1. 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.

2. Recessive Disorders 1.Simple recessive heredity is the cause of most genetic disorders.

3. 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.

4. 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.

5. DisorderOccurrenceCauseAffectCure/Treatment Cystic fibrosis most common in Caucasians 1:3500 Enzyme deficiency Excess mucus in lungs No cure Enzyme supplement Chest percussions Albinism 1:17,000Lack melanin White hair Pale skin Pink pupils No cure Protect from sun Galactosemia 1:50,000-70,000Missing 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 1:10,000-50,000Defective enzyme Mental retardation Restricted diet

6. Albinsim

7. Tay Sachs Disease

8. 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.

9. 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.)

10. 5. Examples of simple dominant traits. hitchhiker’s thumb, tongue rolling, free earlobes

11. Dominant Traits Six fingers or six toes - polydactyly

12. Polydactyly

13. Girl in Burma with many digits Total fingers and toes = 26 6 fingers on each hand = 12 7 toes on each foot = 14

14. Tongue rolling and Ear lobes Free vs. attached

15. Widow’s Peak Hitchhiker’s thumb

16. Cleft chin is a dominant trait

17. 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.

18. Huntington’s Disease

19. 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.

20. Achondroplasia (dwarfism)

21. 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.

22. Making a Pedigree

23. 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.

24. 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.

26. Achondroplasia pedigree Dominant Disorder

27. Analyze the pedigree - Dogs

28. Generation I Generation II Generation III Generation IV White = Tall Dominant Black = Short Recessive Male Female

29. 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?

30. 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

31. 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

32. 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

33. 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

34. Inherited Traits - Chickens

35. 1.How many mating pairs are shown on this pedigree?

36. 2.How many chickens on this pedigree are female?

37. 3. How many chickens on this pedigree are male?

38. 4. How many generations are shown here?

39. 5. How many roosters (males) had the trait being studied?

40. 6. How many roosters (males) lacked the trait being studied?

41. 7. How many hens (females) had the trait being studied?

42. 8. How many hens (females) lacked the trait being studied?

43. 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.

44. 9. Did any inbreeding occur? If so where?

45. 10. If your answer to question 9 is “yes” can you explain the results of the inbreeding? How does this relate to selective breeding?

46. 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.

47. 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.

48. GGgg Gg gg Gg ggGg G = green eyesg = blue eyes John Jill Pamela John Jr.AliceLisa Jessica Shari Mary John III Sean Robin Robin’s father Robin’s mother Nicholas HarryDonnaSean Jr.

49. GGgg Gg gg Gg ggGg Draw a pedigree that traces the trait for green eyes for three generations.

50. GGgg Gg gg Gg ggGg Draw a pedigree that traces the trait for blue eyes for three generations.

51. GGgg Gg gg Gg ggGg

52. Galactosemia Recessive disorder that cannot breakdown galactose

53. 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.

54. G= normal g= galactosemia Paternal grandparents father boy Gg gg Gg Maternal grandparents ½ shaded = carrier

55. Textbook p. 300 View Pedigree Read paragraph Answer question at the end of paragraph.

56. I’m My Own Grampa

57. 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.

58. 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.

59. 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.

60. 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.

61. 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!

62. 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.

63. 1. Incomplete Dominance The heterozygote is an intermediate phenotype between the two homozygotes 2. P= RR = red RR 1 = pink R 1 R 1 = white Ex. Red flower X white flower  pink flower RR X R 1 R 1  RR 1

64. Crosses Involving Incomplete Dominance

65. Alleles: R = red R¹ = white Genotypes: RR = red R 1 R 1 = white; RR¹ = pink 1. In Japanese four-o’ clocks, predict the phenotype of a cross between: a. a red plant and a white plant _____________

66. Alleles: R = red R¹ = white Genotypes: RR = red R 1 R 1 = white; RR¹ = pink 1. In Japanese four-o’ clocks, predict the phenotype of a cross between: b. a white plant and a pink plant _____________

67. Alleles: R = red R¹ = white Genotypes: RR = red R 1 R 1 = white; RR¹ = pink 1. In Japanese four-o’ clocks, predict the phenotype of a cross between: c. a red plant and a pink plant _____________

68. Alleles: R = red R¹ = white Genotypes: RR = red R 1 R 1 = white; RR¹ = pink 1. In Japanese four-o’ clocks, predict the phenotype of a cross between: d. 2 pink plants _____________

69. 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 ___________________

70. b. A long-tail cat and a short-tail cat _____________________

71. c. a short-tail cat and a cat with no tail ______________

72. d. two short-tail cats _____________

73. 3. CODOMINANCE Both alleles are expressed in the heterozygous condition Ex. checkered chicken black chicken X white chicken  checkered chicken BB X WW  BW

74. 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.

75. Ex: Sickle Cell SS - normal Ss - some normal, some sickled sickle cell trait ss - all sickled-sickle cell disease 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.

76. 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

77. 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

78. Type O is recessive and represented by ii; universal donor Type AB is codominant and represented by I A I B; universal recipient.

79. Ex: Blood Groups -3 alleles (A, B, and O) I A Codes for type A blood I B Codes for type B blood i Codes for type O blood PhenotypeGenotype Type O blood ii Type A blood I A i; I A I A Type B blood I B i; I B I B Type AB blood I A I B Type AB blood type is an example of codominance.

80. Blood Types

81. Complex Inheritance and Human Heredity 12. Another example of multiple alleles-- Coat Color of Rabbits hierarchy of dominance (ex. C>C ch > C h > c Presence of multiple alleles increases the possible number of genotypes and phenotypes Light gray Dark gray Himalayan Albino Chinchilla

82. MULTIPLE ALLELES Ex: Coat color in rabbits – C - gray C ch - chinchilla C h - himalayan c - albino (white) 10 possible genotypes 5+ phenotypes

83. Crosses Involving Multiple Alleles

84. I A Codes for type A blood I B Codes for type B blood i Codes for type O blood PhenotypeGenotype Type O blood ii Type A blood I A i; I A I A Type B blood I B i; I B I B Type AB blood I A I B

85. 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?

86. 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?

87. 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?

88. 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?

89. 5. What is the probability that a couple whose blood types are AB and O will have a type A child?

90. 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?

91. 13. Epistasis-- one allele hides the effects of another allele Example: coat color in Labrador retrievers two sets of alleles E and B No dark pigment present in fur Dark pigment present in fur eebb eeB_ E_bbE_B_

92. 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.

93. 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 23 rd pair; X and Y In humans gametes: XX= females XY= males

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

95. Female XX Male XY XX 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.

96. 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

98. 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”)

99. Colorblindness

101. H= normal h= hemophilia X H X H = normal female X H X h = carrier female X h X h = hemophiliac female X H Y = normal male X h 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?

102. H= normal h= hemophilia X H X H = normal female X H X h = carrier female X h X h = hemophiliac female X H Y = normal male X h 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?

103. H= normal h= hemophilia X H X H = normal female X H X h = carrier female X h X h = hemophiliac female X H Y = normal male X h Y =hemophiliac male 3. A hemophiliac woman has a phenotypically normal mother. What are the possible genotypes of her mother and her father?

104. H= normal h= hemophilia X H X H = normal female X H X h = carrier female X h X h = hemophiliac female X H Y = normal male X h 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?

105. H= normal h= hemophilia X H X H = normal female X H X h = carrier female X h X h = hemophiliac female X H Y = normal male X h 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?

106. C = normal c= colorblind X C X C = normal femaleX C X c = carrier female X c X c = colorblind female X C Y = normal male X c 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?

107. C = normal c= colorblind X C X C = normal femaleX C X c = carrier female X c X c = colorblind female X C Y = normal male X c 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?

108. R = normal r= white X R X R = normal femaleX R X r = carrier female X r X r = white eyed female X R Y = normal male X r 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?

109. 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

110. 17. Polygenic Traits Controlled by multiple pairs of genes; results in numerous phenotypes Example of human polygenic traits – skin color, height, eye color, fingerprints

111. 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

112. 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

113. 11.3 Chromosomes and Human Heredity Main Idea – Chromosomes can be studied using karyotypes.

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

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

116. 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.

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

118. 3.Cell division during which sister chromatids fail to separate properly is called nondisjunction. Nondisjunction alters the chromosome number in gametes.

119. 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.

120. 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.

122. Nondisjunction in meiosis I Results in 2 trisomy gametes, and 2 monsomy gametes

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

124. terminal.hu

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

126. 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.

127. 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.

128. Complex Inheritance and Human Heredity Chapter 11

129. 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.

130. 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.

131. Vocabulary 11 Carriersex chromosome Pedigreesex linked traits Autosomekaryotype Codominancenondisjunction Epistasistelomere Incomplete dominance Multiple alleles Polygenic traits

132. 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

133. 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

134. Pedigrees

135. 11.2 Complex patterns of inheritance Incomplete Dominance- Flowers-red pink white- RR, RR 1, R 1 R 1 Codominance-checkered chickens, sickle cells Multiple Alleles-Blood Types, coat color in rabbits Epistasis-coat color in labradors

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

137. 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 Heredity Chapter 11 Chapter Diagnostic Questions

138. Under what circumstances will a recessive trait be expressed? Complex Inheritance and Human Heredity Chapter 11 Chapter Diagnostic Questions 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.

139. 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 Heredity Chapter 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 results in black urine.

140. 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 Heredity Chapter 11 11.1 Formative Questions

141. 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 Heredity Chapter 11 11.1 Formative Questions

142. 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 Heredity Chapter 11 11.2 Formative Questions

143. 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 Heredity Chapter 11 11.2 Formative Questions

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

145. 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 Heredity Chapter 11 11.3 Formative Questions

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

147. 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 Heredity Chapter 11 11.3 Formative Questions

148. 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 Heredity Chapter 11 Chapter Assessment Questions

149. A. I A B. I O C. I B D. i Which is not an allele in the ABO blood group? Complex Inheritance and Human Heredity Chapter 11 Chapter Assessment Questions

150. 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 Heredity Chapter 11 Chapter Assessment Questions

151. 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 Heredity Chapter 11 Standardized Test Practice

152. 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 Heredity Chapter 11 Standardized Test Practice

153. 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 Heredity Chapter 11 Standardized Test Practice

154. 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 Heredity Chapter 11 Standardized Test Practice D. The traits are located on the Y chromosomes.

155. 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 Heredity Chapter 11 Standardized Test Practice