Transcript Complex Inheritance and Human Heredity

Ch 11 Complex Patterns of
Inheritance
11.1 Basic Patterns of Human Inheritance
Main Idea – The inheritance of a trait over
several generations can be shown in a
pedigree.
Recessive Disorders
1. Simple recessive heredity is the cause of
most genetic disorders.
2. A recessive trait is expressed when the
individual is homozygous recessive for
the trait.
Therefore, those individuals with at least one
dominant allele will NOT express the
recessive trait.
An individual who is heterozygous for a
recessive disorder is called a carrier.
• The following chart shows simple
recessive human disorders.
• Remember an individual must inherit a
recessive allele from each parent in order
to have this disease.
Disorder
Occurrence
Cause
Affect
Cure/Treatment
Cystic fibrosis
most common in
Caucasians
1:3500
Enzyme
deficiency
Excess mucus in
lungs
No cure
Enzyme supplement
Chest percussions
1:17,000
Lack
melanin
White hair
Pale skin
Pink pupils
No cure
Protect from sun
Galactosemia
1:50,000-70,000
Missing
enzyme
Mental
deterioration
Liver, kidney
problems
No cure
Restricted diet, avoid milk
Tay-Sachs
disease
Jewish
1:2500
Lack
enzyme
Mental
retardation
No cure or treatment
Death by age 5
PKU
1:10,000-50,000
Defective
enzyme
Mental
retardation
Restricted diet
Albinism
Albinsim
Tay Sachs Disease
Recessive disorders are more common
because carriers (heterozygous alleles) do
not display the disorder so they don’t realize
they could pass it on to offspring.
Dominant Traits/Disorders
4. These disorders are caused by the
presence of a single dominant allele to
be expressed in an individual. (fewer in
number because if the trait interferes
with survival, that individual is less likely
to pass the gene to the next generation.)
5. Examples of simple dominant traits.
hitchhiker’s thumb, tongue rolling, free
earlobes
Dominant Traits
• Six fingers or six toes - polydactyly
Polydactyly
Girl in Burma with many digits
Total fingers and toes = 26
6 fingers on each hand = 12
7 toes on each foot = 14
Tongue rolling and Ear lobes
Free vs. attached
Widow’s Peak
Hitchhiker’s thumb
Cleft chin is a dominant trait
6. Huntington’s Disease is a dominant
inherited disorder that affects the CNS
(central nervous system) and is
fatal/lethal.
It does not occur until between the ages of
30-50.
A person with this disease has a 50%
chance of passing it on to his/her
children.
Huntington’s Disease
7. Achondroplasia (dwarfism) – 75% of
individuals are born to parents of normal
size.
Therefore, the condition occurred because
of a new mutation or genetic change.
Achondroplasia (dwarfism)
Dominant disorders are caused by the
presence of a single dominant allele.
Therefore, those that do not have the
disorder are homozygous recessive for
the trait.
Making a Pedigree
8. A pedigree is a family tree of inheritance
used to predict disorders in future offspring.
9. In a pedigree, a circle represents a female
and a square represents a male.
10. Horizontal connecting lines indicate
parents. Vertical lines that drop down
between the parents represent offspring.
11. Roman numerals (I, II, III, IV) indicate
the generations.
12. Arabic numbers (1, 2, 3, 4) indicate
individuals.
13. The trait being studied is represented by
a shaded circle or square.
14. A carrier is a heterozygous individual
that carries the trait but does not show
the trait phenotypically.
15. In a pedigree, a carrier is represented by
a ½ shaded circle or square.
Achondroplasia pedigree
Dominant Disorder
Analyze the pedigree - Dogs
Generation I
Generation II
Generation III
Generation IV
White = Tall Dominant
Black = Short Recessive
Male
Female
1. How many generations are shown in the pedigree?
Generation I
Generation II
Generation III
Generation IV
White = Tall Dominant
Black = Short Recessive
Male
Female
2. How many offspring did the parents in the first generation have?
Generation I
Generation II
Generation III
Generation IV
White = Tall Dominant
Black = Short Recessive
Male
Female
3. What does the square in generation I stand for? Why is it half shaded?
Generation I
Generation II
Generation III
Generation IV
White = Tall Dominant
Black = Short Recessive
Male
Female
4. Which dog was the first in the family to be short?
Generation I
Generation II
Generation III
Generation IV
White = Tall Dominant
Black = Short Recessive
Male
Female
5. A female dog from generation III has four puppies. How many of these
offspring carry (are carriers for) the short trait? How many of the offspring are
short?
Generation I
Generation II
Generation III
Generation IV
White = Tall Dominant
Black = Short Recessive
Male
Female
Inherited Traits - Chickens
1. How many mating pairs are
shown on this pedigree?
2. How many chickens on this
pedigree are female?
3. How many chickens on this
pedigree are male?
4. How many generations are
shown here?
5. How many roosters (males) had
the trait being studied?
6. How many roosters (males)
lacked the trait being studied?
7. How many hens (females) had
the trait being studied?
8. How many hens (females)
lacked the trait being studied?
• Inbreeding
• May result in a far higher phenotypic
expression of harmful recessive genes
• increases the chances of passing harmful
recessive traits to the next generation.
• Selective breeding is a process to
produce organisms with desired/favorable
traits.
9. Did any inbreeding occur?
If so where?
10. If your answer to question 9 is “yes” can you
explain the results of the inbreeding?
How does this relate to selective breeding?
Making A Pedigree
Draw a pedigree that traces eye color for
three generations.
Assume that green eye is dominant and
the blue-eye trait is recessive.
The mother in generation I is
homozygous recessive, and the father is
homozygous dominant.
Indicate the generation number and
individual number.
John Jones, a green-eyed man, marries Jill
Smith, a blue-eyed woman. John and Jill have
four green-eyed children: John Jr., Alice, Lisa,
and Sean. John Jr. later marries blue-eyed
Pamela, and they have four children: Jessica,
Shari, Mary, and John III. Shari and Mary both
have green eyes, Jessica and John III have blue
eyes.
Sean marries Robin, a blue-eyed woman. Both
of Robin’s parents have blue eyes also. Sean
and Robin have four children: Nicholas, Harry,
Donna, and Sean Jr. Nicholas, Harry, and Donna
all have green eyes. Sean Jr. has blue eyes.
G = green eyes
gg
Pamela
gg
Jessica
Gg
Shari
g = blue eyes
GG
gg
gg
John
Jill
Robin’s
father
Gg
John Jr.
Gg
Alice
Gg
Mary
gg
John III
Gg
Lisa
Gg
Gg
gg
Sean
Robin
Gg
Nicholas Harry
Gg
gg
Robin’s
mother
gg
Donna Sean Jr.
Draw a pedigree that traces the trait for green eyes for three generations.
GG
gg
gg
Gg
Gg
Gg
Gg
gg
gg
Gg
Gg
gg
Gg
Gg
gg
gg
Gg
gg
Draw a pedigree that traces the trait for blue eyes for three generations.
GG
gg
gg
Gg
Gg
Gg
Gg
gg
gg
Gg
Gg
gg
Gg
Gg
gg
gg
Gg
gg
GG
gg
gg
Gg
Gg
Gg
Gg
gg
gg
Gg
Gg
gg
Gg
Gg
gg
gg
Gg
gg
Galactosemia
Recessive disorder that cannot
breakdown galactose
Pedigree
• Draw the pedigree of a boy who has galactosemia.
His father has galactosemia, his paternal
grandparents are phenotypically normal, and his
mother and maternal grandparents are both
phenotypically normal.
G= normal
g= galactosemia
½ shaded = carrier
Maternal
grandparents
Paternal
grandparents
Gg
Gg
Gg
Gg
father
gg
Gg
boy
gg
Textbook p. 300
View Pedigree
Read paragraph
Answer question at the end of paragraph.
I’m My Own Grampa
Many, many years ago
When I was twenty-three,
I was married to a widow
Who was pretty as could be.
This widow had a grown-up daughter
Who had hair of red.
My father fell in love with her,
And soon they too were wed.
This made my dad my son-in-law
And really changed my very life.
My daughter was my mother,
For she was my father’s wife.
To complicate the matters worse,
Even though it brought me joy,
I soon became the father
Of a bouncing baby boy.
My little baby then became
A brother-in-law to dad.
And so became my uncle,
Though it made me very sad.
For if he were my uncle,
then that also made him brother
To the widow’s grown-up daughter
Who was, of course, was my step-mother.
My father’s wife then had a son,
Who kept them on the run.
And he became my grandson,
For he was my daughter’s son.
My wife is now my mother’s mother.
And it surely makes me blue.
Because, although she is my wife,
She is my grandmother, too.
Now, if my wife is my grandmother,
Then I am her grandchild.
And every time I think of it
It nearly drives me wild.
For now I have become
The strangest tale you ever saw.
As the husband of my grandmother,
I am my own grampa!
Ch 11.2 Complex Patterns of Inheritance
COMPLEX
Main Idea – Complex inheritance of traits
does NOT followPATTERNS
inheritance patterns
OF
described by Mendel.
INHERITANCE
1. Incomplete Dominance
The heterozygote is an intermediate
phenotype between the two
homozygotes
2. P= RR = red RR1 = pink R1R1 = white
Ex. Red flower X white flower  pink
flower
RR
X
R 1R 1
 RR1
Crosses Involving
Incomplete Dominance
Alleles: R = red R¹ = white
Genotypes: RR = red R1R1 = white; RR¹ = pink
1. In Japanese four-o’ clocks, predict the phenotype of a
cross between:
a. a red plant and a white plant
_____________
Alleles: R = red R¹ = white
Genotypes: RR = red R1R1 = white; RR¹ = pink
1. In Japanese four-o’ clocks, predict the phenotype of a
cross between:
b. a white plant and a pink plant
_____________
Alleles: R = red R¹ = white
Genotypes: RR = red R1R1 = white; RR¹ = pink
1. In Japanese four-o’ clocks, predict the phenotype of a
cross between:
c. a red plant and a pink plant
_____________
Alleles: R = red R¹ = white
Genotypes: RR = red R1R1 = white; RR¹ = pink
1. In Japanese four-o’ clocks, predict the phenotype of a
cross between:
d. 2 pink plants
_____________
2. In some cats the genes for tail length shows the
incomplete dominance. Cats with long tails and those
with no tails are homozygous for the respective alleles.
Cats with one long-tail allele and one no-tail allele have
short tails. Predict the phenotype ratio of a cross
between:
a. a long-tail cat and a cat with no tail
___________________
b. A long-tail cat and a short-tail cat
_____________________
c. a short-tail cat and a cat with no tail
______________
d. two short-tail cats
_____________
3. CODOMINANCE
Both alleles are expressed
in the heterozygous condition
Ex. checkered chicken
black chicken X white chicken  checkered chicken
BB
X
WW

BW
Sickle Disease
4. Sickle cell disease is also called sickle
cell anemia and is a blood disorder common
in people of African descent
5. Changes in hemoglobin cause the cells to
sickle.
6. Sickle cell trait results from having 1 allele
or is a heterozygous condition.
7. People with sickle-cell trait (heterozygous)
can resist malaria. Death rate due to
malaria is lower where the sickle cell trait is
higher.
8. Because less malaria exists in those
areas, more people live to pass on the
sickle-cell trait.
Ex: Sickle Cell
SS - normal
Ss - some normal, some sickled
sickle cell trait
ss - all sickled-sickle cell disease
Roan cattle another example of
codominance
RR red hair, RR’ roan(pinkish brown),
R’R’ white
Type AB blood type
is an example of
codominance
9. Multiple Alleles--Inheritance involving
more than 2 alleles
10. Blood groups in humans involve
multiple alleles.
11. Blood groups—3 alleles; and 4
phenotypes – type A, type B, type AB,
and type O
• Type O is recessive and represented by ii;
universal donor
• Type AB is codominant and represented
by IAIB; universal recipient.
Ex: Blood Groups -3 alleles (A, B, and O)
IA Codes for type A blood
IB Codes for type B blood
i Codes for type O blood
Phenotype
Type O blood
Type A blood
Type B blood
Type AB blood
Genotype
ii
IAi; IA IA
IBi; IB IB
IA IB
Type AB blood type is an example of
codominance.
Blood Types
Complex Inheritance and Human Heredity
12. Another example of multiple alleles-Coat Color of Rabbits
hierarchy of dominance (ex. C>Cch > Ch > c
Presence of multiple alleles increases the
possible number of genotypes and phenotypes
Chinchilla
Light gray
Dark gray
Albino
Himalayan
MULTIPLE ALLELES
Ex: Coat color in rabbits –
• C - gray
• Cch - chinchilla
• Ch - himalayan
• c - albino (white)
• 10 possible genotypes
• 5+ phenotypes
Crosses Involving Multiple
Alleles
IA Codes for type A blood
IB Codes for type B blood
i Codes for type O blood
Phenotype
Genotype
Type O blood
ii
Type A blood
IAi; IA IA
Type B blood
IBi; IB IB
Type AB blood
IA IB
1. A woman homozygous for type B blood
marries a man who is heterozygous type
A. What will be the possible genotypes
and phenotypes of their children?
2. A man with type O blood marries a woman
with type AB blood. What will be the
possible genotypes and phenotypes of
their children?
3. A type B woman, whose mother was type
O, marries a type O man. What will be the
possible genotypes and phenotypes of their
children?
4. A type A woman, whose father was type B,
marries a type B man whose mother was
type A. What will be their childrens’ possible
phenotypes and genotypes?
5. What is the probability that a couple whose
blood types are AB and O will have a type A
child?
6. A couple has a child with type A blood. If
one parent is type O, what are the possible
genotypes of the other parent?
13. Epistasis-- one allele hides the effects of
another allele
Example: coat color in Labrador retrievers
two sets of alleles E and B
eebb
eeB_
No dark pigment present in fur
E_bb
E_B_
Dark pigment present in fur
EPISTASIS
allele E determines whether the fur will have dark pigment,
alleles
ee = no dark pigment
allele B determines how dark the pigment will be
B = black; b = chocolate (determines how dark pigment will be)
EEBB or EEBb = black
EEbb or Eebb = brown
eeBb, eeBB = yellow with black pigment (black nose)
eebb = yellow with no pigment (pink nose)
The e allele masks the dominant B allele.
Inheritance and Human Heredity
14. SexComplex
DeterminationSex chromosomes
determine an individual’s gender
• Autosomes are the first 22 pairs of
chromosomes
• Sex chromosomes are the 23rd pair; X and Y
In humans gametes: XX= females
• XY= males
The sex of the offspring is determined by
the chromosomes of the sperm cell.
1. What do the letters X and Y stand for?
the sex chromosomes
2. Which chromosome is found only in the male?
Y chromosomes
3. True or false? A person having two X chromosomes is female.
true
4. In the mating shown in the diagram, which statement is true?
Male
a. All the offspring are female.
XY
b. All the offspring are male.
c. One-half the offspring are female.
d. Three of the four offspring are female.
5. What happens to offspring with an extra sex chromosome, such as
XXX or XXY?
some of these individuals exhibit
mental retardation. Others, although
Female
X
X
Y
XX
XY
XX
XY
XX
leading active lives, will be unable
to have children.
X
15. Dosage Compensation
• In females, one X chromosome
deactivates; x-inactivation
• Barr Bodies- are the darkly stained
inactive x chromosome
• Cause calico cats, pigmentation
• gene is located on X
• chromosomes
16.Sex-linked traits – are also
called x linked traits
• Located on X chromosome
• Since males have one X, they are affected
more frequently
• Passed from mother to son because
inherit the x chromosome from her
• Ex: red-green color blindness
• hemophilia (failure of blood to clot; called
“free bleeders”)
Colorblindness
H= normal h= hemophilia
XH XH = normal female XH Xh = carrier female XhXh = hemophiliac female
XH Y = normal male
Xh Y =hemophiliac male
1. A woman who is heterozygous for hemophilia
marries a normal man. What will be the possible
phenotype ratio of the males versus females?
H= normal h= hemophilia
XH XH = normal female XH Xh = carrier female XhXh = hemophiliac female
XH Y = normal male
Xh Y =hemophiliac male
2. A woman who is a carrier for hemophilia marries a
hemophiliac man. What will be their children’s
possible male/female phenotypes?
H= normal h= hemophilia
XH XH = normal female XH Xh = carrier female XhXh = hemophiliac female
XH Y = normal male
Xh Y =hemophiliac male
3. A hemophiliac woman has a phenotypically normal
mother. What are the possible genotypes of her
mother and her father?
H= normal h= hemophilia
XH XH = normal female XH Xh = carrier female XhXh = hemophiliac female
XH Y = normal male
Xh Y =hemophiliac male
4. A phenotypically normal woman has phenotypically normal
parents. However she has a phenotypically hemophiliac
brother. (a) what are the chances of her being a carrier for
hemophilia? (b) If she is a carrier and marries a normal male,
what is the chance of a child being a hemophiliac?
H= normal h= hemophilia
XH XH = normal female XH Xh = carrier female XhXh = hemophiliac female
XH Y = normal male
Xh Y =hemophiliac male
5. A phenotypically normal man (who has a hemophiliac
brother) marries a homozygous normal woman. What
is the probability that any of their children
(male/female) will be a hemophiliac?
C = normal
c= colorblind
XCXC = normal female XC Xc = carrier female XcXc = colorblind female
XC Y = normal male
Xc Y = colorblind male
6. If a normal-sighted woman whose father was colorblind
marries a color-blind man, what is the probability that they will
have a son who is color-blind? What is the probability that
they will have a color-blind daughter?
C = normal
c= colorblind
XCXC = normal female XC Xc = carrier female XcXc = colorblind
female
XC Y = normal male
Xc Y = colorblind male
7. What is the probability that a color-blind woman who
marries a man with normal vision will have a color-blind
child?
R = normal
r= white
XRXR = normal female XR Xr = carrier female
XrXr = white eyed female
XR Y = normal male
Xr Y = white eyed male
8. In fruit flies, white eyes is a sex-linked
recessive trait. Normal eye color is red. If a
white-eyed male is crossed with a
heterozygous female, what proportion of
the offspring will have red eyes?
Sex-Influenced Traits
Gene is located on an autosome not a sex
chromosome
Appears to be dominant in one gender and
recessive in another
• Baldness is dominant in males; recessive in
females
• A male is bald if he is heterozygous for the
trait
• A female is bald if she is homozygous
recessive
17. Polygenic Traits
Controlled by multiple pairs of genes; results
in numerous phenotypes
Example of human polygenic traits – skin
color, height, eye color, fingerprints
18. Environmental Influence on
phenotype -Sunlight and water can influence phenotype;
ex. Leaves droop, flower buds shrivel,
chlorophyll disappears, roots stop growing
Temperature ex. Siamese cat – more
pigment in cooler conditions
Affects the cats
Phenotype
only
19. Twin Studies
A. Focuses on identical twins
B. Identical twins have the same inherited
traits.
C. Influenced by the environment
• Traits appearing frequently are presumed
to be genetic
• Traits expressed differently are probably
influenced by the environment
11.3 Chromosomes and
Human Heredity
Main Idea – Chromosomes can
be studied using karyotypes.
Normal Female Karyotype
www.miscarriage.com.au/.../karyotype_normal.jpg
Normal Male Karyotype
www.contexo.info/DNA_Basics/images/karyotype1.gif
1. Scientists also study whole chromosomes
by using images of chromosomes stained
during metaphase. The pairs of
homologous chromosomes arranged in
decreasing size produce a picture called a
karyotype. 22 autosome pairs are
matched together with 1 pair of
nonmatching sex chromosomes.
2. Chromosomes end in protective caps
called telomeres. These might be
involved in aging and cancer. Made of
DNA and a protein.
Analogy - shoestring
3. Cell division during which sister
chromatids fail to separate properly is
called nondisjunction. Nondisjunction
alters the chromosome number in
gametes.
4. If nondisjuction occurs during meiosis I or
meiosis II, the resulting gametes will not
have the correct number of chromosomes.
5. Trisomy – having a set of three
chromosomes of one kind.
6. Monosomy – have only one of a particular
type of chromosome.
7. If nondisjunction occurs during meiosis I,
2 trisomy gametes result, and 2
monosomy gametes result. All four
gametes would be abnormal.
8. If nondisjuction occurs during meiosis II,
2 normal, 1 trisomy and 1 monosomy
gametes result.
Nondisjunction in meiosis I
Results in 2 trisomy gametes, and 2 monsomy gametes
Nondisjunction
Two pairs of chromosomes
2n = 4
Meiosis I
n=2
Meiosis I I
Nondisjunction
occurs
Abnormal gametes
Results in
trisomy
2n + 1
Results in
monosomy
2n - 1
Normal gametes
• terminal.hu
Karyotype of Down Syndrome
members.aol.com/chrominfo/images/tri21.gif
9. Down syndrome is often called trisomy 21
because it has 3 copies of chromsome 21.
The frequency of Down syndrome increases
with the age of the mother. Characteristics
of Down syndrome are distinctive facial
features, short stature, heart defects, and
mental disability.
10. Nondisjuction occurs in both autosomes
and sex chromosomes.
11. Females with Turner’s syndrome have
only one sex chromosome, XO. This results
from fertilization with a gamete that had no
sex chromosome.
12. Males with Klinefelter’s syndrome, have
XXY. This is a result of nondisjunction,
called trisomy.
Chapter
11
Complex Inheritance and Human Heredity
13. Fetal tests are performed if a couple
suspects they are carriers for a certain
genetic trait.
14. Types of fetal tests are amniocentesis,
chorionic villus sampling, and fetal blood
testing.
15. Many of these tests put the mother or
fetus at risk so they are only used when
absolutely necessary.
8. The three fetal tests are amniocentesis,
chorionic villus sampling, and fetal blood
sampling. Any of these procedures
include a small amount of risk.
Therefore, the health of the mother and
baby (fetus) need to be monitored
closely.
Vocabulary 11
Carrier
sex chromosome
Pedigree
sex linked traits
Autosome
karyotype
Codominance
nondisjunction
Epistasis
telomere
Incomplete dominance
Multiple alleles
Polygenic traits
Section 11.1 Basic patterns of human inheritance
Recessive Genetic Disorders- (cause of
most genetic disorders)
Example rr-trait expressed in homozygous
state
Carrier is heterozygous state Rr
• Cystic fibrosis
• Albinism
• Galactosemia
• Tay-Sach’s
• PKU
Dominant Genetic Disorders
• Only need one dominant allele to express
trait- Aa or AA
• Fewer of these conditions in number
• Simple traits- cleft chin, widows peak,
tongue rolling, earlobes, hitchhikers thumb
• Disorders- Huntington’s, polydactyly,
achondroplasia
Pedigrees
11.2 Complex patterns of
inheritance
• Incomplete Dominance• Flowers-red pink white- RR, RR1, R1R1
• Codominance-checkered chickens, sickle
cells
• Multiple Alleles-Blood Types, coat color in
rabbits
• Epistasis-coat color in labradors
Chapter
11
Complex Inheritance and Human Heredity
Chapter Diagnostic
Questions
Identify the disease characterized by the
absence of melanin.
A. albinism
B. cystic fibrosis
C. galactosemia
D. Tay-Sachs
Chapter
11
Complex Inheritance and Human Heredity
Chapter Diagnostic
Questions
An individual with Tay-Sachs disease would
be identified by which symptom?
A. excessive mucus production
B. an enlarged liver
C. a cherry-red spot on the back of the eye
D. vision problems
Chapter
11
Complex Inheritance and Human Heredity
Chapter Diagnostic
Questions
Under what circumstances will a recessive
trait be expressed?
A. A recessive allele is passed on by both
parents.
B. One parent passes on the recessive allele.
C. The individual is heterozygous for the trait.
D. There is a mutation in the dominant gene.
Chapter
11
Complex Inheritance and Human Heredity
11.1 Formative
Questions
1. Which of Dr. Garrod’s observations about
alkaptonuria was most critical to his determination
that it is a genetic disorder?
A. It appears at birth and runs in families.
B. It is linked to an enzyme deficiency.
C. It continues throughout a patient’s life,
affecting bones and joints.
D. It is caused by acid excretion and results
in black urine.
Chapter
11
Complex Inheritance and Human Heredity
11.1 Formative
Questions
2. Which is the genotype of a person who
is a carrier for a recessive genetic
disorder?
A. DD
B. Dd
C. dd
D. dE
Chapter
11
Complex Inheritance and Human Heredity
11.1 Formative
Questions
3. Albinism is a recessive condition. If an albino
squirrel is born to parents that both have normal
fur color, what can you conclude about the
genotype of the parents?
A. at least one parent is a carrier
B. both parents are carriers
C. both parents are homozygous recessive
D. at least one parent is homozygous
dominant
Chapter
11
Complex Inheritance and Human Heredity
11.2 Formative
Questions
4. When a homozygous male animal with black
fur is crossed with a homozygous female with
white fur, they have offspring with gray fur.
What type of inheritance does this represent?
A. dosage compensation
B. incomplete dominance
C. multiple alleles
D. sex-linked
Chapter
11
Complex Inheritance and Human Heredity
11.2 Formative
Questions
5. Of the 23 pairs of chromosomes in human
cells, one pair is the _______.
A. autosomes
B. Barr bodies
C. monosomes
D. sex chromosomes
Chapter
11
Complex Inheritance and Human Heredity
11.2 Formative
Questions
6. Which is an example of a polygenic trait?
A. blood type
B. color blindness
C. hemophilia
D. skin color
Chapter
11
Complex Inheritance and Human Heredity
11.3 Formative
Questions
7. What does a karyotype show?
A. The blood type of an individual.
B. The locations of genes on a chromosome.
C. The cell’s chromosomes arranged in order.
D. The phenotype of individuals in a
pedigree.
Chapter
11
Complex Inheritance and Human Heredity
11.3 Formative
Questions
8. What is occurring in this
diagram?
A. multiple alleles
B. nondisjunction
C. nonsynapsis
D. trisomy
Chapter
11
Complex Inheritance and Human Heredity
11.3 Formative
Questions
9. What condition occurs when a person’s cells
have an extra copy of chromosome 21?
A. Down syndrome
B. Klinefelter’s syndrome
C. Tay-Sachs syndrome
D. Turner’s syndrome
Chapter
11
Complex Inheritance and Human Heredity
Chapter Assessment
Questions
Use the figure to describe what the top horizontal
line between numbers 1 and 2 indicates.
A. 1 and 2 are siblings
B. 1 and 2 are parents
C. 1 and 2 are offspring
D. 1 and 2 are carriers
Chapter
11
Complex Inheritance and Human Heredity
Chapter Assessment
Questions
Which is not an allele
in the ABO blood group?
A. IA
B. IO
C. IB
D. i
Chapter
11
Complex Inheritance and Human Heredity
Chapter Assessment
Questions
Down Syndrome results from what change
in chromosomes?
A. one less chromosome on pair 12
B. one extra chromosome on pair 21
C. one less chromosome on pair 21
D. one extra chromosome on pair 12
Chapter
11
Complex Inheritance and Human Heredity
Standardized Test
Practice
If a genetic disorder is caused by a dominant
allele, what is the genotype of those who do
not have the disorder?
A. heterozygous
B. homozygous dominant
C. homozygous recessive
Chapter
11
Complex Inheritance and Human Heredity
Standardized Test
Practice
Analyze this pedigree showing the inheritance of
a dominant genetic disorder. Which would be the
genotype of the first generation father?
A. RR
B. Rr
C. rr
Chapter
11
Complex Inheritance and Human Heredity
Standardized Test
Practice
Shorthorn cattle have an allele for both red and
white hair. When a red-haired cow is crossed with
a white-haired bull, their calf has both red and
white hairs scattered over its body. What type of
inheritance does this represent?
A. codominance
B. dosage compensation
C. epistasis
D. sex-linked
Chapter
11
Complex Inheritance and Human Heredity
Standardized Test
Practice
Why are males affected by recessive sexlinked traits more often than are females?
A. Males have only one X chromosome.
B. Males have two X chromosomes.
C. Males have only one Y chromosome.
D. The traits are located on the Y
chromosomes.
Chapter
11
Complex Inheritance and Human Heredity
Standardized Test
Practice
A carrier of hemophilia and her husband, who
is unaffected by the condition, are expecting a
son. What is the probability that their son will
have hemophilia?
A. 25%
B. 50%
C. 75%
D. 100%