Transcript Red kitten
Advanced
Biology
Unit 3
The Monk and his peas
An Austrian monk, Gregor Mendel,
developed the fundamental principles that
would become the modern science of
genetics. Mendel demonstrated that
heritable properties are parceled out in
discrete units, independently inherited.
These eventually were termed genes.
Seed shape and color
Flower color
Pod color and shape
Flower Position
Stem height
• P1: smooth X
wrinkled
• F1 : all smooth
About a 3 to 1 ratio
F2 : 5474 smooth
and 1850 wrinkled
Q
A test cross is used to determine if the
genotype of a plant with the dominant
phenotype is homozygous or
heterozygous. If the unknown is
homozygous, all of the offspring of the test
cross have the __________ phenotype. If
the unknown is heterozygous, half of the
offspring will have the __________
phenotype.
answer
dominant, recessive
The test cross was invented by Mendel to
determine the genotype of plants
displaying the dominant phenotype.
Q
Disappearance of parental phenotypes in
the F1 generation
A genetic cross of inbred snapdragons
with red flowers with inbred snapdragons
with white flowers resulted in F1-hybrid
offspring that all had pink flowers. When
the F1 plants were self-pollinated, the
resulting F2-generation plants had a
phenotypic ratio of 1 red: 2 pink: 1 white.
The most likely explanation is:
answer
Heterozygous plants have a different
phenotype than either inbred parent because
of incomplete dominance of the dominant
allele. The features of crosses involving
incomplete dominance are intermediate
phenotype of heterozygous individuals, and
parental phenotypes reappear in F2 when
heterozygotes are crossed.
ABO’s of human blood
• Human blood type is determined by codominant
alleles. There are three different alleles, known
as IA, IB, and i. The IA and IB alleles are codominant, and the i allele is recessive.
• The possible human phenotypes for blood group
are type A, type B, type AB, and type O. Type A
and B individuals can be either homozygous
(IAIA or IBIB, respectively), or heterozygous (IAi
or IBi, respectively).
• A woman with type A blood and a man with type
B blood could potentially have offspring with
which of the following blood types?
A, B, AB, or O
Q
What are the possible blood types of the
offspring of a cross between individuals
that are type AB and type O? (Hint: blood
type O is recessive)
answer
A, B, AB, O
all of the above
But if the man was type O rather than type
B, offspring of type B and type AB would
not be possible.
answer
In this problem there is no
uncertainty about the
genotype of either parent. A
parent of blood type AB has
the codominant IA and IB
alleles. A parent of blood type
O is homozygous recessive
for the i allele. The Punnett
square for their offspring is
shown to the right. The
genotypes of their offspring
could be either IAi or IBi. Their
children could be in blood
groups A or B, but not AB or
O.
Manx cats are heterozygous for a
dominant mutation that results in no tails
(or very short tails), large hind legs, and a
distinctive gait. The mating of two Manx
cats yields two Manx kittens for each
normal, long-tailed kitten, rather than
three-to-one as would be predicted from
Mendelian genetics. Therefore, the
mutation causing the Manx cat phenotype
is likely a(n) __________ allele.
answer
lethal
The predicted segregation pattern in the F2
generation is 1/4 normal (homozygous),
1/2 Manx phenotype (heterozygous), an
1/4 embryonic lethal (homozygous for the
Manx allele).
Monohybrid test cross
To identify the genotype of yellow-seeded
pea plants as either homozygous
dominant (YY) or heterozygous (Yy), you
could do a test cross with plants of
genotype _______.
YY or Yy x ___?___
Answer:
A cross with the homozygous recessive
(yy) is a test cross. If the parent of
unknown genotype is heterozygous (Yy),
half of the offspring will have the recessive
trait. The unknown genotype could also be
determined by a cross with a known
heterozygote (Yy).
When true-breeding tall stem pea plants
are crossed with true-breeding short stem
pea plants, all of the ________ plants, and
3/4 of the __________ plants had tall
stems. Therefore, tall stems are dominant.
Answers
F1, F2.
The F1 plants are all Tt hybrids. The
recessive trait (tt) reappears in the F2
generation in about 25% of the plants.
Q
A genetic cross between two F1-hybrid
pea plants having yellow seeds will yield
what percent green-seeded plants in the
F2 generation? Yellow seeds are dominant
to green.
Answers
25%
• Among the F2 plants of a Yy x Yy cross,
25% will be yy with the recessive, greenseeded phenotype.
Q
In Mendel's "Experiment 1," true-breeding pea
plants with spherical seeds were crossed with
true-breeding plants with dented seeds.
(Spherical seeds are the dominant
characteristic.) Mendel collected the seeds from
this cross, grew F1-generation plants, let them
self-pollinate to form a second generation, and
analyzed the seeds of the resulting F2
generation. The results that he obtained, and
that you would predict for this experiment are:
Answer
All the F1 and 3/4 of the F2 generation
seeds were spherical.
All of the F1 plants were true hybrids with
a phenotype of Ss. The recessive trait
reappears in the F2 generation.
Q
A phenotypic ratio of 3:1 in the offspring of a
mating of two organisms heterozygous for
a single trait is expected when:
answer
the alleles segregate during meiosis.
Mendel first proposed that alleles
segregate from one another during the
formation of gametes.
Q
In pea plants, spherical seeds (S) are
dominant to dented seeds (s). In a genetic
cross of two plants that are heterozygous
for the seed shape trait, what fraction of
the offspring should have spherical seeds?
answer
3/4
One fourth of the offspring will be
homozygous dominant (SS), one half will
be heterozygous (Ss), and one fourth will
be homozygous recessive (ss).
Q
•
•
•
•
•
The gametes of a plant of genotype SsYy
should have the genotypes:
A. Ss and Yy
B. SY and sy
C. SY, Sy, sY, and sy
D. Ss, Yy, SY and sy
E. SS, ss, YY, and yy
answer
• The gametes will receive
one of each pair (Ss and
Yy) of alleles. All
combinations of alleles
will occur with equal
probability (SY, Sy, sY,
and sy) because alleles
of different genes are
assorted independently
during gamete formation
(meiosis.)
• Each gamete has one
seed shape (S=spherical
or s=dented) and one
color (Y=yellow or
y=green) allele.
Dihybrid cross
A cross that involves
two sets of
characteristics
Instead of 4 possible
genotypes from a
monohybrid cross,
dihybrid crosses have
as many as 16
possible genotypes.
Dihybrid test cross
Which of the following genotypes would
you not expect to find among the offspring
of a SsYy x ssyy test cross:
A. ssyy
B. SsYy
C. Ssyy
D. ssYy
E. SsYY
Answer
SsYY
Offspring could not be homozygous for the
dominant yellow seed color (YY), because
one recessive y allele must be inhereited
from the ssyy parent.
Q
The expected phenotypic ratio of the
progeny of a SsYy x ssyy test cross is:
• A. 9:3:3:1
• B. 3:1
• C. 1:1:1:1
• D. 1:2:1
• E. 3:1:1:3
C. 1:1:1:1.
SsYy, ssYy, Ssyy, ssyy are predicted to
occur in a ratio of 1:1:1:1
Q
•
•
•
•
•
In a dihybrid cross, AaBb x AaBb, what
fraction of the offspring will be
homozygous for both recessive traits?
A. 1/16
B. 1/8
C. 3/16
D. 1/4
E. 3/4
There are four possible
combinations of gametes
for the AaBb parent. Half
of the gametes get a
dominant A and a
dominant B allele; the
other half of the gametes
get a recessive a and a
recessive b allele.
• Both parents produce
25% each of AB, Ab, aB,
and ab.
Answer
There is only one of 16
possible combinations
with this genotype. The
predicted fraction is
therefore 1/16.
What does 3:1 indicate?
A cross between parents who are both
heterozygous produces a phenotypic ratio
of 3:1
• Aa x Aa
Give an example of a monohybrid
test cross
• T is tall in pea plants
• T is short
• Crossing a T plant (could be TT or Tt) with
a short (tt) plant could produce:
• 1:1 tall and short, if Tall parent is Tt
• All tall is tall parent is TT
Sex-linked traits
•
•
•
•
•
In a cross between a pure bred, red-eyed
female fruit fly and a white-eyed male, what
percent of the male offspring will have white
eyes? (white eyes are X-linked, recessive)
A. 100%
B. 75%
C. 50%
D. 25%
E. 0%
A
0%
All of the males and all of the females are
red-eyed.
Q
A white-eyed female fruit fly is crossed with
a red-eyed male. Red eyes are dominant,
and X-linked. What are the expected
phenotypes of the offspring?
All of the females eggs
will contain an X
chromosome with
the white-eye mutation.
A
The sperm will contain
either a normal X chromosome or a Y chromosome.
We use a punnett square to predict the outcome of
this cross. Female offspring receive an X
chromosome from both the sperm and egg. All
females receive the dominant, red-eyed allele from
their fathers and the recessive, white-eyed allele
from their mothers.
Hemophilia
Hemophilia results from a mutated gene
on an X chromosome. Mothers always
contribute an X chromosome to their
offspring, while fathers contribute either an
X or a Y chromosome. If hemophilia is
found on any of these X chromosomes, it
can be passed on to the child.
Hemophilia
Normal Mother + Father
with Hemophilia
Each pregnancy has a 50%
chance of resulting in a
female carrier and a 50%
chance of resulting in a
normal male. Sons of
hemophiliac fathers and
normal mothers will not
have hemophilia.
But…
Carrier Mother + Normal
Father
Each pregnancy has a
25% chance of resulting
in a normal female, a
25% chance of resulting
in a female carrier, a 25%
chance of resulting in a
normal male, and a 25%
chance of resulting in a
male with hemophilia
Carrier Mother + Father with
Hemophilia
Each pregnancy has
a 25% chance of
resulting in a female
carrier, a 25% chance
of resulting in a
female with
hemophilia, a 25%
chance of resulting in
a normal male, and a
25% chance of
resulting in a male
with hemophilia.
Mother with Hemophilia + Father
with Hemophilia
All hemophilia.
Mother with, Dad normal
Each pregnancy has
a 50% chance of
resulting in a female
carrier and a 50%
chance of resulting in
a male with
hemophilia (actual
occurrence is
extremely rare).
In summary, knowledge of genetics
lets us make the following
statements about hemophilia:
• Nearly all affected people are male.
• Hemophilia may represent a new mutation in the
affected male.
• An affected male never transmits the trait to his
sons.
• All daughters of an affected male will be carriers (if
the mom is not a carrier).
• A carrier female transmits the trait to her sons 50
percent of the time.
• No daughters of a carrier female will show the trait,
but a daughter in this case (if the dad is not affected)
will be a carrier 50 percent of the time.
The ABCs of Hemophilia
Type
Hemophilia A
Hemophilia B
Hemophilia C
•Classical hemophilia
•Standard hemophilia
•Factor VIII deficiency
•Christmas disease (named
after Stephen Christmas, a
young British boy who was
the first person diagnosed
with the disorder)
•Factor IX deficiency
•Factor XI deficiency
•About 1 in 5,000 U.S. male
births
•About 80 percent of people
with hemophilia
•About 13,500 people in the
U.S.
•About 1 in 30,000 U.S. male
births
•Up to 20 percent of people
with hemophilia
•More than 3,000 people in the
U.S.
•About 1 in 100,000 U.S. male
births
•About 200 cases reported
worldwide since its discovery
in the 1950s
Gender Affected
Males almost exclusively
Males almost exclusively
Males and females equally
Missing Factor Protein
Factor VIII
Factor IX
Factor XI
Other Names
Frequency
Q
• Hemophilia in humans is due to an Xchromosome mutation. What will be the results
of mating between a normal (non-carrier) female
and a hemophilac male?
• A. half of daughters are normal and half of sons
are hemophilic.
• B. all sons are normal and all daughters are
carriers.
• C. half of sons are normal and half are
hemophilic; all daughters are carriers.
• D. all daughters are normal and all sons are
carriers.
• E. half of daughters are hemophilic and half of
daughters are carriers; all sons are normal.
A
• The eggs of the mother will all contain
the normal X chromosome.
• The sperm of the father will contain
either the X chromosome with the
mutation causing hemophilia or the Y
chromosome.
• All of the daughters inherit an X
chromosome with the mutation from
their father, and will be carriers.
• All the sons inherit a normal X from
the mother
Color blindness
• A human female "carrier" who is heterozygous
for the recessive, sex-linked trait causing redgreen color blindness, marries a normal male.
What proportion of their male progeny will have
red-green color blindness?
• A. 100%
• B. 75%
• C. 50%
• D. 25%
• E. 0%
A
• The eggs of the mother will
contain either a normal X
chromosome or an X
chromosome with the mutation
causing red-green color
blindness.
• The sperm of the father will
contain either the normal X
chromosome or the Y
chromosome.
• None of the female children
would be red-green color blind,
but half would be "carriers."
• Half of the sons would inherit
the allele from their mother
and be afflicted.
Q
Women have sex chromosomes of XX, and men
have sex chromosomes of XY.
Which of a man's grandparents could not be the
source of any of the genes on his Ychromosome?
A. Father's Mother.
B. Mother's Father.
C. Father's Father.
D. Mother's Mother, Mother's Father, and Father's
Mother.
E. Mother's Mother.
A
D. Mother's Mother, Mother's
Father, and Father's Mother.
The Y chromosome is inherited solely
from father to son in each generation.
E. Mother's Mother. The diagram
shows how the X and Y chromosomes
are inherited from the maternal and
paternal grandparents to the parents
to the son.
The Y chromosome is passed strictly
from the father to male children in
each generation.
Neither maternal grandparent nor the
paternal grandmother can be a source
of any genes on the Y chromosome.
Two sex linked traits…
• What offspring would you expect from a cross
between the female Drosophila described in
problem 1 (red eyes and a yellow body,
homozygous recessive for the yellow body color
allele and homozygous dominant for the eye
color allele) and the male described in problem 2
(hemizygous for both the recessive (white) eye
color allele and dominant (tan) body color
allele?) A reminder that the alleles for eye color
and for body color are on the X chromosome of
Drosophila, but not on the Y. Red eye color (w+)
is dominant to white eye color (w), and tan body
color (y+) is dominant to yellow body color (y).
Choices are…
A. Daughters would be yellow-bodied, red-eyed;
the sons would be tan-bodied, white-eyed
B. Daughters would be tan-bodied, red-eyed; the
sons would be yellow-bodied, white-eyed
C. Daughters would be tan-bodied, red-eyed; the
sons would be yellow-bodied, red-eyed
D. Daughters would be yellow-bodied, white-eyed;
the sons would be tan-bodied, white-eyed
E. Daughters would be yellow-bodied, red-eyed;
the sons would be tan-bodied, red-eyed
A
C. Daughters
would be tanbodied, redeyed; the sons
would be
yellow-bodied,
red-eyed
Q
If we mated the F1 female and male flies,
what male phenotype in the F2 generation
would be evidence that crossing over had
occurred during gamete formation?
Daughters were tan-bodied, red-eyed,
heterozygous for both eye and body color.
The sons were yellow-bodied, red-eyed.
A
Evidence for recombination
between X chromosomes in
the F1 female would be a
new combination of alleles
not present on any of the X
chromosomes of the
parents of the F1 female.
Yellow body/white eyes and
tan body/red eyes F2 males
would be evidence of
crossing over.
Karyotyping Activity
• Patient A's Karyotype
Place this chromosome in the partially
completed karyotype below by clicking on its
homologous chromosome. If you match the
chromosome correctly, you will proceed to
the next chromosome. If you match
incorrectly, a page will explain why the
chromosome you chose is not the
unknown's pair and you can choose again.
Common Terms
Gene - a unit of inheritance that usually is directly
responsible for one trait or character.
Allele - an alternate form of a gene. Usually there are two
alleles for every gene, sometimes as many a three or
four.
Homozygous - when the two alleles are the same.
Heterozygous - when the two alleles are different, in such
cases the dominant allele is expressed.
Dominant - a term applied to the trait (allele) that is
expressed irregardless of the second allele.
Recessive - a term applied to a trait that is only expressed
when the second allele is the same (e.g. short plants are
homozygous for the recessive allele).
Phenotype - the physical expression of the allelic
composition for the trait under study.
Genotype - the allelic composition of an organism.
Punnett squares - probability diagram illustrating the
possible offspring of a mating.