Inheritance, Genes, and Chromosomes

Inheritance, Genes, and Chromosomes
8 Inheritance, Genes, and Chromosomes

Chapter 8 Inheritance, Genes, and Chromosomes
Key Concepts 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws 8.2 Alleles and Genes Interact to Produce Phenotypes 8.3 Genes Are Carried on Chromosomes 8.4 Prokaryotes Can Exchange Genetic Material

Chapter 8 Opening Question
How is hemophilia inherited through the mother, and why is it more frequent in males?

Early experiments with genetics yielded two theories:
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Early experiments with genetics yielded two theories: Blending inheritance—gametes contained determinants (genes) that blended when gametes fused during fertilization Particulate inheritance—each determinant was physically distinct and remained intact during fertilization

Mendel used the scientific method and studied garden peas.
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Mendel used the scientific method and studied garden peas. Their flowers have both male and female sex organs, pistils, and stamens, to produce gametes. Male organs can be removed to allow fertilization by another flower.

In-Text Art, Ch. 8, p. 145

Character—observable physical feature (e.g., flower color, seed shape)
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Character—observable physical feature (e.g., flower color, seed shape) Trait—form of a character (e.g., purple flowers or white flowers, wrinkled seeds) Mendel worked with true-breeding varieties—when plants of the same variety were crossed, all offspring plants produced the same seeds and flowers.

Parental generation = P
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Mendel’s crosses: Pollen from one parent was transferred to the stigma of the other parent. Parental generation = P Resulting offspring = first filial generation or F1 If F1 plants self-pollinate, they produce second filial generation or F2.

This produced monohybrids in the F1 generation.
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws In Mendel’s first experiment, he crossed plants differing in just one character (P). This produced monohybrids in the F1 generation. The monohybrids were then allowed to self- pollinate to form the F2 generation—a monohybrid cross. Mendel repeated this for seven characters.

Figure 8.1 Mendel’s Monohybrid Experiments (Part 1)

Figure 8.1 Mendel’s Monohybrid Experiments (Part 2)

The trait that appears in the F1 is the dominant trait.
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws One trait of each pair disappeared in the F1 generation and reappeared in F2 —these traits are recessive. The trait that appears in the F1 is the dominant trait. The ratio of dominant traits to recessive traits in the F2 was about 3:1.

Each plant has two genes for each character, one from each parent.
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Mendel’s observations rejected the blending theory of inheritance and supported the particulate theory. He proposed that the determinants occur in pairs and are segregated in the gametes. Each plant has two genes for each character, one from each parent. Diploid—two copies of a gene Haploid—one copy of a gene LINK Review discussion of haploid and diploid organisms in Concept 7.1

Heterozygous individuals have two different alleles (e.g., Ss).
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Alleles are different forms of a gene, such as smooth or wrinkled seeds. True-breeding individuals have two copies of the same allele—they are homozygous for the allele (e.g., ss). Heterozygous individuals have two different alleles (e.g., Ss).

Phenotype—physical appearance of an organism (e.g., spherical seeds)
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Phenotype—physical appearance of an organism (e.g., spherical seeds) Genotype—the genetic makeup (e.g., Ss) Spherical seeds can be the result of two different genotypes—SS or Ss.

Each gamete receives only one copy.
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Mendel’s first law: The law of segregation states that the two copies of a gene separate when an individual makes gametes. Each gamete receives only one copy.

Figure 8.2 Mendel’s Explanation of Inheritance (Part 1)

Allele combinations can be predicted using a Punnett square.
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws When the F1 self-pollinates, there are three ways to get the dominant trait (e.g., spherical), but only one way to get the recessive (wrinkled)—resulting in the 3:1 ratio. Allele combinations can be predicted using a Punnett square.

Figure 8.2 Mendel’s Explanation of Inheritance (Part 2)

A gene is a short sequence on a longer DNA molecule.
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws A gene is a short sequence on a longer DNA molecule. DNA molecules make up the chromosomes. Different alleles of a gene segregate when chromosomes separate during meiosis I. LINK The process of meiosis is described in Concept 7.4

Figure 8.3 Meiosis Accounts for the Segregation of Alleles (Part 1)

Mendel tested his hypothesis by doing test crosses:
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Mendel tested his hypothesis by doing test crosses: He did this to determine whether an individual is homozygous or heterozygous for a trait by crossing it with a homozygous recessive individual. Mendel crossed the F1 with known homozygotes (e.g., wrinkled or ss).

Figure 8.4 Homozygous or Heterozygous? (Part 1)

Figure 8.4 Homozygous or Heterozygous?

Mendel’s next experiment involved:
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Mendel’s next experiment involved: Crossing peas that differed in two characters—seed shape and seed color True-breeding parents: SSYY—spherical yellow seeds ssyy—wrinkled green seeds

F1 generation is SsYy—all spherical yellow.
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws F1 generation is SsYy—all spherical yellow. Crossing the F1 generation (all identical double heterozygotes) is a dihybrid cross. Mendel asked whether, in the gametes produced by SsYy, the traits would be linked, or segregate independently.

Two possibilities included:
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Two possibilities included: Alleles could maintain associations seen in parental generation—they could be linked If linked, gametes would be SY or sy; F2 would have three times more spherical yellow than wrinkled green. If independent, gametes could be SY, sy, Sy, or sY. F2 would have nine different genotypes; phenotypes would be in 9:3:3:1 ratio.

If independent, gametes could be SY, sy, Sy, or sY in equal numbers.
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Or: The segregation of S from s could be independent of Y from y—the two genes could be unlinked If independent, gametes could be SY, sy, Sy, or sY in equal numbers. The F2 generation would have nine different genotypes; and four phenotypes in a 9:3:3:1 ratio. This prediction was supported.

Figure 8.5 Independent Assortment (Part 1)

Figure 8.5 Independent Assortment (Part 2)

Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws
Mendel’s second law: The law of independent assortment states that alleles of different genes assort independently during gamete formation. This law doesn’t always apply to genes on the same chromosome, but chromosomes do segregate independently. ANIMATED TUTORIAL 8.1 Independent Assortment of Alleles

Figure 8.6 Meiosis Accounts for Independent Assortment of Alleles

Probability calculations and Punnett squares give the same results.
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws One of Mendel’s contributions to genetics was the use of mathematical analyses— the rules of statistics and probability. His analyses revealed patterns that allowed him to formulate his hypotheses. Probability calculations and Punnett squares give the same results.

If an event is certain to happen, probability = 1
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Probability If an event is certain to happen, probability = 1 If an event cannot possibly happen, probability = 0 All other events have a probability between 0 and 1

The probability that both will come up heads is:
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Two coin tosses are independent events, each will come up heads ½ the time. The probability that both will come up heads is: ½ x ½ = ¼ To get the joint probability, multiply the individual probabilities (multiplication rule).

Figure 8.7 Using Probability Calculations in Genetics

Probability in a monohybrid cross
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Probability in a monohybrid cross After self-pollination of an F1 Ss, the probability that the F2 offspring will have the genotype SS is ½ x ½ = ¼; the same for ss offspring. There are two ways to get a heterozygote Ss; the probability is the sum of the individual probabilities (addition rule): ¼ + ¼ = ½ See Figures 8.2 and 8.7

Human pedigrees can show Mendel’s laws.
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Human pedigrees can show Mendel’s laws. Humans have few offspring; pedigrees do not show the clear proportions that the pea plants showed. Geneticists use pedigrees to determine whether a rare allele is dominant or recessive. INTERACTIVE TUTORIAL 8.1 Pedigree Analysis

Pattern of inheritance for a rare dominant allele:
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Pattern of inheritance for a rare dominant allele: Every person with the abnormal phenotype has an affected parent. Either all (if homozygous parent) or half (if heterozygous parent) of offspring in an affected family are affected.

Figure 8.8 Pedigree Analysis and Inheritance (Part 1)

Pattern of inheritance for a rare recessive allele:
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Pattern of inheritance for a rare recessive allele: Affected people often have two unaffected parents. In an affected family, one-fourth of children of unaffected parents are affected.

Figure 8.8 Pedigree Analysis and Inheritance (Part 2)

The great “chestnut vs. black” debate
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws The great “chestnut vs. black” debate Shortly after the rediscovery of Mendel’s pea plant experiments, biologists began investigating whether Mendel’s laws applied to other species besides peas. In England, a debate broke out about whether the color of racehorses might be governed by Mendel’s laws. Using the British Jockey Club registry of racehorse pedigrees, biologists noted: A chestnut (red) stallion, when bred to chestnut mares, always produced chestnut foals. Certain black stallions, when bred to chestnut mares, produced only black foals. Other black stallions, when bred to chestnut mares, produced a mix of black foals and chestnut foals. Discuss with a partner: 1. Which is dominant: black, chestnut, or neither? Why? 2. Would you expect horses with the dominant allele to be superior in some way? (For example, would they be faster, or healthier, or produce more offspring?) 3. Consider a cross of black × chestnut. Is this a monohybrid cross, dihybrid cross, or a test cross? What will the results of this cross tell you? Can one cross (one foal) be informative? Answers: Black is dominant. Though this may seem obvious, often a sizeable fraction of students misinterprets “chestnut × chestnut = chestnut” as evidence that chestnut is dominant. (They are misled by thinking that true-breeding somehow indicates dominance.) It helps to point out that the recessive phenotype will always be true-breeding. Black horses, of course, are not superior; they are simply black. (The point of this question is to unearth common student misconceptions that dominance is related in some way to superiority or fitness.) A black × chestnut cross is a test cross, and it can provide information about whether the black parent is a homozygote or a heterozygote. One chestnut foal would prove the black parent is a heterozygote; but one black foal would not prove that the black parent is a homozygote. Several black foals in a row, with no chestnut foals, would be required to be reasonably confident that the black parent is a homozygote. (If students ask, with only 4 black foals - and no chestnut foals - one could be reasonably confident that the black parent is a homozygote. A heterozygote would have only a ~6% chance of producing 4 black foals in a row.) INSTRUCTOR NOTES: If any students ask about “sorrel” or “bay” colors, sorrel is a synonym for chestnut (red). Bay is produced by a different locus. In equine genetics, the black/chestnut locus is designated as the E or “Extension” locus (E = black, e = chestnut). This locus encodes the melanocortin-1 receptor gene, which enables melanocytes to respond to the hormone alpha-MSH. If an E (black) allele is present, melanocytes respond normally to alpha-MSH, producing black pigment (eumelanin). The e allele is a loss-of-function mutation that disrupts the ability of melanocytes to respond to alpha-MSH; with a homozygous ee genotype, the result is production of reddish pigment (phaeomelanin). Most other mammals have a homologous locus that governs whether the predominant pigment in the fur or hair is black/brown eumelanin vs. red/yellow phaeomelanin. A dog example is presented in textbook Figure 8.12 (the canine E locus is homologous to the equine E locus discussed here). Human hair color is governed by the same locus, called MC1R in humans. Several other loci have epistatic effects to the locus discussed here. Most mammals also have a “brown” locus that can lighten black eumelanin to brown, a “diliution” locus that can lighten red phaeomelanin to yellow, an “agouti” locus that restricts eumelanin to the extremities, and various spotting and striping loci. See the dog example in textbook Figure 8.12, the rabbit example in textbook Figure 8.9, and the palomino example later in this exercise. 45

Which is dominant, black or chestnut? a. Black b. Chestnut
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Which is dominant, black or chestnut? a. Black b. Chestnut c. Neither; black and chestnut are incompletely dominant. d. I don’t know. Answer: a [NOTE TO THE INSTRUCTOR: It can be useful to include an “I don't know” choice with clickers, because it can help you discover how many students really haven’t understood the concept at all. Use of this option may depend on whether you assign participation-only points or performance points (or some combination) to clicker questions in your course. If you only assign participation points, it may be useful to leave the “I don't know” choice in the question, as it gives students a penalty-free way of indicating that more time may be needed on this concept.] 46

Coin tossing for fun and (not much) profit
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Coin tossing for fun and (not much) profit Working in small groups, make predictions about the results of tossing a dime and a penny: What is the probability that both coins will come up heads? What is the probability that one coin (either coin) will come up heads, and the other will come up tails? Are your predictions the same in each case, or different? Why? Now each member of your group will flip a real dime and real penny once. Each member of your team should record his or her own data for the next question. Then tally up the group data. Look at the overall data for your group. Did the results match your predictions? INSTRUCTOR NOTES: This exercise walks students through the steps of Figure 8.7 in the textbook. Rather than presenting textbook Figure 8.7 first, try having students do the exercise first - that is, coming up with their own hypotheses and doing an actual coin-flipping experiment. With a little prompting, students may then be able to derive the multiplication and addition rules by themselves. Then present Figure 8.7 at the end, as a summary. [Note: It is a good idea to bring extra coins to class in case students don't have enough.] 47

b. Dime heads, penny tails. c. Dime tails, penny heads.
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws What were your personal results? (Report just the results of your own dime & penny toss - not the results of your whole group.) a. Both were heads. b. Dime heads, penny tails. c. Dime tails, penny heads. d. Both were tails. Do the overall class data match your predictions? Do they differ from the results of your small group? Why or why not? If your predictions were not supported by the class data, explain why not. If your predictions were supported by the class data, now think about how you derived those predictions. What mathematical procedures were you using when you calculated your two predictions? Why? INSTRUCTOR NOTES: Class data can be tabulated either with clickers (using this slide) or on a board at the front of the room. If using a board, make four columns on the board for the four outcomes, and then have each group send one member to the board to put tick marks in the columns for all the group's data. Then students can add up the overall class data. In addition to prodding students to think about the multiplication rule and addition rule, this is also an excellent exercise for pointing out the effect of sample size, by asking students to compare their small-group data to the overall class data. 48

Testing your probability intuition: The two-child puzzle
Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws Testing your probability intuition: The two-child puzzle Working in pairs or small groups, consider the following brainteaser: A family has two children: a newborn and a toddler. At least one of the children is a boy. What is the probability that the other child is a girl? Discuss with your classmates, and be prepared to explain your reasoning to the rest of the class. INSTRUCTOR NOTES: The two-child question is a famous probability puzzle with a notoriously non-intuitive answer. Students will want to answer 1/2, but in fact the answer is 2/3. This can be an excellent discussion question for prodding students to really think about probability, and in particular, the addition rule for calculating outcomes that can be produced in multiple ways. If students are confused about the answer, have them write out the four equally probable possibilities, which are: male toddler and female newborn female toddler and male newborn male toddler and male newborn female toddler and female newborn See if students agree that the above four combinations are all equally probable. If they can agree on that, then point out that if you know one child is a boy, one of those four possibilities is ruled out (girl-girl), leaving 3 remaining possibilities that are all equally probable. In two of those three cases, the other child is a girl. Thus, if you know that one child is a boy, there is a 2/3 probability that the other child is a girl. The key to understanding this puzzle (and the key to understanding the frequency of heterozygotes in monohybrid crosses) is that there are two different ways to produce a boy-girl combination: girl first and boy second; or boy first and girl second. If students still object, this is an excellent opportunity to have them gather empirical data. If the class is large enough, poll students about their own families. Make four columns on the board: M-M, M-F, F-M, F-F (indicating: sex of first child - sex of second child). Then ask all students who come from two-child families to come up to the board and put a tick mark in the appropriate column for their family. Tally up the results and calculate the observed frequency of: [one boy and one girl]/[at least one boy]. If the class is not large enough to gather sufficient data, ask students to query their friends about the friends’ families and return with more data. With a large enough data set, results will approach 2/3. 49

Concept 8.1 Genes Are Particulate and Are Inherited According to Mendel’s Laws
A family has two children, at least one of which is a boy. What is the probability that the other is a girl? a. 1/4 b. 1/3 c. 1/2 d. 2/3 e. 3/4 Answer: d INSTRUCTOR NOTES: See notes on previous slide for explanation. 50

3/8 walnut : 1/8 pea : 1/8 single
Apply the concept p. 154 In the genetic cross AaBbCcDdEE x AaBBCcDdEe where all the genes are unlinked, what fractions of offspring will be heterozygous for all of these genes? In a plant species, two alleles control flower color, which can be yellow, blue, or white. Crosses of these plants produce the offspring provided on the next slide. What will be the phenotypes of the offspring and the ratios among them from a cross of blue x blue? In chickens, when the dominant alleles of the genes for rose comb (R) and pea comb (A) are present together, the bird has a walnut comb. Birds that are homozygous recessive for both genes have a single comb. A rose-combed bird mated with a walnut-combed bird, and the offspring were: 3/8 walnut : 1/8 pea : 1/8 single What were the genotypes of the parents?

Apply the Concept, Ch. 8, p. 154

Concept 8.2 Alleles and Genes Interact to Produce Phenotypes
Different alleles arise through mutation— rare, stable, inherited changes in the genetic material. The wild type is the allele present in most of the population. Other alleles are mutant alleles. A gene with a wild-type allele that is present less than 99 percent of the time is called polymorphic. VIDEO 8.1 Mutant alleles in Drosophila melanogaster See Chapter 15

A given gene may have more than two alleles. Multiple alleles increase the number of possible phenotypes and may show a hierarchy of dominance in heterozygotes. One example is the coat color in rabbits.

Figure 8.9 Multiple Alleles for Coat Color in Rabbits

Some alleles are neither dominant nor recessive. A heterozygote has an intermediate phenotype in incomplete dominance. Red + white snapdragons = pink in F1 Red and white colors reappear in F2 as well as pink.

Figure 8.10 Incomplete Dominance Follows Mendel’s Laws (Part 1)

Figure 8.10 Incomplete Dominance Follows Mendel’s Laws (Part 2)

Codominance—two alleles of a gene produce phenotypes that are both present in the heterozygote. Example: ABO blood group system has three alleles of the gene: IA, IB, and IO. See Chapter 31

Figure 8.11 ABO Blood Reactions Are Important in Transfusions

Epistasis—phenotypic expression of one gene is influenced by another gene Example: Coat color in Labrador retrievers: Allele B (black) dominant to b (brown) Allele E (pigment deposition) is dominant to e (no pigment deposition—yellow)

Figure 8.12 Genes Interact Epistatically

Hybrid vigor, or heterosis, is a cross between two different true-breeding homozygotes. It can result in offspring with stronger, larger phenotypes. Most complex phenotypes are determined by multiple genes. Quantitative traits conferred by multiple genes are measured, rather than assessed qualitatively.

In-Text Art, Ch. 8, p. 154

Genotype and environment interact to determine the phenotype of an organism. Two parameters describe the effects: Penetrance is the proportion of individuals with a certain genotype that show the phenotype. Expressivity is the degree to which genotype is expressed in an individual. APPLY THE CONCEPT Alleles and genes interact to produce phenotypes

Concept 8.2 Allleles and Genes Interact To Produce Phenotypes
The search for a true-breeding palomino A golden horse with a white mane and tail is known as a palomino. For many years the genetics of this color was a mystery. Suppose you’ve been hired by a horse breeder who wants to produce a line of true-breeding palomino horses—palomino horses that, when crossed with each other, always produce palomino foals. The breeder has 12 palomino stallions that are not related to each other. He tells you that every one of the twelve stallions, when bred to palomino mares, has produced a mix of three kinds of offspring: Some palomino foals Some chestnut (red) foals Some foals that are a very pale cream color (“cremello”) The breeder wants your advice on how to produce more palomino foals. 66

The search for a true-breeding palomino (continued) Working in pairs or small groups: 1. Develop a hypothesis about the genetic relationship between chestnut, palomino, and cremello. [Hint: Only one locus is involved.] How many alleles do you think are involved? Which (if any) are dominant? 2. What breeding experiments could you do to test your hypothesis? (For example, what crosses could you do, or what data would you request from the breeder?) What results would falsify your hypothesis, and what results would support it? 3. Do you think is it possible to develop a strain of horses that are true-breeding palominos? If not, what recommendation will you make to the breeder? Answers: This is a case of incomplete dominance. Palomino horses are heterozygotes at a locus that has two common alleles that are co-dominant, the chestnut allele (usually designated C), and the cremello allele (Cr). The best data to request from the breeder would be the percentage of foals of each color. If palominos are heterozygotes, a palomino × palomino cross should produce 25% homozygous chestnuts, 50% heterozygous palominos, and 25% homozygous cremellos. Palominos can never be true-breeding, because they are heterozygotes. However, the breeder can produce 100% palomino foals if he breeds chestnuts to cremellos; this is exactly what horse breeders do today. (Cremello stallions fetch high stud fees for exactly this reason.) INSTRUCTOR NOTES: The cremello allele, Cr, causes dilution of red (phaeomelanin) pigment as it is deposited in the hair follicles. Thus, CC homozygotes are strongly colored red chestnuts; but C/Cr heterozygotes have partial dilution of the red pigment, resulting in the golden palomino body color. (For unknown reasons, the pigment tends to be diluted even more in the mane and tail, causing a cream-colored mane and tail but a golden body.) Cr/Cr homozygotes have extreme dilution of hair pigment, resulting in a very pale cream color all over the body as well as in the mane and tail. Palomino is also a good example of epistasis, since the C locus interacts with the E locus discussed earlier. The colors discussed above - chestnut, palomino, and cremello - occur only if horses have an ee (chestnut) genotype at the E locus. But if the horse carries at least one E allele at the E locus, the colors produced are entirely different. (If students ask: E/CC horses are black. E/CCr horses are “smoky black,” very like true black but with faint golden highlights; and E/CrCr horses are a very pale ashy color called “smoky cream.”) 67

The search for a true-breeding palomino (continued) Later, the horse breeder returns to you with the following data: When one of his palomino stallions was crossed repeatedly with several palomino mares, the resulting offspring were: 42 palomino foals 23 cremello foals 18 chestnut foals Do these results support your hypothesis? If not, discuss how you could modify your hypothesis. Answer: These data are in (roughly) the 1:2:1 ratio typical of a monohybrid cross with incomplete dominance. 68

Which of the following statements is most likely to be true about the relationship between chestnut, palomino and cremello? a. These colors are caused by three different alleles. b. This is a case of incomplete dominance. c. Palominos are homozygotes. d. Both a and c e. Both b and c Answer: b (This is a case of incomplete dominance. Two alleles, not three, are involved, and palominos are heterozygotes.) 69

Concept 8.3 Genes Are Carried on Chromosomes
Genes are sequences of DNA that reside at a particular site on a chromosome—a locus (plural loci). The genetic linkage of genes on a single chromosome can alter their patterns of inheritance. ANIMATED TUTORIAL 8.2 Alleles That Do Not Assort Independently See Figures 8.3 and 8.6

Genetic linkage was discovered by Thomas Hunt Morgan and students at Columbia University using the fruit fly Drosophila melanogaster. Much genetic research has been done with Drosophila, which is considered a model organism because of its size, ease of breeding, and short generation time.

Some crosses performed with Drosophila did not yield expected ratios according to the law of independent assortment. Instead, some genes for body color and wing shape were inherited together. Morgan theorized that the two loci were linked on the same chromosome and could not assort independently.

Figure 8.13 Some Alleles Do Not Assort Independently (Part 1)

Figure 8.13 Some Alleles Do Not Assort Independently (Part 2)

Some offspring showed recombinant phenotypes, different from their parents. Genes may recombine during prophase I of meiosis by crossing over. Homologous chromosomes exchange corresponding segments. The exchange involves two chromatids of four in the tetrad—both chromatids become recombinant (each ends up with genes from both parents). LINK Review the discussion of crossing over in Concept 7.4

Figure 8.14 Crossing Over Results in Genetic Recombination (Part 1)

Recombinant offspring phenotypes (non- parental) appear in recombinant frequencies. To determine the recombinant frequencies, divide the number of recombinant offspring by the total number of offspring. Recombinant frequencies are greater for loci that are farther apart on the chromosome.

Figure 8.15 Recombination Frequencies (Part 1)

Figure 8.15 Recombination Frequencies (Part 2)

Recombinant frequencies can be used to make genetic maps showing the arrangement of genes along a chromosome. Recombinant frequencies are converted to map units corresponding to distances between genes.

In-Text Art, Ch. 8, p. 157 (1)

The fruit fly genome has four pairs of chromosomes—three pairs are similar in size, called autosomes. The fourth pair are of different size, the sex chromosomes. Many genes on the X chromosome are not present on the Y chromosome.

In-Text Art, Ch. 8, p. 157 (2)

Genes on sex chromosomes don’t follow Mendelian patterns. The Y chromosome carries few genes; the X chromosome carries many. Thus, males have only one copy of these genes—hemizygous. APPLY THE CONCEPT Genes are carried on chromosomes

Sex-linked inheritance—inheritance of a gene that is carried on a sex chromosome One example is the eye color in Drosophila.

Figure 8.16 A Gene for Eye Color Is Carried on the Drosophila X Chromosome (Part 1)

Figure 8.16 A Gene for Eye Color Is Carried on the Drosophila X Chromosome (Part 2)

X-linked recessive phenotypes: They appear much more often in males than females. A male with the mutation can only pass it on to daughters. Daughters who receive one X-linked mutation are heterozygous carriers. Mutant phenotype can skip a generation if it passes from a male to his daughter (normal) and then to her son.

Figure 8.17 Red–Green Color Blindness Is Carried on the Human X Chromosome

Besides the genes in the nucleus, mitochondria and plastids contain small numbers of genes. Mitochondria and plastids are inherited only from the mother. The inheritance of organelles and their genes is non-Mendelian and is called maternal or cytoplasmic inheritance.

Figure 8.18 Cytoplasmic Inheritance

Genes are carried on chromosomes
Apply the concept p. 158 Genes are carried on chromosomes The pedigree on the next slide shows the inheritance pattern of a rare mutant phenotype in humans, congenital cataract (filled in symbols). Are cataracts inherited as an autosomal dominant? Autosomal recessive? Sex-linked dominant? Sex-linked recessive? Person #5 in the second generation marries a man who does not have cataracts. Two of their four children, a boy and a girl, develop cataracts. What is the probability that their next child will be a girl with cataracts?

Apply the Concept, Ch. 8, p. 158

The case of the 800 Martian children Martians can have four legs or eight legs, and can have blue ears or spotted ears. You are interviewing a pair of Martians who are visiting the United Nations. They are a married couple: an eight- legged male with spotted ears, and a four-legged female with blue ears. They tell you that they have 800 children, and they insist on describing every one of the 800 children in great detail. Listening attentively, you notice the following pattern: 311 children have eight legs and spotted ears 329 children have four legs and blue ears 76 children have eight legs and blue ears 84 children have four legs and spotted ears What sort of inheritance pattern is this? Explain to a friend the physical basis of this inheritance pattern. Answer: This is a case of linkage. The leg-number gene and the ear-color gene are on the same chromosome. 96

What sort of inheritance pattern is this? a. Incomplete dominance b. Linkage c. A dihybrid cross involving two loci d. Multiple alleles, all at one locus e. I don’t know. Answer: b [NOTE TO THE INSTRUCTOR: It can be useful to include an “I don't know” choice with clickers, because it can help you discover how many students really haven’t understood the concept at all. Use of this option may depend on whether you assign participation-only points or performance points (or some combination) to clicker questions in your course. If you only assign participation points, it may be useful to leave the “I don't know” choice in the question, as it gives students a penalty-free way of indicating that more time may be needed on this concept.] 97

What is the recombination frequency? a. 0.2% b. 20% c. 25% d. 75% e. 80% Answer: b (160 recombinant offspring / 800 total offspring = 20%) INSTRUCTOR NOTES: Students who pick C divided by the number of offspring that had parental phenotypes, but they should have divided by the total number of offspring (parental + recombinant phenotypes). Students who pick E calculated the percentage of parental phenotypes instead of the percentage of recombinant phenotypes. Students who pick A need a reminder that 0.2% is not the same thing as 0.2. Rather, 0.2% = 98

Concept 8.4 Prokaryotes Can Exchange Genetic Material
Bacteria exchange genes by bacterial conjugation. Sex pilus is a projection that initiates contact between bacterial cells. Conjugation tube is a cytoplasmic bridge that forms between cells. The donor chromosome fragments and some material enters the recipient cell. See Chapters 4 and 7 LINK The evolutionary consequences of lateral gene transfer are discussed in Concepts 15.6 and 19.1

Figure 8.19 Bacterial Conjugation and Recombination (Part 1)

Figure 8.19 Bacterial Conjugation and Recombination (Part 2)

Bacteria have plasmids—small circular DNA molecules—besides the main chromosome. Genes on the plasmids are in categories: Metabolic tasks, breaking down hydrocarbons Involved in conjugation Antibiotic resistance

Plasmids can move between the cells during conjugation. They can: Replicate independently of the main chromosome Add their genes to the recipient cell’s genome

Figure 8.20 Gene Transfer by Plasmids

Answer to Opening Question
In hemophilia, the mutant gene for factor VIII, the clotting factor, is carried on the X chromosome. The affected males inherited their single X chromosome from their mothers—if the mutated form of the gene was present, they would develop the disease. Daughters would inherit a normal X chromosome as well and would not express the recessive trait, though could be carriers.

Figure 8.21 Sex Linkage in Royal Families of Europe

Inheritance, Genes, and Chromosomes

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Inheritance, Genes, and Chromosomes

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