff0lN4kzo/R0nC6wqrwI/AAAAAAAAAGc/XllkIxtFMF4/s320/img_Bio_00129.jpg
12. A woman that is homozygous dominant and a man that is heterozygous have a child.
Calculate the probability that their child will have a widow’s peak.
a. Complete the following Punnet square. Place the female alleles on the first 2 rows, and the
male alleles on the first row of the second and third columns.
Formative feedback
Homozygous (homo=same) means that the woman’s two alleles has to be the same (either capital
or both lower case). We know the woman has the dominant alleles (it said so in the question),
therefore the woman’s alleles are HH. We are halfway there. To figure out the man’s allele, the
question states he is heterozygous. Wait a minute, how come it didn’t tell me if he was dominant
or recessive….does it matter? Heterozygous (hetero=different), so we know there will be one
allele that is capital and one allele that is lower case. The man’s allele is going to be Hh. Now
you do the rest to fill out the Punnet square.
b. Complete the following table by writing the genotypes, genotype proportion annd phenotypes
of the child this couple will have.
Genotype
Genotype Proportion
Formative feedback
a)
b)
c)
d)
c. The possible phenotype(s) of the child with the percentage/ratio.
______% widow’s peak
______% non-widow’s peak
Formative feedback
a)
b)
c)
d)
Phenotype
Now you have accomplished monohybrid (1 trait) crosses, we want to see if you can accomplish
dihybrid (2 traits) crosses. The two traits we are selecting for peas are seed shape and seed
colour.
Image obtained from:
http://bioserv.fiu.edu/~walterm/human_online/inheritance/introduction_to_genetics_files/image0
02.gif
Seed shape:
Peas that are spherical are dominant, and we’ll pick “R” as the dominant allele. Wrinkled seeds
(“r”) are recessive.
Seed colour:
Peas that are yellow are dominant, and we’ll pick “Y” as the dominant allele. Green seeds (“y”)
are recessive.
The two parents we are selecting are YyRr x YyRr.
13. In the space below, fill in and complete the following Punnet square.
YR
YR
Yr
yR
yr
Yr
yR
yr
Formative feedback
Combine the blue parent’s alleles on top with the red parent’s alleles on the left side. There
should be a total of 4 letters (two alleles contributed from each parent) in each Punnet square
box.
14. Complete the following table of the genotypes, genotype proportion and phenotypes
matching the information in the table above. (Y=allele for yellow, y=allele for green, R=allele
for round, r=allele for wrinkled.)
Genotype
Genotype Proportion
Phenotype
Formative feedback
Write down all the possible letter combinations for the genes (from the boxes in the Punnet
square of question 13). Then count out how many of the same genotypes (letter combination)
there are. It will be out of 16 because there are 16 total boxes in the Punnet square. For the
phenotype, we know yellow is a dominant trait and so is round. If there is at least one “Y” and
one “R”, the phenotype will be yellow and round.
15. Write down the phenotype ratio of the offspring resulting from this cross.
______yellow-round: ______yellow-wrinkled: ______green-round: ______green-wrinkled
Formative feedback
Phenotype is the way you look. We know yellow is a dominant trait and so is round. If there is
at least one “Y” and one “R”, the phenotype will be yellow and round. Count up how many
different phenotypes there are in the last column of question 14. Ignore the genotypes, we only
want to know the way they look, not their genes.
Answer Key
1) c
2) c
3) c
4) c
5) a
6) b
7) b
8) d
9) b
10) c
(multiple choice can be 2 points each)
11. In squash, the allele for white fruits (W) is dominant over the allele for yellow (w). If a
white-fruited plant is crossed with a yellow-fruited plant and 50% of the offspring are yellow,
what are the possible genotypes of the parents and the offspring?
a. parents: White fruited squash’s genotype is Ww and yellow fruited squah’s genotype is ww
b. offspring: 50% Ww and 50% ww
(a and b are each 4 points each for a total of 8 points)
12 a)
H
h
H
HH
Hh
H
HH
Hh
(each box is 1 point each for a total of 8 points)
12b)
Genotype
Genotype Proportion
HH
50%
Hh
50%
(each row is 4 points each for a total of 8 points)
12c)
100% widow’s peak
0% non-widow’s peak
Phenotype
Widow’s peak
Widow’s peak
13)
YR
Yr
YYRR
YYRr
YyRR
YR
YYRr
YYrr
YyRr
Yr
YyRR
YyRr
yyRR
yR
YyRr
Yyrr
yyRr
yr
(each box is 1 point each for a total of 16 points)
14)
Genotype
Genotype Proportion
YYRR
1/16
YYRr
2/16 or 1/8
YYrr
1/16
YyRR
2/16 or 1/8
YyRr
4/16 or 1/4
yR
yr
YyRr
Yyrr
yyRr
yyrr
Phenotype
Yellow and Round
Yellow and Round
Yellow and Wrinkled
Yellow and Round
Yellow and Round
Yyrr
2/16 or 1/8
yyRR
1/16
yyRr
2/16 or 1/8
yyrr
1/16
(each box is 1 point each for a total of 27 points)
Yellow and Wrinkled
Green and Round
Green and Round
Green and Wrinkled
15.
9 yellow-round: 3 yellow-wrinkled: 3 green-round: 1 green-wrinkled
(1 points for each ratio, for a total of 4 points)
Checking for Understanding Discussion Questions:
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What are alleles?
How are genotype and phenotypes different? Give an example of a situation in which
two organisms might have identical phenotypes but different genotypes.
What does it mean for a gene to be homozygous dominant, homozygous recessive, and
heterozygous?
Why can’t there be heterozygous dominant/recessive traits?
What is a Punnett square? Explain what the Punnett square shows.
What is a test cross? When would a breeder need to perform a test cross?
How does a dihybrid cross differ from a monohybrid cross? Can a dihybrid cross be
modeled using a Punnett square? How would the Punnett square for a dihybrid cross
differ from one for a monohybrid cross?
What is the principle of independent assortment? Why is the principle of independent
assortment important when we consider dihybrid crosses? Could basic Punnett squares
be used to make predictions about dihybrid crosses if the principle of independent
assortment did not hold true?