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Chapter 23 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 23.1 Mendel’s Laws • Genetics explains the process of inheritance and why there are variations between offspring from one generation to the next. • Virtually every culture in history has attempted to explain observed inheritance patterns. • Understanding inheritance patterns is especially important in agriculture, animal husbandry, and medicine. 2 Gregor Mendel • Understanding of genetics is based on work of Gregor Mendel • Investigated inheritance at the organism level (1860s) • Concluded that plants transmit distinct factors to offspring – Now called genes, found on chromosomes Figure 23.1 3 Gregor Mendel • In peas and humans, chromosomes come in pairs called homologous chromosomes. • One member of the pair is inherited from the mother, and the other is inherited from the father. • Homologous pairs have certain characteristics. – Both members have same length and centromere location. – Both carry similar types of genes. – Alternate forms of a gene for a trait are called alleles. – Alleles are always on the same spot, or locus. 4 Homologous Chromosomes Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. alleles at a gene locus Figure 23.2 G g R r S s t T 5 Gregor Mendel • The Law of Segregation – Mendel had no knowledge of chromosomes, but he decided, based on his observations, that the following were true. • Each pea plant has two factors for each trait. • The factors segregate (separate) during the formation of gametes. • Each gamete contains only one factor from each pair of factors. • Fertilization gives each new individual two factors for each trait. 6 The Inheritance of a Single Trait • Phenotype: an individual’s actual appearance – May include physical characteristic, microscopic or metabolic characteristics • Genotype: alleles carried by the chromosomes that are responsible for a given trait 7 The Inheritance of a Single Trait • In diploid organisms, a pair of homologous chromosomes contains two alleles for each trait. – One allele is on each member of the homologous pair. • Mendel used letters to indicate each allele. – A capital letter symbolizes a dominant allele. – A lowercase letter symbolizes a recessive allele. – Dominant refers to the allele that will mask the expression of the alternate (recessive) allele, when both are present in a given organism. 8 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Trait Characteristics Dominant F2Results Recessive Dominant Recessive Ratio Stem length Tall Short 787 277 2.84:1 Pod shape Inflated Constricted 882 299 2.95:1 Seed shape Round Wrinkled 5,474 1,850 2.96:1 Seed color Yellow Green 6,022 2,001 3.01:1 Flower position Axial Terminal 651 207 3.14:1 Flower color Purple White 705 224 3.15:1 Pod color Green Yellow 428 152 2.82:1 14,949 5,010 2.98:1 Totals: 9 The Inheritance of a Single Trait • Hairline characteristic – The dominant allele of widow’s peak is assigned (W). – The recessive allele of straight hairline is assigned (w). – In the case of a single trait, there are three possible combinations of the two alleles. • WW, Ww, ww – If the two alleles are the same (homo), the individual is said to be homozygous. – If the two alleles are different (hetero), the individual is said to be heterozygous. 10 Two Human Characters Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. a. Widow’s peak: WW or Ww b. Straight hair line: ww Figure 23.3 c. Short fingers: SS or Ss d. Long fingers: ss a: © HFPA, 63rd Golden Globe Awards; b: © Dynamic Graphics/PictureQuest RF; c–d(both): © The McGraw-Hill Companies, Inc. Bob Coyle, photographer 11 Gamete Formation • The genotype has two alleles for each trait, whereas the gamete has one allele for a trait. • Mendel’s law of segregation states: – During meiosis, homologous chromosomes separate so only one member of each pair is in a gamete. – Therefore, only one allele exists for each trait in a haploid gamete. • When solving genetics problems, – No two letters in a gamete can be the same letter of the alphabet. – If genotype is Ww, then gametes from this individual will contain either a W or a w. 12 One-Trait Cross • A sample cross – A homozygous man with a widow’s peak reproduces with a woman with a straight hairline. • What type of hairline will their children have? Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. widow’s peak WW Parents meiosis gametes Offspring straight hairline ww W w Ww widow’s peak 13 One-Trait Cross • Steps used to solve a genetics cross 1. Determine the genotype of each parent. 2. List the possible gametes from each parent. 3. Combine all possible gametes. 4. Determine the genotypes and phenotypes of all offspring. 14 One-Trait Cross Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. • Solving a cross when two individuals are both Ww Ww ♀ Ww Key W = Widow’s peak w = Straight hairline oocytes ♀ W W w WW Ww sperm – Perform a monohybrid cross – A Punnett square is useful to solve this problem Parents Widow’s peak Straight hairline Phenotypic Ratio w Ww ww 3 1 Widow’s peak Straight hairline Offspring Figure 23.4 15 One-Trait Cross • After determining the genotype and phenotypes of the offspring from the Punnett square analysis, a tally is performed. – The genotypic ratio is: 1 WW: 2Ww: 1ww or 1:2:1 – The pheotypic ratio is: 3 Widow’s peak: 1 straight hairline or 3:1 – Three individuals have a widow’s peak because they carry at least one dominant allele 16 One-Trait Crosses and Probability • The chance of two or more independent events occurring together is the product of their chance of occurring separately (product rule of probability). • In the cross Ww X Ww, what is the chance of obtaining either a W or a w from a parent? – Chance of W = ½ and the chance of w = ½ • Probability of having these genotypes is as follows. 1.Chance of WW= ½ X ½ = ¼ 2.Chance of Ww = ½ X ½ = ¼ 3.Chance of wW= ½ X ½ = ¼ 4.Chance of ww = ½ X ½ = ¼ 17 One-Trait Crosses and Probability • The sum rule of probability, states that the chance of an event can occur in more than one way is the sum of the individual chances. • To calculate the chance of an offspring having widow’s peak, add the chances of WW, Ww or wW, or ww. • Chance of widow’s peak: ¼ + ¼ + ¼ = ¾ or 75% – (WW) + (Ww) + (wW) = 3 out of 4 total possibilities • Chance of straight hairline: ¼ or 25% – (ww) = 1 out of 4 total possibilities 18 The One-Trait Testcross • One cannot determine by observation if an individual expressing the dominant trait is homzygous or heterozygous. – In a test cross, an organism with the dominant phenotype is crossed with a recessive individual. – If there are any offspring produced with the recessive phenotype, then the dominant parent must be heterozygous. 19 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Parents ♀ ww WW oocytes ♀ w Ww Ww sperm W w W Figure 23.5a Ww Ww Key W = Widow’s peak w = Straight hairline Widow’s peak Straight hairline Phenotypic Ratio All Widow’s peak Offspring 20 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. ♀ Parents Ww ww oocytes ♀ w Ww Ww sperm W w Key W = Widow’s peak w = Straight hairline Widow’s peak Straight hairline Phenotypic Ratio w ww ww 1 1 Widow’s peak Straight hairline Offspring Figure 23.5b 21 One-Trait Testcross Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. ♀ ww WW oocytes ♀ w Ww Ww sperm W w a. Ww Offspring Figure 23.5 Ww Ww Key W = Widow’s peak w = Straight hairline Widow’s peak Straight hairline Phenotypic Ratio W ♀ Parents All ww oocytes ♀ sperm Parents W b. w Ww Ww Phenotypic Ratio w Widow’s peak w Key W = Widow’s peak w = Straight hairline Widow’s peak Straight hairline ww ww 1 1 Widow’s peak Straight hairline Offspring 22 The Inheritance of Two Traits • Homologous chromosomes are inherited as a pair, one member from each parent. – Each has similar length and similar genes. • During meiosis, each gamete receives one member of each pair of homologues. • The homologues separate independently. • It does not matter which member of a pair goes into which gamete. • We can assume that the alleles for two genes are on these homologues in the following examples. 23 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. one pair S Cell has two pairs of homologues. W Allele Key s W = Widow’s peak w = Straight hairline S = Short fingers s = Long fingers w one pair Figure 23.6 24 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. one pair S Cell has two pairs of homologues. W Allele Key s w either S Ss W W w Figure 23.6 W = Widow’s peak w = Straight hairline S = Short fingers s = Long fingers or s one pair S Ss s w W w W MEIOSIS I w 25 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. one pair S Cell has two pairs of homologues. W Allele Key s w either S S S Ss s W W w w W = Widow’s peak w = Straight hairline S = Short fingers s = Long fingers or one pair S Ss s w W w W MEIOSIS I s s S S s s w w W W MEIOSIS II W Figure 23.6 W w w 26 The Inheritance of Two Traits Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. one pair Figure 23.6 S Cell has two pairs of homologues. W Allele Key W = Widow’s peak w = Straight hairline S = Short fingers s = Long fingers s w either S S Ss s W W w w or one pair S Ss s w W w W MEIOSIS I S s s S S s w w W s MEIOSIS II W W S w s S W W SW w s w S w sw S w s w Sw W s W W sW 27 Independent Assortment • Mendel had no knowledge of meiosis, but realized his results were attainable only if the sperm and egg contains every possible combination of factors. • This gave rise to the Law of Independent Assortment. – Each pair of factors assorts independently (without regard to how the others separate). – All possible combinations of factors can occur in the gametes. 28 Two-Trait Crosses • A two-trait (dihybrid) cross is used to test the law of independent assortment. • Example cross – A person homozygous for widow’s peak and short fingers (WWSS) – A person who has a straight hairline and long fingers (wwss) – The cross is WWSS X wwss 29 Two-Trait Crosses • According to the law of segregation, the gametes for WWSS parent must be WS. • The gametes for the wwss parent must be ws. • The offspring will all have the genotype WwSs. • All offspring will have same phenotype (widow’s peak and short fingers). • This phenotype is called a dihybrid, because the individual is heterozygous for both hairline and fingers. 30 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. ♀ P generation WWSS wwss Allele Key W = Widow’s peak w = Straight hairline S = Short fingers s = Long fingers Figure 23.7 31 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. P generation WWSS P gametes ♀ wwss ws WS Allele Key W = Widow’s peak w = Straight hairline S = Short fingers s = Long fingers Figure 23.7 32 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. ♀ P generation WWSS P gametes wwss ws WS Allele Key W = Widow’s peak w = Straight hairline S = Short fingers s = Long fingers F1 generation WwSs Figure 23.7 33 Two-Trait Crosses • In the next generation, the dihybrid (WwSs) reproduces with another dihybrid; what gametes are possible? – Law of segregation states that each gamete gets only one letter of each kind. – Law of independent assortment states that all combinations are possible. • WS, Ws, wS, ws 34 Two-Trait Crosses (Dihybrid Cross) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. ♀ P generation WWSS wwss ws WS P gametes F1 generation WwSs oocytes ♀ F1 gametes WS Ws wS ws WWSS WWSs WwSS WwSs WWSs WWss WwSs Wwss wwSS wwSs wwSs wwSs WS sperm F2 generation Ws wS WwSS WwSs ws WwSs Wwss Offspring Allele Key Figure 23.7 W = Widow’s peak w = Straight hairline S = Short fingers s = Long fingers 9 3 3 1 Phenotypic Ratio Widow’s peak, short fingers Widow’s peak, long fingers Straight hairline, short fingers Straight hairline, long fingers 35 Two-Trait Crosses • WwSs (X) WwSs – Phenotypic Ratio • 9 widow’s peak, short fingers • 3 widow’s peak, long fingers • 3 straight hairline, short fingers • 1 straight hairline, long fingers • A 9:3:3:1 phenotypic ratio is always expected for a dihybrid cross when simple dominance is present. 36 Two-Trait Crosses and Probability • Probability Laws – – – – Probability of widow’s peak = ¾ Probability of short fingers= ¾ Probability of straight hairline= ¼ Probability of long fingers= ¼ • Product Rule – – – – Probability of widow’s peak and short fingers = ¾ X ¾ = 9/16 Probability of widow’s peak and long fingers = ¾ X ¼ = 3/16 Probability of straight hairline and short fingers = ¼ X ¾ = 3/16 Probability of straight hairline and long fingers = ¼ X ¼ = 1/16 37 Two-Trait Testcross • It is impossible to tell by inspection whether an individual expressing the dominant allele for two traits is homozygous dominant or heterozygous in regard to these traits. • A testcross with an individual homozygous recessive for both traits is necessary. • A recessive phenotype is used because it has a known phenotype. 38 Two-Trait Testcross • If a man who is homozygous dominant for widow’s peak and short fingers reproduces with a women who is homozygous recessive for both traits, then all his children will have dominant phenotypes. • If a man is heterozygous for both traits, then each child has a 25% chance of showing either one or both recessive traits. – 4 phenotypes at a ratio of 1:1:1:1 39 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Parents WwSs wwss Key W = Widow’s peak w = Straight hairline S = Short fingers s = Long fingers Figure 23.8 40 Two-Trait Testcross Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Parents WwSs wwss eggs ♀ ws WS WwSs sperm Ws Wwss wS wwSs ws wwss Offspring Key W = Widow’s peak w = Straight hairline S = Short fingers s = Long fingers Figure 23.8 Phenotypes 9 3 3 1 Widow’s peak, short fingers Widow’s peak, long fingers Straight hairline, short fingers Straight hairline, long fingers 41 23.2 Pedigree Analysis and Genetic Disorders • Many traits and disorders in humans, and other organisms, are genetic in origin. • Most follow Mendel’s laws. • Such traits are controlled by a single allele pair on the autosomal chromosomes-any chromosome other than (X or Y). 42 Patterns of Inheritance • Pedigree – A chart of family’s history with regard to a particular genetic trait – Used to determine whether inherited condition is due to autosomal dominant or an autosomal recessive allele 43 Pedigrees for Autosomal Disorders Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Pattern I Key = affected = unaffected • In this pattern, the child is affected, but neither parent is. • This can happen only if the disorder is recessive and the parents are heterozygotes. • Notice that the parents are carriers because they are unaffected but are capable of having a child with the genetic disorder. 44 Autosomal Recessive Pedigree Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. I ♀ II A? aa A? Aa * III Aa A? Relatives Aa Aa A? IV aa aa Key A? Autosomal recessive disorders • Most affected children have unaffected parents. A? aa = affected Aa = carrier (unaffected) A A = unaffected A? = unaffected (one allele unknown) • Heterozygotes (Aa) have an unaffected phenotype. • Two affected parents will always have affected children. • Affected individuals with homozygous unaffected mates will have unaffected children. • Close relatives who reproduce are more likely to have affected children. • Both males and females are affected with equal frequency. Figure 23.9 45 Autosomal Recessive Disorders • Requires the inheritance of two recessive alleles to display the disorder – Tay-Sachs disease • Usually occurs among Jewish people in the US • Results from a lack of the enzyme hexosaminidase A (Hex A) • The substrate, glycosphingolipid, is stored in lysosomes • Lysosomes build up in brain cells, resulting in progressive deterioration of psychomotor functions 46 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. normal lysosome defective lysosomes Malfunctioning lysosomes in Tay-Sachs disease. Courtesy Robert D. Terry/University of California, San Diego School of Medicine Figure 23.10a 47 Autosomal Recessive Disorders • Cystic Fibrosis (CF) – Most common lethal genetic disorder among Caucasians in the United States – Chloride ions fail to pass through a plasma membrane channel protein in cells – Causes abnormally thick mucus in bronchial tubes and pancreatic ducts 48 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Cl Cl H2O H2O Cl Cl Cl H2O cytoplasm Chloride ions and water are trapped inside cell. Defective chloride ion channel does not allow chloride ions to pass through. Lumen of respiratory tract fills with thick, sticky mucus. Malfunctioning channel protein in cystic fibrosis. Figure 23.10b 49 Genetic Disorders Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. normal lysosome defective lysosomes Cl Cl H2O H2O Cl Cl Cl H2O cytoplasm Chloride ions and water are trapped inside cell. Defective chloride ion channel does not allow chloride ions to pass through. Lumen of respiratory tract fills with thick, sticky mucus. a. Malfunctioning lysosomes in Tay-Sachs disease. b. Malfunctioning channel protein in cystic fibrosis. a: Courtesy Robert D. Terry/University of California, San Diego School of Medicine Figure 23.10 50 Autosomal Recessive Disorders • Phenylketonuria (PKU) – Affected in dividuals lack the enzyme needed for the normal metabolism of the amino acid phenylalanine – Infants will develop normally if placed on a low phenylalanine diet – Severe mental retardation will result otherwise 51 Autosomal Recessive Disorders • Sickle cell disease – Red blood cells are sickle-shaped due to abnormal hemoglobin. – Red blood cells differ from normal hemoglobin by one amino acid. – Immediate symptoms include clogging of blood vessels and breakdown of red blood cells. – Other symptoms include poor circulation, anemia, and low infection resistance. – Internal hemorrhaging leads to further complications. 52 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. normal red blood cells sickled red blood cell Abnormally-shaped red blood cells in sickle cell disease. Figure 23.10 © Eye of Science/Photo Researchers, Inc.; 53 Autosomal Dominant Disorders Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Pattern II Key = affected = unaffected • The child is unaffected, but the parents are both affected. • This is possible if the condition is autosomal dominant and the parents are heterozygotes. • This also illustrates that when both parents are unaffected, all their children are unaffected. – Neither parent has a dominant gene that passes the condition on. 54 Autosomal Dominant Pedigree Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. ♀ I Aa Aa * II III Aa aa Aa Aa A? aa Autosomal dominant disorders Figure 23.11 aa aa aa aa aa aa Key AA = affected Aa = affected A? = affected (one allele unknown) aa = unaffected • Affected children will usually have an affected parent. • Heterozygotes (Aa) are affected. • Two affected parents can produce an unaffected child. • Two unaffected parents will not have affected children. • Both males and females are affected with equal frequency. 55 Autosomal Dominant Disorders • Marfan syndrome – Caused by a defect in an elastic connective tissue protein called fibrillin – Fibrillin is present in the eye lens, bones of limbs, fingers, ribs and aorta – Symptoms-dislocated lens, long limbs, and fingers and caved in chest – Aorta wall is weak-risk for aneurism 56 Autosomal Dominant Disorders • Huntington disease – Neurological disorder that leads to progressive degeneration of brain cells – Caused by a mutated copy of the gene for a protein called huntingtin – Most patients appear normal until they are of middle age and have already had children – Test now exists to detect presence of gene 57 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Many neurons in normal brain. Courtesy Dr. Hemachandra Reddy, The Neurological Science Institute, Oregon Health & Science University Figure 23.12 58 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Loss of neurons in Huntington brain. Figure 23.12 Courtesy Dr. Hemachandra Reddy, The Neurological Science Institute, Oregon Health & Science University 59 23.3 Beyond Simple Inheritance Patterns • Incomplete Dominance – Occurs when the heterozygote has an intermediate phenotype between the two homozygotes – Curly-haired person reproduces with a straighthaired person • Their children have wavy hair – Two wavy-haired persons reproduce with an expected phenotypic ratio of: 1:2:1 • 1 curly-hair: 2 wavy-hair: 1 straight-hair 60 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Heterozygous Parents (H1 H2) H1 H2 ♀ H1 H2 oocytes sperm ♀ H1 Key Straight hair Wavy hair Curly hair H2 H1 H1 H1 H1 H2 H2 H1 H2 H2 H2 Phenotypic Ratio 1 2 1 Offspring Figure 23.13 61 Incomplete Dominance Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A person with straight hair (H1H1) Heterozygous Parents (H1 H2) H1 H2 ♀ Straight hair Wavy hair Curly hair H1 H2 H1 H1 H1 H2 sperm H1 H2 Figure 23.13 H1 H2 Key H1 H2 oocytes ♀ A person with naturally curly hair (H2H2) H2 H2 Phenotypic Ratio 1 2 1 Offspring (man): © Vol. 88/PhotoDisc; (woman): © Larry Williams/Corbis 62 Incomplete Dominance and Codominance • Codominance – Occurs when alleles are equally expressed in a heterozygote • Ex: blood type AB – Red blood cells have both Type A and Type B surface antigens. – We assume both alleles code for a product, and we observe the result of both products being present. 63 Incompletely dominant Disorders • Symptom severity in familial hypercholesterolemia (FH) parallels the number of LDL-cholesterol receptor proteins in the plasma membrane. • A heterozygote has a mild form of disorder. Plasma cholesterol (milligrams/deciliter) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Homozygote 1000 900 800 700 600 Heterozygote 500 400 300 Normal cholesterol deposits 200 100 0 © Mediscan/Medical-On-Line Figure 23.14 64 Multiple Allele Inheritance • When a trait is controlled by multiple alleles, the gene exists in several allelic forms. – Each person has only two of the possible alleles. – An example is demonstrated by the three alleles responsible for the ABO blood type group. 65 ABO Blood Types • ABO Blood Types – – – – IA = A antigens on red blood cells IB = B antigens on red blood cells i = neither A nor B antigens on red blood cells Both IA and IB are dominant over i, IA and IB are codominant Phenotype A B AB O Genotype IAIA or IAi IBIB or IBi IAIB ii 66 ABO Blood Types • ABO Blood Types – Both IA and IB are dominant over I; IA and IB are codominant. • Rh factor – Rh factor is inherited separately from ABO blood types. – An Rh-positive person has the Rh antigen on their red blood cells. – There are multiple recessive alleles for Rh-negative phenotype. • All recessive to the Rh-positive allele 67 Inheritance of Blood Types Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. ♀ Parents IBi IAi oocytes ♀ i IAIB IBi sperm IB IA i IAi ii Key Blood type A Blood type B Blood type AB Blood type O Phenotypic Ratio 1 1 1 1 Offspring Figure 23.15 68 Polygenic Inheritance – Polygenic inheritance occurs when a trait is governed by two or more genes (sets of alleles). – Dominant alleles have a quantitative effect on the phenotype, and these effects are additive. – The result is a continuous variation of phenotypes. – Skin color is an example. 69 Skin Color • Skin color is the result of pigmentation produced by melanocyte cells. • Over 100 different genes influence skin color. • We use three pairs of alleles as a simplified example, (Aa, Bb, and Cc). • Each dominant allele (capital letter) contributes to pigmentation. • When a very dark person reproduces with a very light person, the children have medium-brown skin. 70 Skin Color • When two individuals with the genotype AaBbCc reproduce with one another their offspring may range in skin color from very dark to very light. • The phenotype distribution follows a bell-shaped curve (characteristic of a polygenic trait). – Implies that few individuals have the extreme phenotypes – Most have the phenotype that lies in the middle 71 Polygenic Inheritance Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. AaBbCc aaBbCC AAbbCc AabbCC AABbcc aaBBCc AaBBcc aaBBCC AAbbCC AABBcc AaBbCC AaBBCc AABbCc 0 1 2 3 4 Frequency aabbcc Aabbcc aaBbcc aabbCc AaBbcc AabbCc aaBbCc AAbbcc aaBBcc aabbCC AaBBCC AABbCC AABBCc 5 AABBCC 6 Number of dominant alleles Figure 23.16 72 23.4 Environmental Influences • Environmental factors such as nutrition or temperature can influence the expression of genetic traits. • Polygenic traits are especially influenced by environment. – Examples • Height is influenced by nutrition. • Primrose is influenced by temperature. – Plants have white flowers grown above 32°C. – Plants have red flowers grown at 24°C. 73 23.4 Environmental Influences • Environmental factors can influence the expression of genetic traits. – In the case of height, differences in nutrition are one of the factors that bring about a bell-shaped curve. Figure 23.17 74 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Number of Men most are this height few 62 short 64 few 66 68 Height in Inches 70 72 74 tall 75 23.4 Environmental Influences • Siamese cats and Himalayan rabbits are darker in color at the ears, nose, paws and tail. • Himalayan rabbit are known to be homozygous for the ch allele, involved in producing melanin. • The enzyme coded for by this gene is active only at low temperatures, and that explains why the black color is limited to the extremities, where body heat is lost to the environment. 76 Coat Color in Himalayan Rabbits Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. © Jane Burton/Bruce Coleman, Inc. Temperature can also affect the phenotypes of plants and animals. Figure 23.18 77