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 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 Albinism 1:17,000 Lack melanin White hair Pale skin Pink pupils Protect from sun Galactosemia 1:50,000-70,000 Missing enzyme Mental deterioration Liver, kidney problems 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 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 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 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 = 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)

These disorders are caused by the presence of a single dominant allele 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

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

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 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 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

How many mating pairs are shown on this pedigree?

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 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.

I II III G = green eyes g = blue eyes GG gg gg gg Robin’s mother John Jill Robin’s father gg Gg Gg Gg Gg gg Pamela John Jr. Alice Lisa Sean Robin gg Gg Gg gg Gg Gg Gg gg Mary Jessica Shari Nicholas John III Harry Donna Sean Jr.

Draw a pedigree that traces the trait for green eyes for three generations. GG gg gg gg I II III 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 I II III gg Gg Gg Gg Gg gg gg Gg Gg gg Gg Gg Gg gg

GG gg gg gg I II III gg Gg Gg Gg Gg gg gg Gg Gg gg Gg Gg Gg gg

Recessive disorder that cannot breakdown galactose 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 Paternal grandparents Maternal 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

When I was twenty-three, I was married to a widow 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 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

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: a. a red plant and a white plant _____________

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: b. a white plant and a pink plant _____________

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: c. a red plant and a pink plant _____________

d. 2 pink plants _____________ 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 in the heterozygous condition Ex. checkered chicken 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 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 ss - all sickled

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 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 blood IB Codes for type B blood i Codes for type O blood Phenotype Genotype Type O blood ii (recessive) universal donor Type 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

Coat Color of Rabbits 12. Another example of multiple alleles-- Complex Inheritance and Human Heredity 12. Another example of multiple alleles-- Coat Color of Rabbits Chinchilla Albino Light gray Option 2 Dark gray Himalayan

MULTIPLE ALLELES Ex: 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

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 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 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 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_ E_bb E_B_ No dark pigment present in fur Dark pigment present in fur

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 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.

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.

X Y 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. 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. Female XX Male XY X Y

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

Sex-Linked Traits

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 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 Influence 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 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 Karyotype www. miscarriage. com. au/. /karyotype_normal Normal Female Karyotype www.miscarriage.com.au/.../karyotype_normal.jpg

Normal Male Karyotype www. contexo. info/DNA_Basics/images/karyotype1 Normal Male Karyotype www.contexo.info/DNA_Basics/images/karyotype1.gif

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.

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

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

Nondisjunction ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ) ) ( ( ) ) ( Two pairs of chromosomes ) ( 2n = 4 ) ( ) ( Meiosis I ) ( ) ( n = 2 ) ( ) ( ) ( Meiosis I I Nondisjunction occurs ) ) ) ( ( ) ) ( Abnormal gametes Results in Results in trisomy monosomy 2n + 1 2n - 1 Normal gametes

terminal.hu

Karyotype of Down Syndrome members.aol.com/chrominfo/images/tri21.gif

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.

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

Chapter 11 Complex Inheritance and Human Heredity

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.

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

Identify the disease characterized by the absence of melanin. Chapter 11 Complex Inheritance and Human Heredity Chapter Diagnostic Questions Identify the disease characterized by the absence of melanin. albinism cystic fibrosis galactosemia Tay-Sachs

excessive mucus production an enlarged liver Chapter 11 Complex Inheritance and Human Heredity Chapter Diagnostic Questions An individual with Tay-Sachs disease would be identified by which symptom? excessive mucus production an enlarged liver a cherry-red spot on the back of the eye vision problems

Under what circumstances will a recessive trait be expressed? Chapter 11 Complex Inheritance and Human Heredity Chapter Diagnostic Questions Under what circumstances will a recessive trait be expressed? A recessive allele is passed on by both parents. One parent passes on the recessive allele. The individual is heterozygous for the trait. There is a mutation in the dominant gene.

1. Which of Dr. Garrod’s observations about 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? It appears at birth and runs in families. It is linked to an enzyme deficiency. It continues throughout a patient’s life, affecting bones and joints. 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? DD Dd dd dE

3. Albinism is a recessive condition. If an albino 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? at least one parent is a carrier both parents are carriers both parents are homozygous recessive at least one parent is homozygous dominant

4. When a homozygous male animal with black 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? dosage compensation incomplete dominance multiple alleles sex-linked

5. Of the 23 pairs of chromosomes in human 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 _______. autosomes Barr bodies monosomes sex chromosomes

6. Which is an example of a polygenic trait? Chapter 11 Complex Inheritance and Human Heredity 11.2 Formative Questions 6. Which is an example of a polygenic trait? blood type color blindness hemophilia skin color

7. What does a karyotype show? Chapter 11 Complex Inheritance and Human Heredity 11.3 Formative Questions 7. What does a karyotype show? The blood type of an individual. The locations of genes on a chromosome. The cell’s chromosomes arranged in order. The phenotype of individuals in a pedigree.

8. What is occurring in this diagram? Chapter 11 Complex Inheritance and Human Heredity 11.3 Formative Questions 8. What is occurring in this diagram? multiple alleles nondisjunction nonsynapsis trisomy

Klinefelter’s syndrome Tay-Sachs syndrome Turner’s syndrome 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? Down syndrome Klinefelter’s syndrome Tay-Sachs syndrome Turner’s syndrome

Use the figure to describe what the top horizontal 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. 1 and 2 are siblings 1 and 2 are parents 1 and 2 are offspring 1 and 2 are carriers

Which is not an allele in the ABO blood group? IA IO IB i Chapter 11 Complex Inheritance and Human Heredity Chapter Assessment Questions Which is not an allele in the ABO blood group? IA IO IB i

Down Syndrome results from what change in chromosomes? Chapter 11 Complex Inheritance and Human Heredity Chapter Assessment Questions Down Syndrome results from what change in chromosomes? one less chromosome on pair 12 one extra chromosome on pair 21 one less chromosome on pair 21 one extra chromosome on pair 12

If a genetic disorder is caused by a dominant 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? heterozygous homozygous dominant homozygous recessive

Analyze this pedigree showing the inheritance of 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? RR Rr rr

Shorthorn cattle have an allele for both red and 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? codominance dosage compensation epistasis sex-linked

Why are males affected by recessive sex- Chapter 11 Complex Inheritance and Human Heredity Standardized Test Practice Why are males affected by recessive sex- linked traits more often than are females? Males have only one X chromosome. Males have two X chromosomes. Males have only one Y chromosome. The traits are located on the Y chromosomes.

A carrier of hemophilia and her husband, who 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? 25% 50% 75% 100%