Patterns of Inheritance

Patterns of Inheritance
Chapter 9 Patterns of Inheritance Lecture by Mary C. Colavito

Introduction: Barking Up the Genetic Tree
Dogs are one of man’s longest genetics experiments Dog breeds are the result of artificial selection Populations of dogs became isolated from each other Humans chose dogs with specific traits for breeding Each breed has physical and behavioral traits due to a unique genetic makeup Sequencing of the dog’s genome shows evolutionary relationships between breeds Copyright © 2009 Pearson Education, Inc.

Wolf Chinese Shar-Pei Ancestral canine Akita Siberian Husky Alaskan Malamute Basenji Afghan hound Saluki Rottweiler Sheepdog Retriever

9.1 The science of genetics has ancient roots
Pangenesis was an early explanation for inheritance It was proposed by Hippocrates Particles called pangenes came from all parts of the organism to be incorporated into eggs or sperm Characteristics acquired during the parents’ lifetime could be transferred to the offspring Aristotle rejected pangenesis and argued that instead of particles, the potential to produce the traits was inherited Blending was another idea, based on plant breeding Hereditary material from parents mixes together to form an intermediate trait, like mixing paint Ask students to give examples that contradict these early proposals. Most will agree that if something happens to a parent during his/her lifetime, the children do not inherit the change, as suggested by this real-life example: A man who lost part of his thumb in an accident became the father of five children, each with fully formed thumbs. Skin color may appear as a blended trait, since children of dark- and light-skinned parents are often intermediate between the two. However, two parents with intermediate skin color can have children with darker or lighter skin than either parent. (See Module 9.14.) In England, there was a recent report of two parents with medium skin color having twin daughters, one with light skin and the other with dark skin. Teaching Tips 1. As you begin your lectures on genetics, consider challenging your students to explain why the theories of pangenesis and blending are incorrect. Perhaps just pick one of the two. You might even ask for short responses from everyone at the start of class or as an assignment before the first lectures. In addition to arousing interest in the answers, the responses should reveal the diverse backgrounds of your students entering this discussion and reveal any preexisting confusion on the subject of genetics. 2. The concept of pangenesis is analogous to the structure of United States representation in Congress. Each congressional district sends a congressman or congresswoman (pangene) to the U.S. House of Representatives (gamete). There, all parts of the United States (body) are represented. 3. In this or future lectures addressing evolution, you may mention that pangenesis is a mechanism that permits Lamarckian evolution. Copyright © 2009 Pearson Education, Inc.

9.2 Experimental genetics began in an abbey garden
Gregor Mendel discovered principles of genetics in experiments with the garden pea Mendel showed that parents pass heritable factors to offspring (heritable factors are now called genes) Advantages of using pea plants Controlled matings Self-fertilization or cross-fertilization Observable characteristics with two distinct forms True-breeding strains Student Misconceptions and Concerns 1. The authors note that Mendel’s work was published in 1866, seven years after Darwin published Origin of Species. Consider challenging your students to consider whether Mendel’s findings supported Darwin’s ideas. Some scientists have noted that Darwin often discussed the evolution of traits by matters of degree. Yet, Mendel’s selection of pea plant traits typically showed complete dominance, rather than the possibility for such gradual inheritance. Teaching Tips 1. This early material introduces many definitions that are vital to understanding the later discussions in this chapter. Therefore, students need to be encouraged to master these definitions immediately. This may be a good time for a short quiz to encourage their progress. Copyright © 2009 Pearson Education, Inc.

Figure 9.2A Gregor Mendel.

Petal Figure 9.2B Anatomy of a garden pea flower (with one petal removed to improve visibility). Pollen is usually transferred from the stamens to the egg-bearing carpel of the same flower, since these structures are surrounded by the petals. Stamen Carpel

pollen from stamens of white flower to carpel of purple flower Parents
White 1 Removed stamens from purple flower Stamens Carpel 2 Transferred pollen from stamens of white flower to carpel of purple flower Parents (P) Purple Figure 9.2C Mendel’s technique for cross-fertilization of pea plants. This slide demonstrates how Mendel could control matings between pea plants. He removed the pollen-bearing structures from one parental plant and transferred pollen from another parental plant. The true-breeding parental plants are called the P generation, their offspring are the F1 generation, and offspring of an F1  F1 cross are the F2 generation.

pollen from stamens of white flower to carpel of purple flower Parents
White 1 Removed stamens from purple flower Stamens Carpel 2 Transferred pollen from stamens of white flower to carpel of purple flower Parents (P) Purple 3 Pollinated carpel matured into pod Figure 9.2C Mendel’s technique for cross-fertilization of pea plants. This slide demonstrates how Mendel could control matings between pea plants. He removed the pollen-bearing structures from one parental plant and transferred pollen from another parental plant. The true-breeding parental plants are called the P generation, their offspring are the F1 generation, and offspring of an F1  F1 cross are the F2 generation.

Parents (P) Offspring (F1)
White 1 Removed stamens from purple flower Stamens Carpel 2 Transferred pollen from stamens of white flower to carpel of purple flower Parents (P) Purple 3 Pollinated carpel matured into pod 4 Planted seeds from pod Figure 9.2C Mendel’s technique for cross-fertilization of pea plants. This slide demonstrates how Mendel could control matings between pea plants. He removed the pollen-bearing structures from one parental plant and transferred pollen from another parental plant. The true-breeding parental plants are called the P generation, their offspring are the F1 generation, and offspring of an F1  F1 cross are the F2 generation. Offspring (F1)

Flower color Purple White Flower position Axial Terminal Seed color
Flower color Purple White Flower position Axial Terminal Seed color Yellow Green Seed shape Round Wrinkled Pod shape Inflated Constricted Figure 9.2D The seven pea characters studied by Mendel. Mendel studied seven characteristics for pea plants. Later studies have shown that pea plants have seven pairs of chromosomes, and each of these characteristics is on a different chromosome. This explains why Mendel’s results were not affected by genetic recombination. Pod color Green Yellow Stem length Tall Dwarf

9.3 Mendel’s law of segregation describes the inheritance of a single character
Example of a monohybrid cross Parental generation: purple flowers  white flowers F1 generation: all plants with purple flowers F2 generation: of plants with purple flowers of plants with white flowers Mendel needed to explain Why one trait seemed to disappear in the F1 generation Why that trait reappeared in one quarter of the F2 offspring For the BLAST Animation Single-Trait Crosses, go to Animation and Video Files. Student Misconceptions and Concerns 1. Students using Punnett squares need to be reminded that the calculations are expected statistical probabilities and not absolutes. Just as we would expect that any six playing cards dealt might be half black and half red, we frequently find that this is not true. This might be a good time to show how larger sample sizes increase the likelihood that sampling will reflect expected ratios. Teaching Tips 1. This early material introduces many definitions that are vital to understanding the later discussions in this chapter. Therefore, students need to be encouraged to master these definitions immediately. This may be a good time for a short quiz to encourage their progress. 2. Many students benefit from a little quick practice with a Punnett square. Have them try these crosses for practice: (a) PP pp and (b) Pp pp. 3. For students struggling with basic terminology, an analogy between a genetic trait and a pair of shoes might be helpful. A person might wear a pair of shoes in which both shoes match (homozygous), or less likely, a person might wear shoes that do not match (heterozygous). 4. Another analogy that might help struggling students is a pair of people trying to make a decision about where to eat tonight. One person wants to eat at a restaurant, the other wants to eat a meal at home. This (heterozygous) couple eats at home (the dominant allele “wins”). 5. Figure 9.4 can be of great benefit when introducing genetic terminology of genes. For students struggling to think abstractly, such a visual aid may be essential when describing these features in lecture. Copyright © 2009 Pearson Education, Inc.

P generation (true-breeding parents) Purple flowers White flowers
Purple flowers White flowers Figure 9.3A Crosses tracking one character (flower color).

P generation (true-breeding parents) Purple flowers White flowers
Purple flowers White flowers F1 generation All plants have purple flowers Figure 9.3A Crosses tracking one character (flower color).

of plants have purple flowers of plants have white flowers
P generation (true-breeding parents) Purple flowers White flowers F1 generation All plants have purple flowers Fertilization among F1 plants (F1 ´ F1) Figure 9.3A Crosses tracking one character (flower color). F2 generation of plants have purple flowers 3 – 4 of plants have white flowers 1 – 4

Four Hypotheses Genes are found in alternative versions called alleles; a genotype is the listing of alleles an individual carries for a specific gene For each characteristic, an organism inherits two alleles, one from each parent; the alleles can be the same or different A homozygous genotype has identical alleles A heterozygous genotype has two different alleles There is an allele for purple flower color and a different allele for white flower color. Student Misconceptions and Concerns 1. Students using Punnett squares need to be reminded that the calculations are expected statistical probabilities and not absolutes. Just as we would expect that any six playing cards dealt might be half black and half red, we frequently find that this is not true. This might be a good time to show how larger sample sizes increase the likelihood that sampling will reflect expected ratios. Teaching Tips 1. This early material introduces many definitions that are vital to understanding the later discussions in this chapter. Therefore, students need to be encouraged to master these definitions immediately. This may be a good time for a short quiz to encourage their progress. 2. Many students benefit from a little quick practice with a Punnett square. Have them try these crosses for practice: (a) PP pp and (b) Pp pp. 3. For students struggling with basic terminology, an analogy between a genetic trait and a pair of shoes might be helpful. A person might wear a pair of shoes in which both shoes match (homozygous), or less likely, a person might wear shoes that do not match (heterozygous). 4. Another analogy that might help struggling students is a pair of people trying to make a decision about where to eat tonight. One person wants to eat at a restaurant, the other wants to eat a meal at home. This (heterozygous) couple eats at home (the dominant allele “wins”). 5. Figure 9.4 can be of great benefit when introducing genetic terminology of genes. For students struggling to think abstractly, such a visual aid may be essential when describing these features in lecture. Copyright © 2009 Pearson Education, Inc.

Four Hypotheses If the alleles differ, the dominant allele determines the organism’s appearance, and the recessive allele has no noticeable effect The phenotype is the appearance or expression of a trait The same phenotype may be determined by more than one genotype Law of segregation: Allele pairs separate (segregate) from each other during the production of gametes so that a sperm or egg carries only one allele for each gene In reference to hypothesis 3, in the F1 generation, plants received a purple allele from one parent and a white allele from the other. The purple allele is dominant since all the F1 plants have purple petals. In reference to hypothesis 4, in general, a dominant allele provides instructions for producing a functional protein. A recessive allele may lead to a nonfunctional protein or the complete absence of the protein. Mendel needed to explain: Why one trait seemed to disappear in the F1 generation—This is explained by dominance of the purple allele over the white allele. Why that trait reappeared in one fourth of the F2 offspring—This is explained by the segregation of alleles, the generation of all possible combinations of gametes, and the dominance of the purple allele. Student Misconceptions and Concerns 1. Students using Punnett squares need to be reminded that the calculations are expected statistical probabilities and not absolutes. Just as we would expect that any six playing cards dealt might be half black and half red, we frequently find that this is not true. This might be a good time to show how larger sample sizes increase the likelihood that sampling will reflect expected ratios. Teaching Tips 1. This early material introduces many definitions that are vital to understanding the later discussions in this chapter. Therefore, students need to be encouraged to master these definitions immediately. This may be a good time for a short quiz to encourage their progress. 2. Many students benefit from a little quick practice with a Punnett square. Have them try these crosses for practice: (a) PP pp and (b) Pp pp. 3. For students struggling with basic terminology, an analogy between a genetic trait and a pair of shoes might be helpful. A person might wear a pair of shoes in which both shoes match (homozygous), or less likely, a person might wear shoes that do not match (heterozygous). 4. Another analogy that might help struggling students is a pair of people trying to make a decision about where to eat tonight. One person wants to eat at a restaurant, the other wants to eat a meal at home. This (heterozygous) couple eats at home (the dominant allele “wins”). 5. Figure 9.4 can be of great benefit when introducing genetic terminology of genes. For students struggling to think abstractly, such a visual aid may be essential when describing these features in lecture. Copyright © 2009 Pearson Education, Inc.

Genetic makeup (alleles) P plants PP pp
PP pp Gametes All P All p F1 plants (hybrids) All Pp Gametes 1 – 2 P 1 – 2 p Sperm Figure 9.3B Explanation of the crosses in Figure 9.3A. A Punnett square is used to show all possible fertilization events for parental gametes. In three quarters of the F2 genotypes, there will be at least one dominant allele. P p F2 plants Phenotypic ratio 3 purple : 1 white P PP Pp Eggs Genotypic ratio 1 PP : 2 Pp : 1 pp p Pp pp

9.4 Homologous chromosomes bear the alleles for each character
For a pair of homologous chromosomes, alleles of a gene reside at the same locus Homozygous individuals have the same allele on both homologues Heterozygous individuals have a different allele on each homologue Student Misconceptions and Concerns 1. Students using Punnett squares need to be reminded that the calculations are expected statistical probabilities and not absolutes. Just as we would expect that any six playing cards dealt might be half black and half red, we frequently find that this is not true. This might be a good time to show how larger sample sizes increase the likelihood that sampling will reflect expected ratios. Teaching Tips 1. Figure 9.4 can be of great benefit when introducing genetic terminology of genes. For students struggling to think abstractly, such a visual aid may be essential when describing these features in lecture. Copyright © 2009 Pearson Education, Inc.

Gene loci Dominant allele P a B P a b Recessive allele Genotype: PP aa
Gene loci Dominant allele P a B P a b Recessive allele Figure 9.4 Matching gene loci on homologous chromosomes. Genotype: PP aa Bb Homozygous for the dominant allele Homozygous for the recessive allele Heterozygous

9.5 The law of independent assortment is revealed by tracking two characters at once
Example of a dihybrid cross Parental generation: round yellow seeds  wrinkled green seeds F1 generation: all plants with round yellow seeds F2 generation: of plants with round yellow seeds of plants with round green seeds of plants with wrinkled yellow seeds of plants with wrinkled green seeds Mendel needed to explain Why nonparental combinations were observed Why a 9:3:3:1 ratio was observed among the F2 offspring Student Misconceptions and Concerns 1. Students using Punnett squares need to be reminded that the calculations are expected statistical probabilities and not absolutes. Just as we would expect that any six playing cards dealt might be half black and half red, we frequently find that this is not true. This might be a good time to show how larger sample sizes increase the likelihood that sampling will reflect expected ratios. Teaching Tips 1. Understanding dihybrid crosses may be the most difficult concept in this chapter. Consider spending additional time to make these ideas very clear. As the text indicates, dihybrid crosses are essentially two monohybrid crosses. Copyright © 2009 Pearson Education, Inc.

9.5 The law of independent assortment is revealed by tracking two characters at once
Law of independent assortment Each pair of alleles segregates independently of the other pairs of alleles during gamete formation For genotype RrYy, four gamete types are possible: RY, Ry, rY, and ry For the BLAST Animation Genetic Variation: Independent Assortment, go to Animation and Video Files. For the BLAST Animation Two-Trait Crosses, go to Animation and Video Files. Student Misconceptions and Concerns 1. Students using Punnett squares need to be reminded that the calculations are expected statistical probabilities and not absolutes. Just as we would expect that any six playing cards dealt might be half black and half red, we frequently find that this is not true. This might be a good time to show how larger sample sizes increase the likelihood that sampling will reflect expected ratios. Teaching Tips 1. Understanding dihybrid crosses may be the most difficult concept in this chapter. Consider spending additional time to make these ideas very clear. As the text indicates, dihybrid crosses are essentially two monohybrid crosses. Copyright © 2009 Pearson Education, Inc.

Hypothesized (not actually seen) Actual results (support hypothesis)
Hypothesis: Dependent assortment Hypothesis: Independent assortment P generation RRYY rryy RRYY rryy Gametes RY ry Gametes RY ry F1 generation RrYy RrYy Sperm Sperm 1 – 4 RY 1 4 – rY 1 4 – Ry 1 4 – ry 1 – 2 RY 1 – 2 ry F2 generation 1 – 4 RY 1 – 2 RY RRYY RrYY RRYy RrYy Eggs Figure 9.5A Two hypotheses for segregation in a dihybrid cross. This figure compares dependent assortment to independent assortment. For dependent assortment, allele combinations present in the original parents are the only ones observed in the F2 offspring. Mendel’s results support independent assortment with nonparental combinations of round with green and wrinkled with yellow produced in the F2 generation. Note that both characteristics, seed shape and seed color, show 3:1 ratios when considered separately. Yellow is seen in 12/16 of the offspring and green is observed in 4/16 of the offspring. This confirms that these alleles are segregating from each other as predicted by the law of segregation, without interference from the other allele pair. 1 4 – rY 1 2 – ry RrYY rrYY RrYy rrYy Eggs Yellow round 1 – 4 Ry 16 –– 9 RRYy RrYy RRyy Rryy Green round 16 –– 3 Hypothesized (not actually seen) 1 – 4 ry Yellow wrinkled RrYy rrYy Rryy rryy 16 –– 3 Actual results (support hypothesis) Green wrinkled 1 16 ––

¼ black coat with normal vision ¼ chocolate coat with normal vision
Blind Blind Phenotypes Genotypes Black coat, normal vision B_N_ Black coat, blind (PRA) B_nn Chocolate coat, normal vision bbN_ Chocolate coat, blind (PRA) bbnn Mating of heterozygotes (black, normal vision) BbNn BbNn Phenotypic ratio of offspring Figure 9.5B Independent assortment of two genes in the Labrador retriever. This figure shows the outcome of independent assortment of two genes in Labrador retrievers. Ask the students to predict the allele combinations that would be found in gametes from an individual with the genotype BbNn. What offspring are produced if a dog that is heterozygous for both traits is mated to a dog that is homozygous recessive for both traits? BbNn  bbnn  ¼ black coat with normal vision ¼ chocolate coat with normal vision ¼ black coat with PRA (blind) ¼ chocolate coat with PRA (blind) Students may have the misconception that if there are four possible outcomes, then each will be represented in a litter of four puppies. Emphasizing that Punnett squares represent the probabilities of each outcome would be useful in addressing this misconception. 9 black coat, normal vision 3 black coat, blind (PRA) 3 chocolate coat, normal vision 1 chocolate coat, blind (PRA)

9.6 Geneticists use the testcross to determine unknown genotypes
Testcross Mating between an individual of unknown genotype and a homozygous recessive individual Will show whether the unknown genotype includes a recessive allele Used by Mendel to confirm true-breeding genotypes Student Misconceptions and Concerns 1. Students using Punnett squares need to be reminded that the calculations are expected statistical probabilities and not absolutes. Just as we would expect that any six playing cards dealt might be half black and half red, we frequently find that this is not true. This might be a good time to show how larger sample sizes increase the likelihood that sampling will reflect expected ratios. Teaching Tips 1. Consider challenging your students to explain why a testcross of two black Labs of unknown genotypes might not reveal the genotype of each dog. (If both dogs are heterozygous, or homozygous, the results would reveal the genotypes because the offspring would either be three dark and one brown or all dark. But if one black Lab was homozygous and the other heterozygous, we could not determine which Lab has which genotype.) Copyright © 2009 Pearson Education, Inc.

Two possibilities for the black dog:
Testcross: Genotypes B_ bb Two possibilities for the black dog: BB or Bb Figure 9.6 Using a testcross to determine genotype. If the black Labrador retriever has the Bb genotype, chocolate offspring are expected. To test whether a round-seeded pea plant is true breeding, what plant would you use for the testcross? A wrinkled-seeded plant. What result would show that the round-seeded plant was NOT true breeding? Having wrinkled-seeded plants among the offspring of the testcross. Gametes B B b b Bb b Bb bb Offspring All black 1 black : 1 chocolate

9.7 Mendel’s laws reflect the rules of probability
The probability of a specific event is the number of ways that event can occur out of the total possible outcomes. Rule of multiplication Multiply the probabilities of events that must occur together Rule of addition Add probabilities of events that can happen in alternate ways The probability of a specific event is the number of ways that event can occur out of the total possible outcomes. For example, the probability of rolling a 2 using conventional dice is 1/6. Rule of Multiplication Multiply the probabilities of events that occur together. For example, the probability of rolling a 2, followed by a 4, is 1/6  1/6 = 1/36. Rule of Addition Add probabilities of events that can happen in alternate ways. For example, the probability of rolling either a 2 or 4 is 1/6 + 1/6 = 1/3. Student Misconceptions and Concerns 1. Students using Punnett squares need to be reminded that the calculations are expected statistical probabilities and not absolutes. Just as we would expect that any six playing cards dealt might be half black and half red, we frequently find that this is not true. This might be a good time to show how larger sample sizes increase the likelihood that sampling will reflect expected ratios. Teaching Tips 1. Many students have trouble with the basic statistics that are necessary for many of these calculations. Give your students some practice. Consider having them work in pairs, each with a pair of dice (for large class sizes, this can be done in laboratories). Let them calculate the odds of rolling three sixes in a row and other possibilities. Copyright © 2009 Pearson Education, Inc.

F1 genotypes Bb male Formation of sperm Bb female Formation of eggs B
Formation of sperm Bb female Formation of eggs 1 – 2 B 1 – 2 b B B B b 1 – 2 B Figure 9.7 Segregation and fertilization as chance events. For an additional genetic example: Freckles are inherited with the dominant allele F. Individuals homozygous for the recessive allele f (genotype ff) do not have freckles. A widow’s peak is inherited with the dominant allele W. Individuals who are homozygous for the recessive allele w (genotype ww) have a straight hairline. What is the probability of a child with no freckles for two parents of genotype FfWw? Each parent has ½ chance of passing the F allele and ½ chance of passing the f allele to his or her offspring. A child without freckles has genotype ff. This requires the father to pass the f allele and the mother to pass the f allele, ½  ½ = ¼. What is the probability of a child with freckles and a straight hairline for two parents with genotype FfWw? This can be satisfied by two genotypes, FFww or Ffww. Using the multiplication rule, the probability of FFww is 1/16. The probability of Ffww is ½*  ¼ = 1/8. *Note: there are two ways to produce Ff; either the father gives F and the mother gives f or the father gives f and the mother gives F, ¼ + ¼ = 1/2 . Using the addition rule: 1/16 + 2/16 = 3/16. Sometimes students find it easier to analyze each pair of alleles separately, using 2 x 2 Punnett squares and multiplying the resulting probabilities. If so, the results would be: Ff  Ff  ¼ FF + ½ Ff + ¼ ff = ¾ freckles + ¼ no freckles Ww  Ww  ¼ WW + ½ Ww + ¼ ww = ¾ widow’s peak + ¼ straight P(no freckles) = ¼ P(freckles with straight hairline) = ¾  ¼ = 3/16 1 – 4 1 – 4 b B b b 1 – 2 b 1 – 4 1 – 4 F2 genotypes

9.8 CONNECTION: Genetic traits in humans can be tracked through family pedigrees
A pedigree Shows the inheritance of a trait in a family through multiple generations Demonstrates dominant or recessive inheritance Can also be used to deduce genotypes of family members Student Misconceptions and Concerns 1. Students might think that dominant alleles are naturally (a) more common, (b) more likely to be inherited, and (c) better for an organism. The text notes that this is not necessarily true. However, this might need to be emphasized further in the lecture. Teaching Tips 1. Students also seem to learn much from Figure 9.8b by analyzing the possible genotypes for the people whose complete genotype is not known. Consider challenging your students to suggest the possible genotypes for these people. Copyright © 2009 Pearson Education, Inc.

Dominant Traits Recessive Traits Freckles No freckles Widow’s peak
Freckles No freckles Figure 9.8A Examples of single-gene inherited traits in humans. This figure shows three traits that are determined by single genes with alternate alleles. The role of the dominant allele in influencing the phenotype when at least one copy is present can be emphasized here. Widow’s peak Straight hairline Free earlobe Attached earlobe

Figure 9.8A Examples of single-gene inherited traits in humans. Freckles No freckles

Widow’s peak Straight hairline
Figure 9.8A Examples of single-gene inherited traits in humans. Widow’s peak Straight hairline

Free earlobe Attached earlobe
Figure 9.8A Examples of single-gene inherited traits in humans. Free earlobe Attached earlobe

First generation (grandparents) Ff Ff ff Ff Second generation
First generation (grandparents) Ff Ff ff Ff Second generation (parents, aunts, and uncles) FF ff ff Ff Ff ff or Ff Third generation (two sisters) Figure 9.8B Pedigree showing inheritance of attached versus free earlobe in a hypothetical family. ff FF or Female Male Ff Affected Unaffected

9.9 CONNECTION: Many inherited disorders in humans are controlled by a single gene
Inherited human disorders show Recessive inheritance Two recessive alleles are needed to show disease Heterozygous parents are carriers of the disease-causing allele Probability of inheritance increases with inbreeding, mating between close relatives Dominant inheritance One dominant allele is needed to show disease Dominant lethal alleles are usually eliminated from the population Students may assume that dominant alleles are “better” and most common. The role of dominant alleles in causing some diseases shows that dominant alleles are not always advantageous. In addition, the incidence of these disease-causing dominant alleles is low, showing that recessive alleles can be more common than dominant ones. In general, a dominant allele leads to the production of a functional protein. In the case of a homozygous recessive genotype, a nonfunctional protein may be produced or the protein may be entirely absent. Cystic fibrosis is caused by a mutation in the gene for a membrane protein (CFTR, cystic fibrosis transmembrane conductance regulator) that controls chloride ion balance across cell membranes. Individuals with at least one copy of the dominant allele produce sufficient amounts of CFTR and are unaffected by the disease. In the homozygous recessive genotype, the CFTR is produced in an altered form that is not inserted in the cell membrane. The resulting ion imbalance leads to the buildup of sticky mucus that causes cystic fibrosis symptoms. Teaching Tips 1. The 2/3 fraction noted in the discussion of carriers of a recessive disorder for deafness often catches students off guard as they are expecting odds of 1/4, 1/2, or 3/4. However, when we eliminate the dd (deaf) possibility, as it would not be a carrier, we have three possible genotypes. Thus, the odds are based out of the remaining three genotypes Dd, dD, and DD. 2. As a simple test of comprehension, ask students to explain why lethal alleles are not eliminated from a population. Several possibilities exist: The lethal allele might be recessive, persisting in the population due to the survival of carriers, or the lethal allele might be dominant, but is not expressed until after the age of reproduction. 3. Ask your class (a) what the odds are of developing Huntington’s disease if a parent has this disease (50%) and (b) whether they would want this genetic test if they were at risk. The Huntington Disease Society website, offers many additional details. It is a good starting point for those who want to explore this disease in more detail. Copyright © 2009 Pearson Education, Inc.

Parents Normal Dd Normal Dd ´ Sperm D d Dd Normal (carrier) DD Normal
Parents Normal Dd Normal Dd Sperm D d Dd Normal (carrier) DD Normal D Offspring Eggs Figure 9.9A Offspring produced by parents who are both carriers for a recessive disorder. This diagram shows the inheritance of deafness, a recessive trait, from two heterozygous parents. Dd Normal (carrier) dd Deaf d

Figure 9.9B Dr. Michael C. Ain, a specialist in the repair of bone defects caused by achondroplasia and related disorders. This physician is afflicted with a dominant inherited disease, achondroplasia.

Table 9.9 Some Autosomal Disorders in Humans.

Video: Ultrasound of Human Fetus
9.10 CONNECTION: New technologies can provide insight into one’s genetic legacy Genetic testing of parents Fetal testing: biochemical and karyotype analyses Amniocentesis Chorionic villus sampling Maternal blood test Fetal imaging Ultrasound Fetoscopy Newborn screening An article in the Los Angeles Times detailed a maternal blood test to determine the sex of the child. The mother sends a few drops of dried blood to a commercial lab for analysis. The test is said to detect the presence of genes on the Y chromosome, indicating that the fetus is a boy. However, the test seemed to have a high failure rate. The amount of blood tested may be the cause of inaccuracy since scientists have shown that this type of test can be done with larger amounts of maternal blood. ( K. Kaplan, 2009, “Accuracy of Gender Test Kits in Question,” Los Angeles Times, February 24, 2009.) Teaching Tips 1. Medical technology raises many ethical issues. Consider asking your students this practical question. How much routine fetal testing do we want our insurance companies to cover and at what cost for insurance? Ultrasound, for example, is routinely performed on pregnant women as a normal part of prenatal care. What other tests should be standard? Who should decide? Who should pay? Video: Ultrasound of Human Fetus Copyright © 2009 Pearson Education, Inc.

Chorionic villus sampling (CVS)
Amniocentesis Chorionic villus sampling (CVS) Ultrasound monitor Needle inserted through abdomen to extract amniotic fluid Suction tube inserted through cervix to extract tissue from chorionic villi Ultrasound monitor Fetus Fetus Placenta Placenta Chorionic villi Uterus Cervix Cervix Uterus Amniotic fluid Centrifugation Fetal cells Fetal cells Figure 9.10A Testing a fetus for genetic disorders. Amniocentesis involves collection of fetal cells, which are cultured and tested. It occurs after 14–16 weeks and requires a few additional weeks to obtain test results. Chorionic villus sampling uses a small amount of placental tissue, occurs earlier in pregnancy, and achieves results within a day. Biochemical tests Several weeks Several hours Karyotyping

Figure 9.10B Ultrasound scanning of a fetus. Ultrasound is a noninvasive method for detecting abnormalities. A low value for the alpha fetoprotein test may indicate Down syndrome. Amniocentesis can determine whether the chromosomal abnormality exists. There is also a high-resolution ultrasound test for fluid that has been collecting under the neck (nuchal translucency) or abnormal characteristics of organs and bones (shortened femur and humerus) that are correlated with a Down syndrome fetus.

VARIATIONS ON MENDEL’S LAWS
Copyright © 2009 Pearson Education, Inc.

9.11 Incomplete dominance results in intermediate phenotypes
Incomplete dominance Neither allele is dominant over the other Expression of both alleles is observed as an intermediate phenotype in the heterozygous individual Student Misconceptions and Concerns 1. After reading the preceding modules, students might expect all traits to be governed by a single gene with two alleles, one dominant over the other. Modules 9.11–9.15 describe deviations from this simplistic model of inheritance. 2. As these variations of Mendel’s laws are introduced, students are likely to get confused and become uncertain about the prior definitions. Consider keeping a clear definition of these different patterns of inheritance available for the class to refer to as new patterns are discussed (perhaps as a handout for student reference). 3. As your class size increases, the chances increase that at least one student will have a family member with one of the genetic disorders discussed. Some students may find this embarrassing, but others might have a special interest in learning more about these topics, and may even be willing to share some of their family’s experiences with the class. Teaching Tips 1. Incomplete dominance is analogous to a compromise, or a gray shade. The key concept is that both “sides” have input. Complete dominance is more analogous to an authoritarian style, overruling others and insisting on things being a certain way. Although these analogies might seem obvious to instructors, many students new to genetics appreciate them. 2. Another analogy for cholesterol receptors is fishing poles. The more fishing poles you use, the more fish you can catch. Heterozygotes for hypercholesterolemia have fewer “fishing poles” for cholesterol. Thus, fewer “fish” are caught and more “fish” remain in the water. Copyright © 2009 Pearson Education, Inc.

P generation Red RR White rr Gametes R r F1 generation Pink Rr Gametes
Red RR White rr Gametes R r F1 generation Pink Rr Gametes 1 – 2 R 1 – 2 r Figure 9.11A Incomplete dominance in snapdragon color. This figure shows incomplete dominance in flower color. The difference in color between the RR and Rr genotypes is proposed to be a dosage effect, where the presence of one allele allows the production of half as much pigment as the presence of two alleles. Sperm 1 – 2 R 1 – 2 r F2 generation RR rR 1 – 2 R Eggs Rr rr 1 – 2 r

Genotypes: HH Hh hh Homozygous Heterozygous Homozygous
Genotypes: HH Homozygous for ability to make LDL receptors Hh Heterozygous hh Homozygous for inability to make LDL receptors Phenotypes: LDL LDL receptor Figure 9.11B Incomplete dominance in human hypercholesterolemia. This figure emphasizes that phenotype can involve characteristics at the molecular level. The intermediate phenotype results from the production of fewer receptors for low-density lipoproteins. Heterozygous individuals have a mild hypercholesterolemia, while individuals that are homozygous for the lack of receptors have severe disease. The gene for apolipoprotein E (APOE) also shows incomplete dominance. APOE is a protein found in lipid carriers called very-low-density lipoproteins that are responsible for lipid transport in the body. The E4 variant of the APOE gene is associated with Alzheimer’s disease. One copy of E4 increases the risk for Alzheimer’s disease, and two copies cause a greater increase in risk. Cell Normal Mild disease Severe disease

9.12 Many genes have more than two alleles in the population
Multiple alleles More than two alleles are found in the population A diploid individual can carry any two of these alleles The ABO blood group has three alleles, leading to four phenotypes: type A, type B, type AB, and type O blood An example for blood type inheritance: Mr. Jones has type A blood. His wife has type AB blood. Their first child has type B blood. What are the possible phenotypes for future offspring and the probabilities for each one? The type B child has the genotype IBi, receiving the IB allele from Mrs. Jones and the i allele from Mr. Jones. Therefore Mr. Jones has a heterozygous genotype, IAi. If he were homozygous for the IA allele, then he could only have children with type A or type AB blood. Mrs. Jones has the genotype IAIB. IAi  IAIB  ¼ type AB, ¼ type B, ½ type A Student Misconceptions and Concerns 1. After reading the preceding modules, students might expect all traits to be governed by a single gene with two alleles, one dominant over the other. Modules 9.11–9.15 describe deviations from this simplistic model of inheritance. 2. As these variations of Mendel’s laws are introduced, students are likely to get confused and become uncertain about the prior definitions. Consider keeping a clear definition of these different patterns of inheritance available for the class to refer to as new patterns are discussed (perhaps as a handout for student reference). 3. As your class size increases, the chances increase that at least one student will have a family member with one of the genetic disorders discussed. Some students may find this embarrassing, but others might have a special interest in learning more about these topics, and may even be willing to share some of their family’s experiences with the class. Teaching Tips 1. Students can think of blood types as analogous to socks on their feet. You can have socks that match, a sock on one foot but not the other, you can wear two socks that do not match, or you can even go barefoot (type O blood)! Developed further, think of Amber (A) and Blue (B) socks. Type A blood can have an Amber sock with either another Amber sock or a bare foot (or “zero” sock). Blue socks work the same way. One amber and one blue sock is the AB blood type. No socks, as already noted, represent type O. 2. Consider specifically comparing the principles of codominance (expression of both alleles) and incomplete dominance (expression of one intermediate trait). Students will likely benefit from this direct comparison. Copyright © 2009 Pearson Education, Inc.

9.12 Many genes have more than two alleles in the population
Codominance Neither allele is dominant over the other Expression of both alleles is observed as a distinct phenotype in the heterozygous individual Observed for type AB blood An example for blood type inheritance: Mr. Jones has type A blood. His wife has type AB blood. Their first child has type B blood. What are the possible phenotypes for future offspring and the probabilities for each one? The type B child has the genotype IBi, receiving the IB allele from Mrs. Jones and the i allele from Mr. Jones. Therefore Mr. Jones has a heterozygous genotype, IAi. If he were homozygous for the IA allele, then he could only have children with type A or type AB blood. Mrs. Jones has the genotype IAIB. IAi  IAIB  ¼ type AB, ¼ type B, ½ type A Student Misconceptions and Concerns 1. After reading the preceding modules, students might expect all traits to be governed by a single gene with two alleles, one dominant over the other. Modules 9.11–9.15 describe deviations from this simplistic model of inheritance. 2. As these variations of Mendel’s laws are introduced, students are likely to get confused and become uncertain about the prior definitions. Consider keeping a clear definition of these different patterns of inheritance available for the class to refer to as new patterns are discussed (perhaps as a handout for student reference). 3. As your class size increases, the chances increase that at least one student will have a family member with one of the genetic disorders discussed. Some students may find this embarrassing, but others might have a special interest in learning more about these topics, and may even be willing to share some of their family’s experiences with the class. Teaching Tips 1. Students can think of blood types as analogous to socks on their feet. You can have socks that match, a sock on one foot but not the other, you can wear two socks that do not match, or you can even go barefoot (type O blood)! Developed further, think of Amber (A) and Blue (B) socks. Type A blood can have an Amber sock with either another Amber sock or a bare foot (or “zero” sock). Blue socks work the same way. One amber and one blue sock is the AB blood type. No socks, as already noted, represent type O. 2. Consider specifically comparing the principles of codominance (expression of both alleles) and incomplete dominance (expression of one intermediate trait). Students will likely benefit from this direct comparison. Copyright © 2009 Pearson Education, Inc.

Figure 9.12 Multiple alleles for the ABO blood groups.
Blood Group (Phenotype) Antibodies Present in Blood Reaction When Blood from Groups Below Is Mixed with Antibodies from Groups at Left Genotypes Red Blood Cells O A B AB Anti-A Anti-B O ii IAIA or IAi A Carbohydrate A Anti-B IBIB or IBi B Carbohydrate B Anti-A Figure 9.12 Multiple alleles for the ABO blood groups. This figure emphasizes the phenotypic manifestation of codominance. Alleles IA and IB are codominant. Allele IA causes the production of carbohydrate A, while allele IB causes the production of carbohydrate B. For the genotype IAIB, both types of carbohydrates are found on the cell surface. The phenotype differs from either of the IAIA or IBIB homozygotes and does not appear as an intermediate as would be seen for intermediate dominance. Another human example of codominance involves the allele for sickle cell anemia (see Module 9.13). HbS codes for an altered form of the beta-hemoglobin molecule; HbA is the nonmutant form. Individuals with the HbS HbS genotype have only the mutant form of beta-hemoglobin on their red blood cells and suffer from sickle cell anemia. Individuals with HbA HbS have both types of hemoglobin on their red blood cells and have a milder condition called sickle cell trait. Individuals with HbA HbA have only the nonmutant form of beta-hemoglobin and neither the anemia nor the trait. Sometimes the distinction between incomplete dominance and codominance depends on the level at which observations are made. One example of incomplete dominance is coat color in cattle, where roan is the heterozygous intermediate to homozygotes red and white. But roan cattle have distinct white hairs overlaying distinct dark hairs so this could be viewed as a codominant phenotype. Students are often interested in the universal donor, universal recipient descriptions for blood type. Type O is called the universal donor since the cells do not have either carbohydrate A or carbohydrate B so that antibodies in the other blood types will not cause a reaction with these cells. But type O blood contains antibodies to type A and type B. In general, the patient is given his or her own blood type in a transfusion in order to avoid adverse reactions. In an emergency, the universal donor type may be used. AB IAIB —

Blood Group (Phenotype) Genotypes Red Blood Cells O ii IAIA or IAi A
Blood Group (Phenotype) Genotypes Red Blood Cells O ii IAIA or IAi A Carbohydrate A IBIB or IBi Figure 9.12 Multiple alleles for the ABO blood groups. This figure emphasizes the phenotypic manifestation of codominance. Alleles IA and IB are codominant. Allele IA causes the production of carbohydrate A, while allele IB causes the production of carbohydrate B. For the genotype IAIB, both types of carbohydrates are found on the cell surface. The phenotype differs from either of the IAIA or IBIB homozygotes and does not appear as an intermediate as would be seen for intermediate dominance. Another human example of codominance involves the allele for sickle cell anemia (see Module 9.13). HbS codes for an altered form of the beta-hemoglobin molecule; HbA is the nonmutant form. Individuals with the HbS HbS genotype have only the mutant form of beta-hemoglobin on their red blood cells and suffer from sickle cell anemia. Individuals with HbA HbS have both types of hemoglobin on their red blood cells and have a milder condition called sickle cell trait. Individuals with HbA HbA have only the nonmutant form of beta-hemoglobin and neither the anemia nor the trait. Sometimes the distinction between incomplete dominance and codominance depends on the level at which observations are made. One example of incomplete dominance is coat color in cattle, where roan is the heterozygous intermediate to homozygotes red and white. But roan cattle have distinct white hairs overlaying distinct dark hairs so this could be viewed as a codominant phenotype. Students are often interested in the universal donor, universal recipient descriptions for blood type. Type O is called the universal donor since the cells do not have either carbohydrate A or carbohydrate B so that antibodies in the other blood types will not cause a reaction with these cells. But type O blood contains antibodies to type A and type B. In general, the patient is given his or her own blood type in a transfusion in order to avoid adverse reactions. In an emergency, the universal donor type may be used. B Carbohydrate B AB IAIB

Reaction When Blood from Groups Below Is Mixed
Blood Group (Phenotype) Antibodies Present in Blood Reaction When Blood from Groups Below Is Mixed with Antibodies from Groups at Left O A B AB Anti-A Anti-B O A Anti-B Figure 9.12 Multiple alleles for the ABO blood groups. This figure emphasizes the phenotypic manifestation of codominance. Alleles IA and IB are codominant. Allele IA causes the production of carbohydrate A, while allele IB causes the production of carbohydrate B. For the genotype IAIB, both types of carbohydrates are found on the cell surface. The phenotype differs from either of the IAIA or IBIB homozygotes and does not appear as an intermediate as would be seen for intermediate dominance. Another human example of codominance involves the allele for sickle cell anemia (see Module 9.13). HbS codes for an altered form of the beta-hemoglobin molecule; HbA is the nonmutant form. Individuals with the HbS HbS genotype have only the mutant form of beta-hemoglobin on their red blood cells and suffer from sickle cell anemia. Individuals with HbA HbS have both types of hemoglobin on their red blood cells and have a milder condition called sickle cell trait. Individuals with HbA HbA have only the nonmutant form of beta-hemoglobin and neither the anemia nor the trait. Sometimes the distinction between incomplete dominance and codominance depends on the level at which observations are made. One example of incomplete dominance is coat color in cattle, where roan is the heterozygous intermediate to homozygotes red and white. But roan cattle have distinct white hairs overlaying distinct dark hairs so this could be viewed as a codominant phenotype. Students are often interested in the universal donor, universal recipient descriptions for blood type. Type O is called the universal donor since the cells do not have either carbohydrate A or carbohydrate B so that antibodies in the other blood types will not cause a reaction with these cells. But type O blood contains antibodies to type A and type B. In general, the patient is given his or her own blood type in a transfusion in order to avoid adverse reactions. In an emergency, the universal donor type may be used. B Anti-A AB —

9.13 A single gene may affect many phenotypic characters
Pleiotropy One gene influencing many characteristics The gene for sickle cell disease Affects the type of hemoglobin produced Affects the shape of red blood cells Causes anemia Causes organ damage Is related to susceptibility to malaria The incidence of heterozygosity for the sickle cell allele has frequently been reported as 1 in 10 for African Americans. Since malaria is not acting as a selective force in America, the incidence is declining for African Americans. An estimate of 1 in 12 has been presented by the Department of Energy at the Gene Gateway site: genomics.energy.gov. ( Student Misconceptions and Concerns 1. After reading the preceding modules, students might expect all traits to be governed by a single gene with two alleles, one dominant over the other. Modules 9.11–9.15 describe deviations from this simplistic model of inheritance. 2. As these variations of Mendel’s laws are introduced, students are likely to get confused and become uncertain about the prior definitions. Consider keeping a clear definition of these different patterns of inheritance available for the class to refer to as new patterns are discussed (perhaps as a handout for student reference). 3. As your class size increases, the chances increase that at least one student will have a family member with one of the genetic disorders discussed. Some students may find this embarrassing, but others might have a special interest in learning more about these topics, and may even be willing to share some of their family’s experiences with the class. Teaching Tips 1. The American Sickle Cell Anemia Association’s website, is a good place to find additional details. Copyright © 2009 Pearson Education, Inc.

Individual homozygous for sickle-cell allele
Sickle-cell (abnormal) hemoglobin Abnormal hemoglobin crystallizes, causing red blood cells to become sickle-shaped Sickle cells Clumping of cells and clogging of small blood vessels Breakdown of red blood cells Accumulation of sickled cells in spleen Figure 9.13 Sickle-cell disease, multiple effects of a single human gene. Physical weakness Heart failure Pain and fever Brain damage Damage to other organs Spleen damage Anemia Impaired mental function Pneumonia and other infections Kidney failure Paralysis Rheumatism

9.14 A single character may be influenced by many genes
Polygenic inheritance Many genes influence one trait Skin color is affected by at least three genes Student Misconceptions and Concerns 1. After reading the preceding modules, students might expect all traits to be governed by a single gene with two alleles, one dominant over the other. Modules 9.11–9.15 describe deviations from this simplistic model of inheritance. 2. As these variations of Mendel’s laws are introduced, students are likely to get confused and become uncertain about the prior definitions. Consider keeping a clear definition of these different patterns of inheritance available for the class to refer to as new patterns are discussed (perhaps as a handout for student reference). 3. As your class size increases, the chances increase that at least one student will have a family member with one of the genetic disorders discussed. Some students may find this embarrassing, but others might have a special interest in learning more about these topics, and may even be willing to share some of their family’s experiences with the class. Teaching Tips 1. Polygenic inheritance makes it possible for children to inherit genes to be taller or shorter than either parent. Similarly, skin tones can be darker or lighter than either parent. The environment also contributes significantly to the final phenotype for both of these traits. Copyright © 2009 Pearson Education, Inc.

Figure 9.14 A model for polygenic inheritance of skin color.
P generation aabbcc (very light) AABBCC (very dark) F1 generation AaBbCc AaBbCc Sperm 8 – 1 8 – 1 8 – 1 8 – 1 8 – 1 8 – 1 8 – 1 8 – 1 F2 generation 8 – 1 8 – 1 8 – 1 64 –– 20 8 – 1 Figure 9.14 A model for polygenic inheritance of skin color. Polygenic inheritance is implicated in susceptibility to certain diseases including insulin-dependent diabetes mellitus, multiple sclerosis, psoriasis, and schizophrenia. Eggs 8 – 1 64 –– 15 8 – 1 8 – 1 Fraction of population 8 – 1 64 –– 6 64 –– 1 64 –– 1 64 –– 6 64 –– 15 64 –– 20 64 –– 15 64 –– 6 64 –– 1 Skin color

aabbcc (very light) AABBCC (very dark) AaBbCc AaBbCc
P generation aabbcc (very light) AABBCC (very dark) F1 generation AaBbCc AaBbCc Sperm F2 generation 8 – 1 8 – 1 8 – 1 8 – 1 8 – 1 8 – 1 8 – 1 8 – 1 8 – 1 8 – 1 8 – 1 Figure 9.14 A model for polygenic inheritance of skin color. Polygenic inheritance is implicated in susceptibility to certain diseases including insulin-dependent diabetes mellitus, multiple sclerosis, psoriasis, and schizophrenia. 8 – 1 Eggs 8 – 1 8 – 1 8 – 1 8 – 1 64 –– 1 64 –– 6 64 –– 15 64 –– 20 64 –– 15 64 –– 6 64 –– 1

Fraction of population
20 –– 64 15 –– 64 Fraction of population 6 –– 64 Figure 9.14 A model for polygenic inheritance of skin color. Polygenic inheritance is implicated in susceptibility to certain diseases including insulin-dependent diabetes mellitus, multiple sclerosis, psoriasis, and schizophrenia. 1 –– 64 Skin color

9.15 The environment affects many characters
Phenotypic variations are influenced by the environment Skin color is affected by exposure to sunlight Susceptibility to diseases, such as cancer, has hereditary and environmental components The Himalayan allele in rabbits codes for a temperature-sensitive enzyme that influences pigmentation. A rabbit homozygous for this allele has a white coat with dark pigment on the feet, ears, and nose. The heat of the animal’s body prevents pigment formation over most of the body, so only the coolest areas darken. If the rabbit is reared at high temperatures, however, the color is completely white because no area is cool enough for the enzyme to function. The environmental temperature influences the phenotype. Student Misconceptions and Concerns 1. After reading the preceding modules, students might expect all traits to be governed by a single gene with two alleles, one dominant over the other. Modules 9.11–9.15 describe deviations from this simplistic model of inheritance. 2. As these variations of Mendel’s laws are introduced, students are likely to get confused and become uncertain about the prior definitions. Consider keeping a clear definition of these different patterns of inheritance available for the class to refer to as new patterns are discussed (perhaps as a handout for student reference). 3. As your class size increases, the chances increase that at least one student will have a family member with one of the genetic disorders discussed. Some students may find this embarrassing, but others might have a special interest in learning more about these topics, and may even be willing to share some of their family’s experiences with the class. Teaching Tips 1. Polygenic inheritance makes it possible for children to inherit genes to be taller or shorter than either parent. Similarly, skin tones can be darker or lighter than either parent. The environment also contributes significantly to the final phenotype for both of these traits. Copyright © 2009 Pearson Education, Inc.

Figure 9.15 Varying phenotypes due to environmental factors in genetically identical twins.

THE CHROMOSOMAL BASIS OF INHERITANCE

9.16 Chromosome behavior accounts for Mendel’s laws
Mendel’s Laws correlate with chromosome separation in meiosis The law of segregation depends on separation of homologous chromosomes in anaphase I The law of independent assortment depends on alternative orientations of chromosomes in metaphase I Student Misconceptions and Concerns 1. This section of the chapter relies upon a good understanding of the chromosome sorting process of meiosis. If students were not assigned Chapter 8, and meiosis has not otherwise been addressed, it will be difficult for students to understand the chromosomal basis of inheritance or linked genes. Teaching Tips 1. Figure 9.16 requires an understanding of meiosis and the general cell cycle from Chapter 8. Students may need to be reminded that chromosomes are duplicated in the preceding interphase, as indicated in the first step. Furthermore, students may not initially notice that this diagram represents four possible outcomes, not stages of any one meiotic cycle. Copyright © 2009 Pearson Education, Inc.

All round yellow seeds (RrYy) F1 generation Metaphase I of meiosis
F1 generation R y r Y R r r R Metaphase I of meiosis (alternative arrangements) Y y Y y Figure 9.16 The chromosomal basis of Mendel’s laws.

All round yellow seeds (RrYy) F1 generation Metaphase I of meiosis
F1 generation R y r Y R r r R Metaphase I of meiosis (alternative arrangements) Y y Y y R r r R Anaphase I of meiosis Y y Y y R r r R Metaphase II of meiosis Y y Y y Figure 9.16 The chromosomal basis of Mendel’s laws.

Fertilization among the F1 plants
All round yellow seeds (RrYy) F1 generation R y r Y R r r R Metaphase I of meiosis (alternative arrangements) Y y Y y R r r R Anaphase I of meiosis Y y Y y R r r R Metaphase II of meiosis Y y Y y Figure 9.16 The chromosomal basis of Mendel’s laws. Gametes y Y Y y Y Y y y R R r r r r R R 1 – 4 ry 1 – 4 rY 1 4 – Ry 1 4 – RY Fertilization among the F1 plants F2 generation 9 :3 :3 :1

9.17 Genes on the same chromosome tend to be inherited together
Linked Genes Are located close together on the same chromosome Tend to be inherited together Example studied by Bateson and Punnett Parental generation: plants with purple flowers, long pollen crossed to plants with red flowers, round pollen The F2 generation did not show a 9:3:3:1 ratio Most F2 individuals had purple flowers, long pollen or red flowers, round pollen Student Misconceptions and Concerns 1. This section of the chapter relies upon a good understanding of the chromosome sorting process of meiosis. If students were not assigned Chapter 8, and meiosis has not otherwise been addressed, it will be difficult for students to understand the chromosomal basis of inheritance or linked genes. Teaching Tips 1. Building on the shoe analogy developed in Chapter 8, linked genes are like a shoe and its shoelaces. The two are usually transferred together but can be moved separately under special circumstances. Copyright © 2009 Pearson Education, Inc.

Not accounted for: purple round and red long
Experiment Purple flower PpLl PpLl Long pollen Observed offspring Prediction (9:3:3:1) Phenotypes Purple long Purple round Red long Red round 284 21 55 215 71 24 Explanation: linked genes Parental diploid cell PpLl PL pl Meiosis Most gametes PL pl Figure 9.17 Experiment involving linked genes in the sweet pea. Fertilization Sperm PL pl PL PL PL Most offspring PL pl Eggs pl pl pl PL pl 3 purple long : 1 red round Not accounted for: purple round and red long

Observed offspring Prediction (9:3:3:1) Phenotypes
Experiment Purple flower PpLl PpLl Long pollen Observed offspring Prediction (9:3:3:1) Phenotypes Figure 9.17 Experiment involving linked genes in the sweet pea. Purple long Purple round Red long Red round 284 21 55 215 71 24

Not accounted for: purple round and red long
Explanation: linked genes Parental diploid cell PpLl PL pl Meiosis Most gametes PL pl Fertilization Sperm PL pl Figure 9.17 Experiment involving linked genes in the sweet pea. PL PL PL Most offspring PL pl Eggs pl pl pl PL pl 3 purple long : 1 red round Not accounted for: purple round and red long

9.18 Crossing over produces new combinations of alleles
Linked alleles can be separated by crossing over Recombinant chromosomes are formed Thomas Hunt Morgan demonstrated this in early experiments Geneticists measure genetic distance by recombination frequency For the Bateson-Punnett experiment described in Module 9.17, the parental chromosomes have alleles P-----L and p------l. Crossing over between the P and L loci would yield recombinant chromosomes P-----l and p------L. Genetic distance was measured in map units where 1% recombination = 1 map unit. The map unit is also called the centimorgan, in honor of Thomas Hunt Morgan. Student Misconceptions and Concerns 1. This section of the chapter relies upon a good understanding of the chromosome sorting process of meiosis. If students were not assigned Chapter 8, and meiosis has not otherwise been addressed, it will be difficult for students to understand the chromosomal basis of inheritance or linked genes. 2. The nature of linked genes builds upon our natural expectations that items that are closely together are less likely to be separated. Yet, students may find such concepts initially foreign. Whether it is parents holding the hands of children or people and their pets, we generally know that separation is more likely when things are farther apart. Teaching Tips 1. Crossing over (from Chapter 8) is like randomly editing out a minute of film from two movies and swapping them. Perhaps the fifth minute of Bambi is swapped for the fifth minute of Texas Chainsaw Massacre. Clearly, the closer together two frames of film are, the more likely they are to move or remain together. 2. Challenge students to explain why Sturtevant and Morgan studied the genetics of fruit flies. As the text notes, their small size, ease of care, and ability to produce several generations in a matter of weeks or months were important factors. Copyright © 2009 Pearson Education, Inc.

A B A b A B a b a B a b Tetrad Crossing over Gametes
A B A b A B a b a B a b Tetrad Crossing over Figure 9.18A Review: Production of recombinant gametes. Gametes

Figure 9.18B Drosophila melanogaster.

Experiment Gray body, long wings (wild type) Black body, vestigial wings GgLl ggll Female Male Offspring Gray long Black vestigial Gray vestigial Black long 965 944 206 185 Parental phenotypes Recombinant phenotypes 391 recombinants 2,300 total offspring Recombination frequency = = 0.17 or 17% Explanation Figure 9.18C Fruit fly experiment demonstrating the role of crossing over in inheritance. This figure shows Morgan’s observations of crossing over. Parental types are the most frequent offspring, and the percentage of recombinants can be used to estimate the genetic distance. G L g l GgLl (female) ggll (male) g l g l G L g l G l g L g l Eggs Sperm G L g l G l g L g l g l g l g l Offspring

Recombination frequency = = 0.17 or 17%
Experiment Gray body, long wings (wild type) Black body, vestigial wings GgLl ggll Female Male Offspring Gray long Black vestigial Gray vestigial Black long Figure 9.18C Fruit fly experiment demonstrating the role of crossing over in inheritance. This figure shows Morgan’s observations of crossing over. Parental types are the most frequent offspring, and the percentage of recombinants can be used to estimate the genetic distance. 965 944 206 185 Parental phenotypes Recombinant phenotypes 391 recombinants 2,300 total offspring Recombination frequency = = 0.17 or 17%

GgLl (female) ggll (male)
Explanation G L g l GgLl (female) ggll (male) g l g l G L g l G l g L g l Eggs Sperm G L g l G l g L Figure 9.18C Fruit fly experiment demonstrating the role of crossing over in inheritance. This figure shows Morgan’s observations of crossing over. Parental types are the most frequent offspring, and the percentage of recombinants can be used to estimate the genetic distance. g l g l g l g l Offspring

9.19 Geneticists use crossover data to map genes
Genetic maps Show the order of genes on chromosomes Arrange genes into linkage groups representing individual chromosomes Example: The distance between gene A and gene B is 30 map units. The distance between genes B and C is 22 map units. The distance between A and C is 8 map units. Which gene is in the middle, between the other two? C is the middle gene. The map is A C B Student Misconceptions and Concerns 1. This section of the chapter relies upon a good understanding of the chromosomesorting process of meiosis. If students were not assigned Chapter 8, and meiosis has not otherwise been addressed, it will be difficult for students to understand the chromosomal basis of inheritance or linked genes. 2. The nature of linked genes builds upon our natural expectations that items that are closely together are less likely to be separated. Yet, students may find such concepts initially foreign. Whether it is parents holding the hands of children or people and their pets, we generally know that separation is more likely when things are farther apart. Teaching Tips 1. Challenge students to explain why Sturtevant and Morgan studied the genetics of fruit flies. As the text notes, their small size, ease of care, and ability to produce several generations in a matter of weeks or months were important factors. Copyright © 2009 Pearson Education, Inc.

g c l 17% 9% 9.5% Recombination frequencies
Chromosome g c l 17% 9% 9.5% Figure 9.19A Mapping genes from crossover data. Recombination frequencies

Mutant phenotypes Short aristae Black body (g) Cinnabar eyes (c)
Mutant phenotypes Short aristae Black body (g) Cinnabar eyes (c) Vestigial wings (l) Brown eyes Figure 9.19B A partial genetic map of a fruit fly chromosome. Long aristae (appendages on head) Gray body (G) Red eyes (C) Normal wings (L) Red eyes Wild-type phenotypes

SEX CHROMOSOMES AND SEX-LINKED GENES

9.20 Chromosomes determine sex in many species
X-Y system in mammals, fruit flies XX = female; XY = male X-O system in grasshoppers and roaches XX = female; XO = male Z-W in system in birds, butterflies, and some fishes ZW = female, ZZ = male Chromosome number in ants and bees Diploid = female; haploid = male While fruit flies have X and Y chromosomes, sex determination is influenced by the X chromosome to autosome ratio. Females with XX genotype have two X chromosomes and two sets of autosomes for an X:A ratio of 1. Males with XY genotype have one X chromosome and two sets of autosomes for an X:A ratio of 0.5. Teaching Tips 1. In certain animals, such as crocodilians and many turtles, sex is not genetically determined. Instead, the incubation temperature of the eggs determines an animal’s sex. Students may enjoy researching this unique form of sex determination, often identified as TSD (temperature-dependent sex determination). Copyright © 2009 Pearson Education, Inc.

X Y Figure 9.20A The human sex chromosomes.
X Y Figure 9.20A The human sex chromosomes. The human X chromosome is larger than the Y chromosome. According to Human Genome Project data, the X chromosome is estimated to have 1336 genes, while the Y chromosome is projected to have 307 genes.

(male) (female) 44 + XY Parents’ diploid cells 44 + XX 22 + X 22 + X
(male) (female) 44 + XY Parents’ diploid cells 44 + XX 22 + X 22 + X 22 + Y Sperm Egg Figure 9.20B The X-Y system. 44 + XX Offspring (diploid) 44 + XY

22 + XX 22 + X Figure 9.20C The X-O system.

76 76 + + ZW ZZ Figure 9.20D The Z-W system.
76 + ZW 76 + ZZ Figure 9.20D The Z-W system. Sex-linked traits affect female birds to a greater extent than males. A female would need to inherit only one copy of a Z-linked recessive allele to show a specific trait, while a male would need to inherit two copies. Bird populations can become endangered if the numbers of females decline due to harmful Z-linked traits.

32 16 Figure 9.20E Sex determination by chromosome number.

9.21 Sex-linked genes exhibit a unique pattern of inheritance
Sex-linked genes are located on either of the sex chromosomes Reciprocal crosses show different results White-eyed female  red-eyed male red-eyed females and white-eyed males Red-eyed female  white-eyed male red-eyed females and red-eyed males X-linked genes are passed from mother to son and mother to daughter X-linked genes are passed from father to daughter Y-linked genes are passed from father to son The X and Y chromosomes separate during meiosis I so that the alleles linked to them segregate from each other. The Y chromosome does not have the same genes as the X chromosome so it is not able to exert dominance over X-linked alleles. XY males will show both dominant and recessive x-linked traits even though the alleles are present in only one copy. XX females must inherit two copies of an X-linked recessive allele to express the trait. Let R = red and r = white White-eyed female  red-eyed male XrXr  XRY  XRXr and XrY (red-eyed females and white-eyed males) Red-eyed female  white-eyed male XRXR  XrY  XRXr and XRY (red-eyed females and red-eyed males) Student Misconceptions and Concerns 1. The prior discussion of linked genes addresses a different relationship than the use of the similar term sex-linked genes. Consider emphasizing this distinction for your students. Teaching Tips 1. An analogy can be drawn between sex-linked genes and the risk of not having a backup copy of a file on your computer. If you only have one copy, and it is damaged, you have to live with the damaged file. Females, who have two X chromosomes, thus have a “backup copy” that can function if one of the sex-linked genes is damaged. Copyright © 2009 Pearson Education, Inc.

Figure 9.21A Fruit fly eye color, a sex-linked characteristic.

Female Male XR XR Xr Y Sperm Xr Y Eggs XR XR Xr XR Y
XR XR Xr Y Sperm Xr Y Figure 9.21B Homozygous, red-eyed female  white-eyed male. Since the female fly is homozygous for the dominant R allele, both female and male offspring will have red eyes. Eggs XR XR Xr XR Y R = red-eye allele r = white-eye allele

Female Male XR Xr XR Y Sperm XR Y XR XR XR XR Y Eggs Xr Xr XR Xr Y
XR Xr XR Y Sperm XR Y Figure 9.21C Heterozygous female  red-eyed male. All of the female offspring will have red eyes, since the father will pass the X-linked R allele to all of his daughters. For the sons, there is a one-half probability of receiving the R allele or the r allele from the mother fly. XR XR XR XR Y Eggs Xr Xr XR Xr Y

Female Male XR Xr Xr Y Sperm Xr Y XR XR XR XR Y Eggs Xr Xr Xr Xr Y
XR Xr Xr Y Sperm Xr Y Figure 9.21D Heterozygous female  white-eyed male. White-eyed female offspring receive an r allele from the mother fly and an r allele from the father fly. XR XR XR XR Y Eggs Xr Xr Xr Xr Y

9.22 CONNECTION: Sex-linked disorders affect mostly males
Males express X-linked disorders such as the following when recessive alleles are present in one copy Hemophilia Colorblindness Duchenne muscular dystrophy Colorblindness is due to the X-linked recessive allele b, while the X-linked dominant allele B leads to full color vision. Predict the ratio of offspring phenotypes for each of the following: Colorblind female x male with full color vision XbXb  XBY  ½ XBXb and ½ XbY = ½ females with full color vision + ½ colorblind males Female heterozygous for colorblindness  colorblind male XBXb  XbY  ¼ XBXb, ¼ XbXb, ¼ XBY, ¼ XbY = ¼ females with full color vision + ¼ colorblind females + ¼ males with full color vision + ¼ colorblind males Female homozygous for full color vision  colorblind male XBXB  XbY  ½ XBXb and ½ XBY = ½ females with full color vision + ½ males with full color vision Student Misconceptions and Concerns 1. The likelihood that at least some students in larger classes are color-blind is very high. Some of these students might find this interesting and want to discuss it further. However, others might be embarrassed by what might be perceived as a defect. Teaching Tips 1. For additional information about hemophilia, consider visiting the website of the National Hemophilia Foundation at Copyright © 2009 Pearson Education, Inc.

Queen Victoria Albert Alice Louis Alexandra Czar Nicholas II of Russia
Queen Victoria Albert Alice Louis Alexandra Czar Nicholas II of Russia Figure 9.22 Hemophilia in the royal family of Russia (half-filled symbols represent heterozygous carriers). There is no prior history of hemophilia in Queen Victoria’s family, so it is thought that the allele arose by spontaneous mutation in the sex cells of one of her parents. Alexis

9.23 EVOLUTION CONNECTION: The Y chromosome provides clues about human male evolution
Similarities in Y chromosome sequences Show a significant percentage of men related to the same male parent Demonstrate a connection between people living in distant locations In Asia, a significant percentage of men are related to Genghis Khan. In Ireland, a significant percentage of men are related to a fifth-century warlord. The Lemba people in southern Africa are connected with the ancient Jews. Teaching Tips 1. Like the Y chromosome, mitochondrial DNA (mtDNA) can be used to trace maternal ancestry (because mitochondria are characteristically inherited from the egg). For a fee, several commercial groups offer to provide information about a person’s ancestry based upon genetic samples. Such groups can be found by searching the Internet using the keywords “genetic ancestry.” Copyright © 2009 Pearson Education, Inc.

Homologous chromosomes Alleles, residing at the same locus Fertilization Meiosis Gamete from other parent Diploid zygote (containing paired alleles) Paired alleles, alternate forms of a gene Haploid gametes (allele pairs separate)

Incomplete dominance Red RR White rr Pink Rr Pleiotropy Single gene Multiple characters Polygenic inheritance Multiple genes Single characters (such as skin color)

inheritance when phenotype
Genes located on alternative versions called chromosomes (a) at specific locations called if both same, genotype called if different, genotype called (b) (c) heterozygous expressed allele called unexpressed allele called (d) (e) inheritance when phenotype In between called (f)

You should now be able to
Explain and apply Mendel’s laws of segregation and independent assortment Distinguish between terms in the following groups: allele—gene; dominant—recessive; genotype—phenotype; F1—F2; heterozygous—homozygous; incomplete dominance—codominance Explain the meaning of the terms locus, multiple alleles, pedigree, pleiotropy, polygenic inheritance Copyright © 2009 Pearson Education, Inc.

You should now be able to
Describe the difference in inheritance patterns for linked genes and explain how recombination can be used to estimate gene distances Describe how sex is inherited in humans and identify the pattern of inheritance observed for sex-linked genes Solve genetics problems involving monohybrid and dihybrid crosses for autosomal and sex-linked traits, with variations on Mendel’s laws Copyright © 2009 Pearson Education, Inc.

Patterns of Inheritance

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