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Inheritance Overview
23-1
23.1 Mendel’s laws
• Gregor Mendel
– An Austrian monk, who combined his farmer’s skills with his
training in mathematics and developed certain laws of
heredity in 1860 after doing crosses between
garden pea plants.
– Published a paper stating that parents pass discrete heritable
factors on to their offspring
• Factors retain individuality generation after generation
• Identified that each trait is inherited by a pair of factors, one
from each parent
– One form of a factor may be dominant over an alternative form
– Reasoned that each egg and sperm must contain only 1 copy
of a factor for each trait
23-2
Mendel’s laws cont’d.
•His research and his findings led to develop what
we call Mendel’s law of segregation;
–Each individual has two factors (genes) for
each trait
–The factors segregate (separate) during the
formation of gametes
–Each gamete contains only one factor from
each pair of factors
–Fertilization gives each new individual 2
factors for each trait
23-3
Mendel’s laws cont’d.
Mendel’s law of segregation
1. Each individual has two factors (genes) for
each trait.
We now know that these factors are what we call
genes and that genes are sections of
chromosomes. We have a pair of each
chromosome, known as homologous pairs which
have genes controlling the same traits located at
the same point, or locus, on each member of the
pair
• Information contained within the homologous genes
is not necessarily the same (ex: widow’s peak vs.
straight hairline)
– Alternative forms of a gene for a specific trait
are called alleles
23-4
Mendel’s laws cont’d.
2. The factors segregate (separate) during the
formation of gametes.
This is due to Independent Assortment of
Homologous Chromosomes during Meiosis.
23-5
Mendel’s laws cont’d.
3. Each gamete contains only one factor from
each pair of factors.
Gametes are haploid, which means they only
have 1 chromosome of each pair, therefore each
gamete has only one allele of the pair of alleles
for a trait
23-6
Mendel’s laws cont’d.
4. Fertilization gives each new individual 2
factors for each trait.
2 Haploid Gametes combine to form a
Diploid Zygote. Diploid means that it has a
pair of each chromosome and therefore a
pair of alleles for a given trait.
23-7
Gene locus
23-8
Mendel’s laws cont’d.
Inheritance of a single trait
• In simple dominance, one form of the allele is
dominant over the other form which is recessive.
– Capitol letter- represents the dominant allele
– Small letter- represents the recessive allele
• Dominant-a certain trait will result if the
individual has at least 1 dominant allele
• Recessive- for a recessive trait to result the
individual must have 2 copies of the recessive
allele
23-9
•The inheritance of a single trait cont’d.
–Phenotype- physical appearance of the
individual with regard to a trait.
–Genotype- genetic composition of an individual
with regard to a specific trait
Homozygous - 2 copies of the same allele for a
specific trait
Heterozygous – 1 of each of the 2 copies of the
allele for that trait
• 2 copies of the dominant allele- homozygous
dominant
•1 copy of the dominant allele and 1 of the
recessive- heterozygous
•2 copies of the recessive allele- homozygous
recessive
23-10
Mendel’s laws cont’d.
• The inheritance of a single trait cont’d.
Phenotype vs. Genotype
– Homozygous dominant individual and heterozygous individual.
These 2 individuals will have the same phenotype but different
gentoypes.
– Homozygous recessive individual will have a different phenotype
than the other 2 individuals.
– Example- inheritance of a widow’s peak vs. a straight hairline in
humans
• Alternative forms of alleles for hairline shape
• Widow’s peak is dominant to straight
– W=allele for widow’s peak
– w= allele for straight hairline
23-11
Widow’s peak
23-12
Mendel’s laws cont’d.
• Inheritance of a single trait cont’d.
– Gamete formation
• During meiosis, homologous chromosomes
separate so there is only 1 member of each pair
in a gamete
• There is one allele for each trait, such as
hairline, in each gamete
• No two letters in a gamete can be the same letter of the alphabet
– If genotype is Ww, then gametes from this individual will
contain either a W or a w
– If the genotype is WwLl (looking at 2 traits), gametes can
contain any of the following combinations
» WL, Wl, wL, or wl
23-13
Mendel’s laws cont’d.
• Practice problems
– For each of the following genotypes, give all possible gametes
•
•
•
•
•
WW
WWSs
Tt
Ttgg
AaBb
– For each of the following, state whether the genotype or a
gamete is represented
•
•
•
•
D
Ll
Pw
LlGg
23-14
Mendel’s laws cont’d.
• Practice problems
– For each of the following genotypes, give all possible gametes
•
•
•
•
•
WW
Gametes – All W
WWSs
Tt
Ttgg
AaBb
– For each of the following, state whether the genotype or a
gamete is represented
•
•
•
•
D
Ll
Pw
LlGg
23-15
Mendel’s laws cont’d.
• Practice problems
– For each of the following genotypes, give all possible gametes
•
•
•
•
•
WW
Gametes – All W
WWSs Gametes – WS or Ws
Tt
Ttgg
AaBb
– For each of the following, state whether the genotype or a
gamete is represented
•
•
•
•
D
Ll
Pw
LlGg
23-16
Mendel’s laws cont’d.
• Practice problems
– For each of the following genotypes, give all possible gametes
•
•
•
•
•
WW Gametes – All W
WWSs Gametes – WS or Ws
Tt
Gametes – T or t
Ttgg
AaBb
– For each of the following, state whether the genotype or a
gamete is represented
•
•
•
•
D
Ll
Pw
LlGg
23-17
Mendel’s laws cont’d.
• Practice problems
– For each of the following genotypes, give all possible gametes
•
•
•
•
•
WW Gametes – All W
WWSs Gametes – WS or Ws
Tt
Gametes – T or t
Ttgg Gametes – Tg or tg
AaBb
– For each of the following, state whether the genotype or a
gamete is represented
•
•
•
•
D
Ll
Pw
LlGg
23-18
Mendel’s laws cont’d.
• Practice problems
– For each of the following genotypes, give all possible gametes
•
•
•
•
•
WW Gametes – All W
WWSs Gametes – WS or Ws
Tt
Gametes – T or t
Ttgg Gametes – Tg or tg
AaBb Gametes – AB or aB or Ab or ab
– For each of the following, state whether the genotype or a
gamete is represented
•
•
•
•
D
Ll
Pw
LlGg
23-19
Mendel’s laws cont’d.
• Practice problems
– For each of the following genotypes, give all possible gametes
•
•
•
•
•
WW Gametes – All W
WWSs Gametes – WS or Ws
Tt
Gametes – T or t
Ttgg Gametes – Tg or tg
AaBb Gametes – AB or aB or Ab or ab
– For each of the following, state whether the genotype or a
gamete is represented
•
•
•
•
D
Ll
Pw
LlGg
Gamete (has one allele for a trait)
23-20
Mendel’s laws cont’d.
• Practice problems
– For each of the following genotypes, give all possible gametes
•
•
•
•
•
WW Gametes – All W
WWSs Gametes – WS or Ws
Tt
Gametes – T or t
Ttgg Gametes – Tg or tg
AaBb Gametes – AB or aB or Ab or ab
– For each of the following, state whether the genotype or a
gamete is represented
•
•
•
•
D
Ll
Pw
LlGg
Gamete (has one allele for a trait)
Genotype (has 2 alleles for a trait)
23-21
Mendel’s laws cont’d.
• Practice problems
– For each of the following genotypes, give all possible gametes
•
•
•
•
•
WW Gametes – All W
WWSs Gametes – WS or Ws
Tt
Gametes – T or t
Ttgg Gametes – Tg or tg
AaBb Gametes – AB or aB or Ab or ab
– For each of the following, state whether the genotype or a
gamete is represented
•
•
•
•
D
Ll
Pw
LlGg
Gamete (has one allele for a trait)
Genotype (has 2 alleles for a trait)
Gamete (has one allele for each of 2 traits)
23-22
Mendel’s laws cont’d.
• Practice problems
– For each of the following genotypes, give all possible gametes
•
•
•
•
•
WW Gametes – All W
WWSs Gametes – WS or Ws
Tt
Gametes – T or t
Ttgg Gametes – Tg or tg
AaBb Gametes – AB or aB or Ab or ab
– For each of the following, state whether the genotype or a
gamete is represented
•
•
•
•
D
Ll
Pw
LlGg
Gamete (has one allele for a trait)
Genotype (has 2 alleles for a trait)
Gamete (has one allele for each of 2 traits)
Genotype (has 2 alleles for each of 2 traits)
23-23
Genotype related to phenotype
23-24
Mendel’s laws cont’d.
• One trait crosses-monohybrid cross
• In one-trait crosses, only one trait such as type of
hairline is being considered.
– Let’s consider a specific cross
• If a homozygous woman with a widow’s peak reproduces with a
man with a straight hairline, what kind of hairline will their children
have?
– Use W= widow’s peak, w=straight hairline
– figure out the genotype of each parent:
» Woman has a widow’s peak and we are told she is
homozygous so she is WW
» Man has a straight hairline so he must be ww
– Determine their gametes: her eggs will all have W and his
sperm will all have w
– Put together all possible combinations: all offspring will receive
a W from her and a w from him so all will have Ww
23-25
Mendel’s laws cont’d.
• One-trait cross cont’d.
– In this example there was only 1 combination of eggs and sperm
possible. The children are all monohybrids, they are
heterozygous for one pair of alleles.
– What if the parents were each Monohybrids (Ww)?
• Then sperm could have either W or w, and eggs also could have W
or w
– One way to figure out all of the possible
combinations of eggs and sperm that can occur is
to use a Punnett square
– This is illustrated on the following slide
23-26
23-27
Mendel’s laws cont’d.
• One-trait cross cont’d.
– After the genotypes and phenotypes of offspring are determined, we can
determine the ratios
• In our previous example, in the Punnett square we had the following offspring:
WW, Ww, Ww, and ww
– The genotypic ratio is 1 WW:2 Ww:1 ww
– The phenotypic ratio is 3 individuals with a widow’s peak to 1 individual with
a straight hairline.
– When a monohybrid reproduces with a
monohybrid (a monohybrid cross) the ratio of
expected phenotypes expressed in the
offspring is 3 : 1.
– Another way to phrase the phenotypic ratio is in terms of probability
– This couple has a 75% chance of producing a child with a widow’s peak
and a 25% chance of producing a child with a straight hairline
• The probability will be the same for each pregnancy between this couple
23-28
Mendel’s laws cont’d.
• One-trait crosses and probability
– Product rule of probability
• The chance of 2 or more independent events occurring together is
the product of their chance of occurring separately
– In the cross Ww X Ww, what is the chance of obtaining either a
W or a w from a parent?
» Chance of W = ½ and the chance of w = ½
» Therefore the probability of having these genotypes is as
follows
» Chance of WW= ½ X ½ = ¼
» Chance of Ww = ½ X ½ = ¼
» Chance of wW= ½ X ½ = ¼
» Chance of ww = ½ X ½ = ¼
23-29
Mendel’s laws cont’d.
• One-trait crosses and probability cont’d.
– Sum rule of probability-the chance of an event that
can occur in more than one way occurring a certain
way is the sum of the individual chances of it
occurring that way.
• To calculate the chance of an offspring having a
widow’s peak, add the chances of WW, Ww, or
wW from the preceding slide
¼ + ¼ + ¼ = ¾ or 75%
23-30
Mendel’s laws cont’d.
• The one-trait test cross
– You cannot distinguish between a homozygous dominant
individual and a heterozygous individual just by looking because
they are phenotypically the same
– Breeders of plants and animals may do a test cross to
determine the likely genotype of an individual with the
dominant phenotype
• Cross with a recessive individual which has a known
genotype - homozygous recessive.
• If there are any offspring produced with the recessive
phenotype, then the dominant parent must be
heterozygous
23-31
One-trait testcross
When crossed with
Recessive all
resulting offspring
show the dominant
trait. Therefore the
dominant parent
must be
Homozygous
Dominant
23-32
One-trait testcross
When crossed with
Recessive some
resulting offspring show
the dominant trait and
some show the
recessive trait.
Therefore the dominant
parent must be
Heterozygous Dominant
23-33
Mendel’s laws cont’d.
• Practice problems
Both a man and a woman are heterozygous for freckles. Freckles
are dominant over no freckles. What is the chance that their child
will have freckles?
Mother - _____________
Father - _____________
23-34
Both a man and a woman are heterozygous for freckles.
Freckles are dominant over no freckles. What is the chance
that their child will have freckles?
Ff________
Mother - _____
Ff__________
Father - ___
23-35
Both a man and a woman are heterozygous for freckles.
Freckles are dominant over no freckles. What is the chance
that their child will have freckles?
Ff________
Mother - _____
F
f
F
Ff__________
Father - ___
f
23-36
Both a man and a woman are heterozygous for freckles.
Freckles are dominant over no freckles. What is the chance
that their child will have freckles?
Ff________
Mother - _____
F
F
Ff__________
Father - ___
f
FF
f
23-37
Both a man and a woman are heterozygous for freckles.
Freckles are dominant over no freckles. What is the chance
that their child will have freckles?
Ff________
Mother - _____
F
Ff__________
Father - ___
F
f
FF
Ff
f
23-38
Both a man and a woman are heterozygous for freckles.
Freckles are dominant over no freckles. What is the chance
that their child will have freckles?
Ff________
Mother - _____
Ff__________
Father - ___
F
f
F
FF
Ff
f
fF
23-39
Both a man and a woman are heterozygous for freckles.
Freckles are dominant over no freckles. What is the chance
that their child will have freckles?
Ff________
Mother - _____
Ff__________
Father - ___
F
f
F
FF
Ff
f
fF
ff
23-40
Both a man and a woman are heterozygous for freckles.
Freckles are dominant over no freckles. What is the chance
that their child will have freckles?
Ff________
Mother - _____
F
Ff__________
Father - ___
3:1 Ratio
Freckles:No Freckles
75% Chance child will have Freckles
f
F
f
FF
Ff
Freckles
Freckles
fF
ff
Freckles
NO Freckles
23-41
Both you and your sibling have attached ear lobes, but your
parents have unattached lobes. Unattached earlobes (E) are
dominant over attached (e). What are the genotypes of your
parents?
Mother - _____________
Father - _____________
23-42
Both you and your sibling have attached ear lobes, but your
parents have unattached lobes. Unattached earlobes (E) are
dominant over attached (e). What are the genotypes of your
parents?
Mother - _____________
Father - _____________
ee
Attached
23-43
Both you and your sibling have attached ear lobes, but your
parents have unattached lobes. Unattached earlobes (E) are
dominant over attached (e). What are the genotypes of your
parents?
Mother - _____________
Unattached Earlobes
e
Father - _____________
Unattached Earlobes
e
ee
Attached
23-44
Both you and your sibling have attached ear lobes, but your
parents have unattached lobes. Unattached earlobes (E) are
dominant over attached (e). What are the genotypes of your
parents?
Mother - _____________
Unattached Earlobes
E
e
E
Father - _____________
Unattached Earlobes
e
ee
Attached
23-45
Both you and your sibling have attached ear lobes, but your
parents have unattached lobes. Unattached earlobes (E) are
dominant over attached (e). What are the genotypes of your
parents?
Mother - ____Ee_________
Unattached Earlobes
E
Father - ___Ee__________
Unattached Earlobes
e
Parents had to both be (Ee) Heterozygous
Dominant to have a recessive offspring
E
e
EE
Ee
unattached
unattached
eE
ee
unattached
Attached
23-46
A father has dimples, the mother of his children does not, and all 5 of
their children have dimples. Dimples (D) are dominant over no dimples
(d). Give the probable genotypes of all persons concerned.
Mother - _____________
Father - _____________
23-47
A father has dimples, the mother of his children does not, and all 5 of
their children have dimples. Dimples (D) are dominant over no dimples
(d). Give the probable genotypes of all persons concerned.
Mother - _____________
Dimples
Dimples
Dimples
Dimples
Father - _____________
23-48
A father has dimples, the mother of his children does not, and all 5 of
their children have dimples. Dimples (D) are dominant over no dimples
(d). Give the probable genotypes of all persons concerned.
Mother - _____________
No Dimples
Dimples
Dimples
Dimples
Dimples
Father - _____________
Dimples
23-49
A father has dimples, the mother of his children does not, and all 5 of
their children have dimples. Dimples (D) are dominant over no dimples
(d). Give the probable genotypes of all persons concerned.
Mother - __dd_________
No Dimples
Dimples
Dimples
Dimples
Dimples
Father - _____________
Dimples
23-50
A father has dimples, the mother of his children does not, and all 5 of
their children have dimples. Dimples (D) are dominant over no dimples
(d). Give the probable genotypes of all persons concerned.
Mother - __dd_________
No Dimples
d
d
Dimples
Dimples
Dimples
Dimples
Father - _____________
Dimples
23-51
A father has dimples, the mother of his children does not, and all 5 of
their children have dimples. Dimples (D) are dominant over no dimples
(d). Give the probable genotypes of all persons concerned.
Mother - __dd_________
No Dimples
Father - _____________
Dimples
d
d
Dd
Dimples
Dd
Dimples
Dd
Dimples
Dd
Dimples
23-52
A father has dimples, the mother of his children does not, and all 5 of
their children have dimples. Dimples (D) are dominant over no dimples
(d). Give the probable genotypes of all persons concerned.
Mother - __dd_________
No Dimples
Father - _____________
Dimples
d
d
D
Dd
Dimples
Dd
Dimples
D
Dd
Dimples
Dd
Dimples
23-53
A father has dimples, the mother of his children does not, and all 5 of
their children have dimples. Dimples (D) are dominant over no dimples
(d). Give the probable genotypes of all persons concerned.
Mother - __dd_________
No Dimples
Father - _____DD________
Dimples
d
d
D
Dd
Dimples
Dd
Dimples
D
Dd
Dimples
Dd
Dimples
Father has to be Homozygous Dominant or
approx. ½ of the children won’t have dimples
23-54
Mendel’s laws cont’d.
• The inheritance of two traits
– Mendel reasoned from the results of his pea plant crosses
that each pair of factors assorts independently
into gametes
• If one pair of homologous chromosomes has a W on one and
w on the other, and a second pair has an R on one and an r
on the other, when gametes form all possible combinations of
these letters are possible: WR, Wr, wR, and wr
• Called the Law of Independent Assortment
23-55
Segregation and independent assortment
• Fig. 23.6
23-56
Mendel’s laws cont’d.
• The law of independent assortment states :
– Each pairs of factors assorts independently
(without regard to how the others separate)
– All possible combinations of factors can occur in
the gametes
23-57
Mendel’s laws cont’d.
• Two-trait crosses-dihybrid cross
– Let’s consider two traits, hairline and finger length
• We’ll use the W and w as before, and now S=short fingers and
s=long fingers
– A person who is WwSs widow’s peak and short fingers and a
person who is also WwSs have children
• Figure the gametes for each parent
– WS, Ws, wS, and ws for both
• Make a big Punnett square as illustrated on the following slides
• Figure the genotypes of the offspring
23-58
• In two-trait crosses, genotypes of the
parents require four letters because
there is an allelic pair for each trait.
• Gametes will contain one letter of each
kind in every possible combination.
23-59
Dihybrid cross
23-60
Dihybrid
cross
23-61
When a dihybrid (heterozygous for 2
traits) reproduces with a dihybrid the
ratio of expected phenotypes expressed
in the offspring are 9 : 3 : 3 : 1.
Phenotypic ratio:
9 widow’s peak, short fingers:
3 widow’s peak, long fingers;
3 straight hairline, short fingers;
1 straight hairline, long fingers
23-62
Mendel’s laws cont’d.
• Two-trait crosses and probability
– Probability laws
• Probability of widow’s peak = ¾
Probability of short fingers= ¾
• Probability of straight hairline= ¼
Probability of long fingers= ¼
– Using the product rule
•
•
•
•
Probability of widow’s peak and short fingers = ¾ X ¾ = 9/16
Probability of widow’s peak and long fingers = ¾ X ¼ = 3/16
Probability of straight hairline and short fingers = ¼ X ¾ = 3/16
Probability of straight hairline and long fingers = ¼ X ¼ = 1/16
23-63
Mendel’s laws cont’d.
• The two-trait testcross
– Cross an individual with the dominant
phenotype for each trait with an
individual with the recessive phenotype
of both traits
• If the dominant individual is homozygous for both traits, all offspring
will show the dominant phenotypes
• If the dominant individual is heterozygous, then a Punnett square
shows the following phenotypic ratio of offspring
– 1 dominant for both: 1 dominant for first trait recessive for
second: 1 recessive for first trait dominant for second: and 1
recessive for both
23-64
Two-trait testcross
23-65
Mendel’s laws
• Practice problems
– Attached earlobes are recessive, What genotype do children
have if one parent is homozygous for earlobes and homozygous
dominant for hairline, and the other is homozygous dominant for
unattached earlobes and homozygous recessive for hairline?
– If an individual from this cross reproduces with another of the
same genotype, what are the chances that they will have a child
with a straight hairline and attached earlobes?
– A child who does not have dimples or freckles is born to a man
who has dimples and freckles (both dominant) and a woman
who does not. What are the genotypes of all persons
concerned?
23-66
23.2 Beyond simple inheritance
patterns
• Polygenic inheritance
– Controlled by 2 or more sets of alleles
– Each dominant allele codes for a product and
effects are additive
– Result is a continuous range of phenotypes
• Distribution resembles a bell curve
• The more gene pairs involved, the more continuous the pattern of
variation
• Ex: human height, skin pigmentation
23-67
Skin Color
• The inheritance of skin color, determined by an
unknown number of gene pairs, is a classic example
of polygenic inheritance.
• A range of phenotypes exist and several possible
phenotypes fall between the two extremes of very dark
and very light.
• The distribution of these phenotypes follows a bellshaped curve.
23-68
Polygenic inheritance
• Fig. 23.9
23-69
Beyond simple inheritance patterns
cont’d.
• Polygenic inheritance cont’d.
– Skin color
• Controlled by many gene pairs and many alleles
• Assuming the simplest model, 2 alleles at loci
– A and B
– When an AaBb person has children with another AaBb person,
children can range from very light to very dark
AABB
Very dark
AABb or AaBB
Dark
AaBb, AAbb,
aaBB
Medium brown
Aabb or aaBb
Light
aabb
Very light
23-70
Beyond simple inheritance patterns
cont’d.
• Environmental influences
– Environment can influence gene expression and
therefore phenotype
• Ex: sunlight exposure on skin; coat color in
Himalayan rabbits
– Human twin studies
• Polygenic traits are most influenced
• “nature vs. nurture”
• Identical twins separated at birth are studied
– If they share a trait in common even though raised in different
environments, it is likely genetic
23-71
Coat color in Himalayan rabbits
• Fig. 23.10
23-72
Beyond simple inheritance patterns
cont’d.
• Incomplete dominance and codominance
– Incomplete dominance
• Heterozygous individual has a phenotype
intermediate to the two homozygous
individuals
• Ex: curly-haired Caucasian woman and a
straight-haired Caucasian man produce
wavy-haired children
– When 2 wavy-haired people have children, the phenotypic ratio is
1 curly: 2 wavy: 1 straight
• Such a cross would produce a phenotypic ratio of 1 : 2 : 1.
23-73
Incomplete dominance
23-74
ABO Blood Types
– Codominance
• Occurs when both alleles are equally
expressed
• Ex: type AB blood has both A antigens and B
antigens on red blood cells
• A person can have an allele for an A antigen (blood type A) or a B
antigen (blood type B), both A and B antigens (blood type AB), or no
antigen (blood type O) on the red blood cells.
• Human blood types can be type A (IAIA or IA i), type B
(IBIB or IBi), type AB (IAIB), or type 0 (ii).
23-75
Beyond simple inheritance patterns cont’d.
• Multiple allele inheritance
– The gene exists in several allelic forms, but each
person still has only 2 of the possible alleles
– ABO blood types
• IA = A antigens on RBCs
• IB = B antigens on RBCs
• i = has neither A nor B antigens on RBCs
– Both IA and IB are dominant over I, IA and IB are
codominant
Phenotype
Genotype
A
IAIA or IAi
B
IBIB or IBi
AB
I AI B
O
ii
23-76
Inheritance of blood type
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Beyond simple inheritance patterns
cont’d.
• ABO blood types cont’d.
– Paternity testing- ABO blood groups often used
• Can disprove paternity but not prove it
– Rh factor- another antigen on RBCs
• Rh positive people have the antigen
• Rh negative people lack it
– There are multiple alleles for Rh negative, but
all are recessive to Rh positive
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Beyond simple inheritance patterns
cont’d.
• Practice problems
– A polygenic trait is controlled by three pairs of alleles. What are
the two extreme genotypes for this trait?
– What is the genotype of the lightest child that could result from a
mating between two medium-brown individuals?
– A child with type O blood is born to a mother with type A blood.
What is the genotype of the child? The mother? ‘what are the
possible genotypes of the father?
– From the following blood types determine which baby belongs to
which parents:
• Baby 1 type O
• Baby 2 type B
Mrs. Doe type A
Mr. Doe type A
Mrs. Jones type A
Mr. Jones type AB
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23.3 Sex-linked inheritance
• Sex chromosomes
– 22 pairs of autosomes, 1 pair of sex
chromosomes
• X and y - X chromosome has many genes, the
Y chromosome does not
– In females, the sex chromosomes are XX
– In males, the sex chromosomes are XY
» Note that in males the sex chromosomes are not
homologous
– Traits controlled by genes in the sex
chromosomes are called sex-linked traits
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Sex-linked inheritance cont’d.
• X-linked traits
– Red-green colorblindness is X-linked
• The X chromosome has genes for normal color vision
– XB = normal vision
– Xb – colorblindness
– Genotypes
XBXB
XBXb
XbXb
XBY
XbY
Phenotypes
female with normal color vision
carrier female with normal color vision
colorblind female
male with normal color vision
colorblind male
A female would have to have two recessive genes to
express the trait; a male would only need one.
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X-Linked Alleles
• The key for an X-linked problem shows the allele
attached to the X as in:
• XB = normal vision
• Xb = color blindness.
• Females with the genotype XBXb are carriers
because they appear to be normal but each son has
a 50% chance of being color blind depending on
which allele the son receives.
• XbXb and XbY are both colorblind.
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Cross involving an X-linked allele
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Color Blindness and X- Linked Recessive Disorders
• Three types of cones are in the retina detecting red,
green, or blue.
• Genes for blue cones are autosomal; those for red
and green cones are on the X chromosome.
• Males are much more likely to have red-green color
blindness than females.
• About 8% of Caucasian men have red-green color
blindness.
• In pedigree charts that show the inheritance pattern for
X-linked recessive disorders, more males than females
have the trait because recessive alleles on the X
chromosome are expressed in males.
• A grandfather passes an X-linked recessive disorder
to a grandson through a carrier daughter.
• X-linked recessive disorders include red-green color
blindness, muscular dystrophy, and hemophilia. 23-84
Sex-linked inheritance-practice problems
• Both the mother and the father of a colorblind male appear to be
normal. From whom did the son inherit the allele for colorblindness?
What are the genotypes of the mother, father, and the son?
• A woman is colorblind. What are the chances that her son will be
colorblind? If she is married to a man with normal vision, what are
the chances that her daughters will be colorblind? Will be carriers?
• Both the husband and the wife have normal vision. The wife gives
birth to a colorblind daughter. Is it more likely the father had normal
vision or was colorblind? What does this lead you to deduce about
the girl’s parentage?
• What is the genotype of a colorblind male with long fingers is s=long
fingers? If all his children have normal vision and short fingers, what
is the likely genotype of the mother?
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23.4 Inheritance of linked genes
• A single chromosome has many genes
– The sequence is fixed because each allele has a
specific locus
• All genes on a single chromosome form a
linkage group
• When linkage is complete, a dihybrid produces only 2 types of
gametes
• Any time traits are inherited together, a linkage group is suspected
• Or, if very few recombined phenotypes appear in offspring, linkage
is also suspected
– Recall that during meiosis crossing over sometimes occurs
• If crossing over occurs between 2 alleles of interest, then 4 types of
gametes are formed instead of 2
– Fig. 23.14 on the following slide illustrates linkage
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Linkage group
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Inheritance of linked genes cont’d.
• Ocurrence of crossing over can indicate the sequence of
genes on a chromosome
– Crossing over occurs more frequently between distant genes
than between those close together on a chromosome
• Practice problems
– When AaBb individuals reproduce, the phenotypic ratio is about
3:1. What ratio was expected? What may have caused the
observed ratio?
– The genes for ABO blood type and for fingernails are on the
same homologous pair of chromosomes. In an actual family,
45% of offspring have type B blood and no fingernails, and 45%
have type O blood and fingernails; 5% have type B blood and
fingernails, and 5% have type O blood and no fingernails. What
process accounts for the recombinant phenotypes?
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