Genetic Crosses Dominant and
Recessive Genes Incomplete Dominance
Pedigree Studies Sex Determination Gregor Mendel
Mendel’s 1st Law Mendel’s 2nd
Law Summary
of Mendel’s Laws
Genetics
is the science of heredity and how new life changes and varies in
their characteristics. Sexual reproduction in humans involves 2 gametes; one male gamete, the sperm,
and one female gamete, the egg.
As was discussed in the cell
division webpage the gametes are haploid
(n). When fertilisation occurs the resulting fertilised egg is diploid (2n) Genetic variations occur
as a result of this union.
As was discussed in the heredity
webpage the physical characteristics of organisms are developed as the
protein builds up their bodies. These proteins are formed as a result of genes
carried on chromosomes.
In genetics genes are
represented by letters. There are usually 2 different types of the same
gene; one is dominant and one is recessive. An example of this
is: T is a gene for tall and t is a gene for short. The two versions
of the same gene are called alleles.
The two alleles are formed at the same position, or locus, on the chromosome.
REMEMBER: THE 2 ALLELES ARE GOTTEN FROM
THE 2 GAMETES OF SEXUAL REPRODUCTION. THAT IS WHY THEY CAN DIFFER.
IN ASSEXUAL REPRODUCTION THE NEW LIFE
IS ALWAYS IDENTICAL TO THE PARENT CELL BECAUSE THERE IS ONLY 1 PARENT.
Dominant
genes will always prevent the recessive gene from working. If a person
has 2 dominant alleles for tallness: T T then he will be tall. If a person has 1 dominant and
I recessive allele for tallness: T t then he will be tall. The only way the recessive gene
will be expressed is if he has 2 recessive alleles for short: t
t then he will be short.
If the pair of genes controlling the characteristic has
identical alleles, TT or tt, we call it a homozygous pair of genes. If the pair
of genes controlling the characteristic are different alleles, Tt or Cr, or Cw we call it a heterozygous pair of genes.
When we express the genotype
of a characteristic we state the pair of alleles. Genotype examples are TT, Tt, tt, BB, Bb, bb, etc.
When we state the physical characteristic expressed by the
genotype we are stating is phenotype.
The phenotype for TT is tall while
the phenotype for tt is short. These phenotypes could vary
because of environment effects. This
is especially true in terms of genotypes and phenotypes for intelligence. The
upbringing, environment, and education experiences greatly affect the
phenotype.
REMEMBER: THE 2 ALLELES ARE GOTTEN FROM
THE 2 GAMETES OF SEXUAL REPRODUCTION. THAT IS WHY THEY CAN DIFFER.
When working on genetic crosses you must state the capital letter which represents the
dominant allele and the small letter
that represents the recessive allele.
The following is an example of how to
work out genetic crosses:
|
A is a dominant characteristic. |
a is a recessive characteristic. |
|
This bird has two genes for red feathers. Its genotype is AA. Its phenotype is RED |
This bird has two genes for blue feathers. Its genotype is aa. Its phenotype is BLUE |
This Punnett Square shows how we can diagram
the genes.
The orange bird has two
dominant A genes. We put two A s along the top of the square.
The blue bird has two
recessive a genes. We put two a s down along the left side of the
square.
All the offspring have
the genes Aa.
They will all have orange feathers (phenotype), but will carry a recessive gene for blue
feathers (genotype).The progeny are the offspring produced.
This is called
the F1 generation progeny.
Now suppose that two individuals from the F1
generation become parents. Here they are!
The baby birds are called the F2 generation. You can see how their genes work
out. The offspring are coded in the squares. One bird will be orange with two AA genes.
Two birds will be orange with genes coded Aa.
One bird will be blue and will have two recessive aa genes.
Individual nests of birds may not turn out exactly like this, but if there are
many baby birds, they will work out genetically with the ratios 1:2:1.
Genotype: 1 AA, 2 Aa,
1 aa
Phenotype: 3 orange feathers, 1 blue feathers
Punnett
Square Animated Example
Incomplete
dominance is the situation where two different alleles are equally
dominant. When this occurs the heterozygous genotype
that is produced is an intermediate phenotype (blend) between the two respective homozygous genotypes. This is
also called codominance.
In this
example AA genotypes have red, Aa genotypes pink and aa genotypes whitish flowers. Note that the heterozygous genotype Aa is a blend of red and white.
Note: This
is the F1 generation.
In the F2 generation two pink flowers will
produce 2 pink phenotypes, 1 red phenotype, and 1 white phenotype.
A pedigree is a diagram showing the genetic history of a group of related organisms.
A pedigree showing the occurrence of a recessive
trait in three generations of a family
.
Circles in a pedigree represent females and squares
represent males. A horizontal line between a circle and a
square indicates a marriage or
partnership. Vertical lines indicate
the children from the marriage
or partnership. In this activity, a
filled-in circle or square shows that the individual has both alleles for the
trait. A half-filled-in circle or square indicates that the individual has one
recessive allele for the trait.
Our cells have 46 chromosomes. There
are 23 pairs. Remember, we get 23 from our mother and 23 from our father. We have
22 pairs of chromosomes called autosomes.
These are the chromosomes that control our body growth, enzymes, etc. We have 1 pair of sex chromosomes.
If the person is a male that pair is composed of an X and a Y chromosome. The male’s genotype is XY.
If the person is a female she will have a sex
chromosome composed of 2 Y chromosomes. The
female’s genotype is XX.
The
If
the child has a genotype of XX then it
becomes a girl. If the child has a genotype of XY then it becomes a boy.
As you can see the ratio of males to females is
1:1.
Gregor Mendel studied 7 characteristics of pea plants. He studied:
As a result of Mendel’s work the study of genetics
began. He discovered that, although an organism may have genotypes for 2
different physical traits (phenotypes)
the organism will only exhibit one of those traits.
The work he did for the height of the plants is a
follows:
He
discovered although the parents of the first generation had TT and tt genotypes all the progeny of the F1 generation were
tall.
He
then discovered that in the F2 generation there was a 3:1 ratio between tall
and short.
The
The
Note
that in the F1 generation the parents were homozygous. One was TT and one was tt. All the progeny were tall because all the progeny had Tt genotypes with T being the dominant characteristic.
In
the F2 generation a Tt plant was used to self-pollinate itself. All of the
phenotypes in this combination were Tt. As a result
the F2 progeny were 3 Tt and1 tt.
Mendel’s Laws
First Law- The Law of Segregation
1.
This
law states:
a.
In
diploid organisms, chromosomes occur in matching pairs. These pairs are called homologous chromosomes. These are
chromosomes that pair at meiosis, have the same length and banding pattern plus
carrying the same genes at the same locus. They have the same sequence of
genes.
Notice that these two chromosomes are homologous
because they have alleles at the same position on the chromosome but one allele
is for purple flowers and the other for white flowers.
b.
For each characteristic or trait organisms inherit
two alternative forms of that gene, one from each parent. These alternative
forms of a gene are called alleles.
c.
When
gametes (sex cells) are produced, allele pairs separate or segregate leaving
them with a single allele for each trait.
d.
When the two alleles of a pair are different, one is
dominant and the other is recessive.
Example:
If your eyes are blue, green or grey
you have two alleles for blue eyes (bb),
then your gametes must have a blue allele (b); if your eyes are brown you might have two brown
allele (BB), then
your gametes have one allele for brown (B)
or you might have one
allele of each kind (Bb),
in which case you make two kinds of gametes some contain the brown allele (B) and some contain the blue allele
(b).
Second Law- Law of Independent Assortment
Mendel’s Second
Law involves dihybrid crosses.
Dihybrid crossing involves the study of 2 characteristics at the same time.The Law of
Independent Assortment states that alleles
for different traits are
distributed to sex cells (& offspring) independently of one another.
Mendel noticed during all his work that the height of the plant and the
shape of the seeds and the color of the pods had no impact on one
another. In other words, being tall didn't automatically mean the plants
had to have green pods, nor did green pods have to be filled only with wrinkled
seeds, the different traits
seem to be inherited independently.
The genotypes of our parent pea plants will be:
RrGg x RrGg where
"R" = dominant allele for
round seeds
"r" = recessive allele for
wrinkled seeds
"G" = dominant allele for
green pods
"g" = recessive allele for
yellow pods
Notice that we are dealing with two different traits: (1) seed texture (round or
wrinkled) & (2) pod color (green or yellow). Notice also that
each parent is hybrid for each trait (one dominant & one recessive allele
for each trait).
|
|
RG |
Rg |
rG |
rg |
|
|
RG |
RRGG |
RRGg |
RrGG |
RrGg |
|
|
Rg |
RRGg |
RRgg |
RrGg |
Rrgg |
|
|
rG |
RrGG |
RrGg |
rrGG |
rrGr |
|
|
Rg |
RrGg |
Rrgg |
rrGg |
rrgg |
|
We need to
"split" the genotype letters & come up with the possible gametes
for each parent. Keep in mind that a gamete (sex cell) should get half as
many total letters (alleles) as the parent and only one of each letter.
So each gamete should have one "R
or r" and one "G or g"
for a total of two letters. There are four possible letter combinations:
RG, Rg, rG, and rg. So, when the two parents’ gametes form a new organism
the punnett square will look like this:
The results from a dihybrid cross
are always the same:
9/16 boxes (offspring) show dominant phenotype
for both traits (round & green),
3/16 show dominant phenotype for first
trait & recessive for second (round & yellow),
3/16 show recessive phenotype for first
trait & dominant form for second (wrinkled & green), &
1/16 show recessive form of both traits
(wrinkled & yellow).
So, as you can
see from the results, a green pod can have round or wrinkled seeds, and the
same is true of a yellow pod. The different
traits do not influence the inheritance of each other. They are inherited
INDEPENDENTLY.
Interesting to
note is that if you consider one trait at a time, we get "the usual"
3:1 ratio of a single hybrid cross (like we did for the Law of Segregation).
For example, just compare the color trait in the offspring; 12 green & 4
yellow (3:1 dominant: recessive). The same deal with the seed texture; 12
round & 4 wrinkled (3:1 ratio). The traits are inherited
INDEPENDENTLY of each other.
Animation
Explaining Independent Assortment of Genes
|
LAW |
PARENT CROSS |
OFFSPRING |
|
DOMINANCE |
TT x tt |
100% Tt |
|
SEGREGATION |
Tt x Tt |
75% tall
|
|
INDEPENDENT ASSORTMENT |
RrGg x RrGg |
9/16 round seeds & green pods |
Click
here to go to practice questions using Punnett squares
Click
here to go to practice questions about genetics and Mendel’s Laws
Interactive
animation with Mendel’s peas
TOP
Linkage (Gene Linkage)
Gene
linkage occurs when traits for 2 separate characteristics occur on the same
chromosome.
The characters Mendel examined happened to be on separate chromosomes. That is
why he observed independent assortment. If,
however, the genes are on the same chromosomes, they will be inherited together.
For example, consider the following parental nuclei. Both father and mother
have a pair of chromosomes with alleles for two different genes:
If we look at this with a punnet square what is going
to happen in the next generation:
The phenotype ratios are still 3:1,
but there are fewer genotype combinations than in the usual cross involving two
alleles.
Remember: With independent assortment
the phenotypes resulted in a 9:3:3:1 ratio.
View
this animation of crossing over
Sex-linked Genes or
Sex linkage
Genes
or traits whose controlling genes are on the X sex chromosome but not on the Y
sex chromosome. As a result recessive
phenotype occurs more often in males than in females.
There is yet another, unrelated, special
case that means something totally different, yet has a similar-sounding name.
This is sex-linked genes, genes located on one of the sex chromosomes (X or Y) but not the other. Since,
typically the X chromosome is longer, it bears a lot of genes not found on the
Y chromosome, and thus most sex-linked genes are X-linked genes. One
example of a sex-linked gene is fruit fly eye colour. An X chromosome carrying
a normal, dominant, red-eyed allele would be symbolized by a plain X,
while the recessive, mutant, white-eyed allele would be symbolized by X'
or Xw. A fly with genotype XX'
would normally be a female with red eyes, yet be a carrier for the
white-eyed allele. Because a male typically only has one X chromosome, he would
normally be either XY and have normal, red eyes or X'Y and have white eyes. The
only way a female with two X chromosomes could have white eyes is if she would
get an X' allele from both parents making her X'X' genotype. The cross between
a female carrier and a red-eyed male would look like this:
|
|
X |
Y |
||||
|
X |
|
|||||
|
X' |
||||||
Notice that while there is a “typical”
ratio of ¾ red-eyed to ¼ white-eyed, all of the white-eyed flies are males.
Sex-linked traits act just like recessive
ones except they also bow to the will of the sex of the child. Genes are
carried on things called chromosomes. Most people already know that in
humans the man has an X and Y chromosome and the female has two X chromosomes.
This is the reason that only the man can determine the sex of the child. Women
can only provide X chromosomes while a man can provide either. Now, the X
chromosome is bigger and can carry more genetic information on it than the Y
can.
This is where sex-linked traits come in.
Because the X is bigger it means that some genes carried on it are not carried
on the Y chromosome. These genes can be expressed even without a corresponding
partner on an X chromosome. They also cannot be blocked out unless there is
another X chromosome carrying a dominant partner.
In humans, two well-known X-linked traits
are haemophilia and red-green colour-blindness. Haemophilia is the
failure (lack of genetic code) to produce certain substance needed for proper
blood-clotting, so a haemophiliac’s blood doesn’t clot, and (s)he could bleed
to death from an injury that a normal person might not even notice.
One human sex-linked trait is
Haemophilia. Haemophilia is a disease that keeps a person's blood from clotting
when he is cut. Haemophiliacs can easily bleed to death and must be very
careful not to injure themselves. Many also take daily injections to help the
problem. Because haemophilia is a sex-linked disease most of the people who
have it are men. Women can carry the gene for haemophilia but will not be
affected by it because their second X chromosome will block it out with a
healthy gene. They must have two copies of the defective gene to display the
disease. Inheriting two copies is highly unlikely. Men carrying haemophilia do
not have another X chromosome so they will have haemophilia with only one gene
for it. Mothers carrying one gene for haemophilia take a great risk with having
children because there is a 50% chance their sons will end up with the disease.
Here's how it works:
As you can see, at least half of her
children (boxes 1 and 2) will inherit the defective gene. One, a daughter, will
only carry the gene. The other, a son, will have the disease haemophilia. The
last two children (boxes 3 and 4) will carry healthy genes. Of course these are
only the possibilities of what her children could end up with. She could very
well end up giving it to all her children or none at all. It's just a matter of
chance. Now let's take a look at what will happen if this woman's haemophiliac
son has children with a healthy woman:
In this case half the children will
still inherit the gene. All the sons will be safe but all the daughters will
end up carrying haemophilia and could end up passing it on to their children.
It is in this way that sex-linked genes can disappear and reappear from
generation to generation.
1.
Which of the following is a possible abbreviation for a genotype?
A. BC
B. Pp
C. Ty
D. fg
2. What is the best way to determine the phenotype of the feathers on a
bird?
A. analyze the bird's DNA
(genes)
B. look at the bird's feathers
C. look at the bird's beak
d. examine the bird's droppings
3. Which of the following pairs is not correct?
A. kk
= hybrid
B. hybrid = heterozygous
C. heterozygous = Hh
D. homozygous = RR
4. The genes present in an organism represent the organism's __________.
A. genotype
B. phenotype
C. physical traits
5. Which choice represents a possible pair of alleles?
A. k & t
B. K & T
C. K & k
D. K & t
6. How many alleles for one trait are normally found in the genotype of
an organism?
A. 1
B. 2
C. 3
D. 4
7. Which statement is not true?
A. genotype determines
phenotype
B. phenotype determines genotype
C. a phenotype is the physical appearance of a trait in an organism
D. alleles are different forms of the same gene
8. Which cross would best illustrate Mendel's Law of
Segregation?
A. TT x tt
B. Hh x hh
C. Bb x Bb
D. rr
x rr
9. In the cross Yy x Yy, what percent of
offspring would have the same phenotype as the parents?
A. 25%
B. 50%
C. 75%
D. 100%
10. In a certain plant,
purple flowers are dominant to red flowers. If the cross of two
purple-flowered plants produces some purple-flowered and some red-flowered
plants, what is the genotype of the parent plants?
A. PP x Pp
B. Pp x Pp
C. pp x PP
D. pp x pp
Base questions #11-14 on
the following information:
A
white-flowered plant is crossed with a pink-flowered plant. All of the F1
offspring from the cross are white.
11. Which phenotype is dominant?
12. What are the genotypes of the original parent
plants?
13. What is the genotype of all the F1 offspring?
14. What would be the percentages of genotypes &
phenotypes if one of the white F1 plants is crossed with a pink-flowered plant?
15. Which of Mendel's Laws is/are illustrated in this
question?
16. Crossing two dihybrid organisms results in
which phenotypic ratio?
A. 1:2:1
B. 9:3:3:1
C. 3:1
D. 1:1
17. The outward appearance (gene expression) of a
trait in an organism is referred to as:
A. genotype
B. phenotype
C. an allele
D. independent assortment
18. In the homologous
chromosomes shown in the diagram, which is a possible allelic pair?
A.
cD
B. Ee
C. AB
D. ee
19. The phenotype of a pea plant can best be determined by:
A. analyzing its genes
B. looking at it
C. crossing it with a recessive plant
D. eating it
20. Mendel formulated his Law of Segregation after he had:
A. studied F1 offspring
B. studied F2 offspring
C. produced mutations
D. produced hybrids
21. Which cross would produce phenotypic ratios that would illustrate
the Law of Dominance?
A. TT x tt
B. TT x Tt
C. Tt x Tt
D. tt x tt
22. The mating of two curly-haired brown guinea pigs results in some
offspring with brown curly hair, some with brown straight hair, some with white
curly hair, and even some with white straight hair. This mating
illustrates which of Mendel's Laws?
A. Dominance
B. Segregation
C. Independent Assortment
D. Sex-Linkage
Click here to go to the answers
Click here to go back to
Genetics Page
1.
Which of the following is a possible abbreviation for a genotype?
A. BC
B. Pp
- genotypes are made up of 2 of the same letter (either 2 capital, 2
lowercase, or one of each)
C. Ty
D. fg
2. What is the best way to determine the
phenotype of the feathers on a bird?
A. analyze
the bird's DNA (genes)
B. look at the bird's feathers
- "phenotype of the feathers" means what the feathers look like, so
look at them
C. look at the bird's beak
d. examine the bird's droppings
3. Which of the following pairs is not
correct?
A. kk = hybrid - Kk
would be hybrid (one capital, one lowercase of the same letter)
B. hybrid = heterozygous
C. heterozygous = Hh
D. homozygous = RR
4. The genes present in an organism represent
the organism's __________.
A.
genotype
B. phenotype
C. physical traits
5. Which choice represents a possible pair of
alleles?
A. k &
t
B. K & T
C. K & k - allele means 2 forms of the same
gene. so this choice shows 2 forms of the same letter
K or k
D. K & t
6. How many alleles for one trait are normally
found in the genotype of an organism?
A. 1
B. 2 - one
allele is inherited from each parent
C. 3
D. 4
7. Which statement is not true?
A. genotype
determines phenotype - (note that the environment does play a role in
influencing phenotype too)
B. phenotype determines genotype
C. a phenotype is the physical appearance of a trait
in an organism
D. alleles are different forms of the same gene - (see
question #5)
8. Which cross would best
illustrate Mendel's Law of Segregation?
A.
TT x tt
B. Hh x hh
C. Bb x Bb - both parent show dominant trait, but some recessive offspring will be
produced (each parent carries a "b")
D. rr
x rr
9. In the cross Yy x Yy, what percent of offspring would have the same phenotype as the
parents?
A.
25%
B. 50%
C. 75%
- in the completed p-square, 3 of 4 boxes will have at least 1 "Y",
producing the dominant phenotype (same as parents)
D. 100%
10. In a
certain plant, purple flowers are dominant to red flowers. If the cross
of two purple-flowered plants produces some purple-flowered and some red-flowered
plants, what is the genotype of the parent plants?
A.
PP x Pp
B. Pp x Pp - for any offspring to be recessive, each parent MUST have at least one
"p"
C. pp x PP - only one parent is purple, this CAN'T be an answer
D. pp x pp - neither parent is purple, this CAN'T be an answer
Base
questions #11-15 on the following information:
A
white-flowered plant is crossed with a pink-flowered plant. All of the F1
offspring from the cross are white.
11. Which phenotype is dominant? white
12. What are the genotypes of the
original parent plants? WW (pure white) x ww
(pink)
13. What is the genotype of all
the F1 offspring? Ww (white)
14. What would be the percentages
of genotypes & phenotypes if one of the white F1 plants is crossed with a
pink-flowered plant?
50% heterozygous
white & 50% homozygous recessive pink.
The cross for this question would be "Ww (white F1) x ww (pink)".
The alleles of the white
parent are above the columns & those of the pink parent are in front of the
rows. 2 of 4 boxes (50%) are "Ww", which is
heterozygous & would have the dominant trait (white). The other 2 of
4 boxes (50%) are "ww", which is homozygous
recessive & would have the recessive trait (pink).
15. Which of Mendel's Laws is/are illustrated in this
question? Dominance is illustrated by the original cross (WW x ww).
16. Crossing two dihybrid organisms
results in which phenotypic ratio?
A.
1:2:1 - genotype ratio of a hybrid
cross, ex: Tt x Tt
B. 9:3:3:1- dihybrid means hybrid for two different traits. An example could be GgYy x GgYy.
C. 3:1 - phenotype ratio of a hybrid cross
D. 1:1
17. The outward appearance (gene
expression) of a trait in an organism is referred to as:
A.
genotype
B. phenotype
C. an allele
D. independent assortment
18. In the
homologous chromosomes shown in the diagram, which is a possible allelic pair?
A. cD
B. Ee- a possible allelic pair but NOT SHOWN IN THE DIAGRAM, so
this CAN'T be an answer
D. ee
- an "allelic pair" is always two forms of the same letter. In
this example they are two lowercase "e's".
19. The phenotype of a pea plant can best be
determined by:
A. analyzing
its genes
B. looking at it
C. crossing it with a recessive plant
D. eating it
20. Mendel formulated his Law of Segregation
after he had:
A. studied
F1 offspring -
B. studied F2 offspring
- he crossed two hybrids (F1's) and got a second generation --- the F2.
C. produced mutations - Mendel
knew NOTHING about mutations so this CAN'T be an answer
D. produced hybrids
21. Which cross would produce phenotypic
ratios that would illustrate the Law of Dominance?
A. TT x tt - one parent tall, the other short,
all offspring would be tall
B. TT x Tt
C. Tt x Tt
- illustrates Segregation
D. tt x tt
22. The mating of two curly-haired brown
guinea pigs results in some offspring with brown curly hair, some with brown
straight hair, some with white curly hair, and even some with white straight
hair. This mating illustrates which of Mendel's Laws?
A.
Dominance
B. Segregation
C. Independent Assortment
- the question involves two different traits (hair color &
hair texture), this is the only law that deals with two different traits
D. Sex-Linkage - Mendel
knew NOTHING about sex-linkage so this CAN'T be an answer
Click here to go back to Genetics Page
The
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) 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 horn is recessive.
Draw a
QUESTION #3
A green-leafed luboplant (not a real plant) 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.
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 (seed appearance)?
Click here to go back to
Genetics Page
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
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
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: |
In purple people eaters, one-horn is dominant and no horn is recessive.
Draw a
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!
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 punnett
square should be:
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 boxes 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).
A green-leafed luboplant 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 |
First write down the 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 = recessive 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? 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".
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 of 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 referred to as the "first filial" or "F1" generation.
Now, our question has to do with crossing two members
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 |
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 punnett
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.