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

Albinism1:17,000 Lack

melaninWhite 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

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 of these conditions 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

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 = 127 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)

These 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.Disorder Occurrence in

the U. S.Cause Effect Cure/Treatment

Huntington’s Disease

1:10,000 Defective gene Decline of mental functions

Ability to move deteriorates

NONE

Achondroplasia 1:25,000 Defective gene that affects bone growth

Short arms and legs

Large head

NONE

Making a Pedigree

8. A pedigree is a family tree of inheritance.

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

Analyze the pedigree - Dogs

Generation I

Generation II

Generation III

Generation IV

White = Tall Dominant

Black = Short Recessive

Male

Female

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?

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.

GG gg

Gg GgGgGg gggg

gg gg

gg ggGg Gg ggGgGg Gg

G = green eyes g = blue eyes

John Jill

Pamela John Jr. Alice Lisa

Jessica Shari MaryJohn III

Sean Robin

Robin’sfather

Robin’s mother

Nicholas Harry Donna Sean Jr.

GG gg

Gg GgGgGg gggg

gg gg

gg ggGg Gg ggGgGg Gg

Draw a pedigree that traces the trait for green eyes for three generations.

GG gg

Gg GgGgGg gggg

gg gg

gg ggGg Gg ggGgGg Gg

Draw a pedigree that traces the trait for blue eyes for three generations.

GG gg

Gg GgGgGg gggg

gg gg

gg ggGg Gg ggGgGg 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

Paternal grandparent

s

father

boy

Gg Gg

gg

gg

Gg Gg

Gg

Maternal grandparents

½ shaded = carrier

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!

COMPLEX PATTERNS OF INHERITANCE

Ch 11.2 Complex Patterns of Inheritance

Main Idea – Complex inheritance of traits does NOT follow inheritance patterns described by Mendel.

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 R1R1 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. CODOMINANCEBoth 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.

Ex: Sickle Cell SS - normalSs - some normal, some sickledss - all sickled

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.

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; A, B and O; 4 phenotypes – type A, type B, type AB, and type O

Ex: Blood Groups -3 alleles (A, B, and O)IA Codes for type A bloodIB Codes for type B bloodi Codes for type O blood

Phenotype GenotypeType O blood ii (recessive) universal

donorType A blood IAi; IA IA

Type B blood IBi; IB IB

Type AB blood IA IB (codominant) universal recipient

Type AB blood type is an example of codominance.

Blood Types

Complex Inheritance and Human Heredity12. Another example of multiple alleles--Coat Color of Rabbits

Light gray

Dark gray Himalayan

Albino

Chinchilla

MULTIPLE ALLELESEx: Coat color in rabbits – hierarchy of dominance

Presence of multiple alleles increases the

possible number of genotypes and phenotypes• C - gray• Cch - chinchilla• Ch - himalayan• c - albino (white)

C is dominant to Cch > Ch > c• 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 retrieverstwo sets of alleles E and B

No dark pigment present in fur Dark pigment present in fur

eebb

eeB_ E_bb E_B_

EPISTASIS

allele E determines whether the fur will have dark pigment, alleles

ee = inability to express dark pigment or coat color

allele B determines how dark the pigment will beB = black; b = chocolate (determines how dark pigment will be)

EEBB or EEBb = blackEEbb or Eebb = browneeBb, eeBB = yellow with black pigment (black nose)eebb = yellow with no pigment (pink nose)

The e allele masks the dominant B allele.

Complex Inheritance and Human Heredity

14. Sex Determination- Sex chromosomes determine an individual’s gender

• Autosomes are the first 22 pairs of chromosomes

• Sex chromosomes are the 23rd pair; X and Y• XX= Human females • XY=Human males

The sex of the offspring is determined by the chromosomes of the sperm cell.

Female

XX

Male

XY

XX

XX

XY

XY

X

X

X

Y

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?

a. All the offspring are female.

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

leading active lives, will be unable

to have children.

15. Dosage Compensation• In females, one X deactivates

• Barr Bodies- 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= hemophiliaXH 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= hemophiliaXH 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= hemophiliaXH 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= hemophiliaXH XH = normal female XH Xh = carrier female XhXh = hemophiliac femaleXH 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= hemophiliaXH 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= colorblindXCXC = 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= colorblindXCXC = normal female XC Xc = carrier female XcXc = colorblind

femaleXC 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= whiteXRXR = normal female XR Xr = carrier female XrXr = white eyed femaleXR 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

• Baldness is dominant in males; recessive in females.

17. Polygenic Traits

Controlled by multiple pairs of genes; results in numerous phenotypes

Human ex. – skin color, height, eye color, fingerprints

18. Environmental InfluenceSunlight 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

• Focuses on identical twins

• Identical twins have the same inherited traits.

• Influenced by environment

11.3 Chromosomes and Human Heredity

Main Idea – Chromosomes can be studied using a karyotype.

Normal Female Karyotypewww.miscarriage.com.au/.../karyotype_normal.jpg

Normal Male Karyotypewww.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 for a total of 46 chromosomes.

2. Chromosomes end in protective caps called telomeres. These might be involved in aging and cancer. Made of

Analogy - shoestring

3. Cell division during which sister chromatids fail to separate properly is called nondisjunction. When a set of three chromosomes results it is called trisomy. A set having only one particular chromosome is called monosomy. Nondisjunction alters the chromosome number in gametes.

Nondisjunction in meiosis I

Results in 2 trisomy gametes, and 2 monsomy gametes

NondisjunctionTwo pairs of chromosomes

Meiosis I

Meiosis I INondisjunctionoccurs

Abnormal gametes Results in Results in trisomy monosomy 2n + 1 2n - 1

Normal gametes

2n = 4

n = 2

• terminal.hu

Karyotype of Down Syndromemembers.aol.com/chrominfo/images/tri21.gif

4. Down syndrome is often called trisomy 21 because it has 3 copies of chromsome 21. This results in the production of gametes with one duplicate chromosome. 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.

5. Females with XO have Turner’s syndrome, a result of nondisjunction, called monosomy.

6. Males with XXY have Klinefelter’s syndrome, a result of nondisjunction, called trisomy. Problem in egg production.

Complex Inheritance and Human HeredityChapter 11

7. Couples who suspect they might be carriers for certain genetic disorders might want to have a fetal test performed. Older couples also might want the chromosome status of their developing baby called a fetus.

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

A. albinism

B. cystic fibrosis

C. galactosemia

D. Tay-Sachs

Identify the disease characterized by the absence of melanin.

Complex Inheritance and Human HeredityChapter 11

Chapter Diagnostic Questions

A. excessive mucus production

B. an enlarged liver

C. a cherry-red spot on the back of the eye

D. vision problems

An individual with Tay-Sachs disease would be identified by which symptom?

Complex Inheritance and Human HeredityChapter 11

Chapter Diagnostic Questions

Under what circumstances will a recessive trait be expressed?

Complex Inheritance and Human HeredityChapter 11

Chapter Diagnostic Questions

A. A recessive allele is passed on by bothparents.

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.

A. It appears at birth and runs in families.

1. Which of Dr. Garrod’s observations about alkaptonuria was most critical to his determination that it is a genetic disorder?

Complex Inheritance and Human HeredityChapter 11

11.1 Formative Questions

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 resultsin black urine.

A. DD

B. Dd

C. dd

D. dE

2. Which is the genotype of a person who is a carrier for a recessive genetic disorder?

Complex Inheritance and Human HeredityChapter 11

11.1 Formative Questions

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

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?

Complex Inheritance and Human HeredityChapter 11

11.1 Formative Questions

A. dosage compensation

B. incomplete dominance

C. multiple alleles

D. sex-linked

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?

Complex Inheritance and Human HeredityChapter 11

11.2 Formative Questions

A. autosomes

B. Barr bodies

C. monosomes

D. sex chromosomes

5. Of the 23 pairs of chromosomes in human cells, one pair is the _______.

Complex Inheritance and Human HeredityChapter 11

11.2 Formative Questions

A. blood type

B. color blindness

C. hemophilia

D. skin color

6. Which is an example of a polygenic trait?

Complex Inheritance and Human HeredityChapter 11

11.2 Formative Questions

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.

7. What does a karyotype show?

Complex Inheritance and Human HeredityChapter 11

11.3 Formative Questions

A. multiple alleles

B. nondisjunction

C. nonsynapsis

D. trisomy

8. What is occurring in this diagram?

Complex Inheritance and Human HeredityChapter 11

11.3 Formative Questions

A. Down syndrome

B. Klinefelter’s syndrome

C. Tay-Sachs syndrome

D. Turner’s syndrome

9. What condition occurs when a person’s cells have an extra copy of chromosome 21?

Complex Inheritance and Human HeredityChapter 11

11.3 Formative Questions

A. 1 and 2 are siblings

B. 1 and 2 are parents

C. 1 and 2 are offspring

D. 1 and 2 are carriers

Use the figure to describe what the top horizontal line between numbers 1 and 2 indicates.

Complex Inheritance and Human HeredityChapter 11

Chapter Assessment Questions

A. IA

B. IO

C. IB

D. i

Which is not an allele in the ABO blood

group?

Complex Inheritance and Human HeredityChapter 11

Chapter Assessment Questions

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

Down Syndrome results from what change in chromosomes?

Complex Inheritance and Human HeredityChapter 11

Chapter Assessment Questions

A. heterozygous

B. homozygous dominant

C. homozygous recessive

If a genetic disorder is caused by a dominant allele, what is the genotype of those who do not have the disorder?

Complex Inheritance and Human HeredityChapter 11

Standardized Test Practice

A. RR

B. Rr

C. rr

Analyze this pedigree showing the inheritance of a dominant genetic disorder. Which would be the

genotype of the first generation father?

Complex Inheritance and Human HeredityChapter 11

Standardized Test Practice

A. codominance

B. dosage compensation

C. epistasis

D. sex-linked

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?

Complex Inheritance and Human HeredityChapter 11

Standardized Test Practice

A. Males have only one X chromosome.

B. Males have two X chromosomes.

C. Males have only one Y chromosome.

Why are males affected by recessive sex-linked traits more often than are females?

Complex Inheritance and Human HeredityChapter 11

Standardized Test Practice

D. The traits are located on the Ychromosomes.

A. 25%

B. 50%

C. 75%

D. 100%

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?

Complex Inheritance and Human HeredityChapter 11

Standardized Test Practice