Lecture - Observable Patterns Of Inheritance

Lecture - Observable Patterns Of Inheritance - Observable...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Observable Patterns of Inheritance Inheritance Earlobe Variation Whether a person has attached or Whether detached earlobes depends on a single gene gene Attached earlobes: two copies of the Attached recessive allele for this gene recessive Detached earlobes: either one or two Detached copies of the dominant allele copies Early Ideas about Heredity People knew that sperm and eggs People transmitted information about traits transmitted Blending theory Problem: Would expect variation to disappear Variation in traits persists Gregor Mendel Gregor Strong background Strong in plant breeding and mathematics mathematics Using pea plants, Using found indirect but observable evidence of how parents transmit genes to offspring offspring Comments: Gene Comments: Mendel never used the term, gene, gene although he clearly made a distinction between, genotypes and phenotypes. between, The term, gene, was coined by Wilhelm The Johannsen (1909) based on the term, Pangenisis which Hugo de Vries (1889) Pangenisis used to describe Mendel’s concept of inherited units that determine phenotypes. inherited Genes Genes Units of information about specific traits Passed from parents to offspring Each has a specific location (locus) on a Each chromosome chromosome Alleles Alleles Different molecular forms of a gene Different Arise by mutation Dominant allele masks a recessive Dominant allele that is paired with it allele Allele Combinations Allele Homozygous Homozygous having two identical alleles at a locus AA or aa aa Heterozygous Heterozygous having two different alleles at a locus Aa Genetic Terms Genetic A pair of homologous chromosomes Figure 11.4 Page 179 A gene locus A pair of alleles Three pairs of genes Figure 11.4 Page 179 Genotype & Phenotype Genotype Genotype refers to particular genes an Genotype individual carries individual Phenotype refers to an individual’s Phenotype observable traits observable Cannot always determine genotype by Cannot observing phenotype observing Parental generation P mates to produce First-generation offspring F1 mate to produce Second-generation offspring F2 Monohybrid Crosses Monohybrid Experimental intercross between Experimental two F1 heterozygotes AA X aa Aa (F1 monohybrids) Aa X Aa ? Mendel’s Monohybrid Cross Results Cross 5,474 round 6,022 yellow 299 wrinkled 428 green Figure 11.5 Page 180 2,001 green 882 inflated F2 plants showed dominant-torecessive ratio that averaged 3:1 1,850 wrinkled 152 yellow 705 purple 224 white 651 long stem 207 at tip 787 tall 277 dwarf Probability Probability The chance that each outcome of a given The event will occur is proportional to the number of ways that event can be reached number Monohybrid Cross Cross Illustrated True-breeding homozygous recessive parent plant F1 PHENOTYPES aa True-breeding homozygous dominant a parent plant Aa Aa Aa a A AA Aa Aa Aa A Aa Aa An F1 plant self-fertilizes and produces gametes: F2 PHENOTYPES Aa AA A Aa Aa aa a A AA Aa Figure 11.7 Page 181 a Aa aa Monohybrid Cross #1 Monohybrid Long hair is dominant to short hair in Long guinea pigs. Show the results of a cross between a homozygous short haired male and a pure long haired female. and Monohybrid Cross #1 Monohybrid What do we know? Both parents are homologous. Both Short haired male hh Long haired female HH (“H” gene is Long dominant) dominant) hh x HH Draw the Punnett Square #1 Punnett Square #1 hh x HH Meiosis 2n -> n H h Hh Hh H Ova Hh zygotes h Hh Hh Hh Hh sperm All of the F1’s are hybrids, or heterogenous and they are long haired. Monohybrid #2 Monohybrid What is the probability that a cross What between two guinea pigs both heterogeneous for long would produce a short haired guinea pig? short Monohybrid #2 Monohybrid Hh x Hh Draw the Punnett square H h H h Complete the problem (p = ¼ or 25%) Monohybrid #3 Monohybrid In Holstein cattle the spotting of the coat is In due to a recessive allele while the solid colored coat is controlled by a dominant allele. What types of offspring might be produced by a cross between two spotted animals? animals? Monohybrid #3 Monohybrid Cross: spotted cow x spotted cow Cross: ss x ss Cross: (where S = solid and s = spotted) Is it possible to have a solid calf? Monohybrid #4 Monohybrid In lemurs brown eyes are dominant over In blue eyes. If a heterozygous brown eyed female mates with a blue eyed male, what would be the expected genotypic and phenotypic ratios of their offspring? phenotypic Monohybrid #4 Monohybrid Cross: Brown eyed heterozygous female Cross: and a blue eyed male (homozygous) and B = brown and b = blue Bb x bb = 2 Bb and 2 bb Genotypic ratio Bb:bb 1:1 Phenotypic ratio Brown: blue 1:1 Monohybrid Problem Set Monohybrid Form cooperative groups of 2 to 4 and Solve the problems in the Monohybrid Solve Problem Set. Problem We’ll check the answers in class. Mendel’s Theory of Segregation of An individual inherits a unit of information An (allele) about a trait from each parent (allele) During gamete formation, the alleles During segregate from each other Test Cross Test Individual that shows dominant phenotype Individual is crossed with individual with recessive phenotype phenotype Examining offspring allows you to Examining determine the genotype of the dominant individual individual BB or Bb ? Both brown Punnett Squares of Test Crosses Test Homozygous recessive a a Homozygous recessive a a A Aa Aa A Aa Aa a aa aa A Aa Aa Two phenotypes 1:1 All dominant phenotype Dihybrid Cross Experimental cross between individuals that are homozygous for different versions of two traits two Dihybrid Cross: F1 Results purple flowers, tall TRUEBREEDING PARENTS: AABB GAMETES: AB white flowers, dwarf x aabb AB ab ab AaBb F1 HYBRID OFFSPRING: Figure 11.9 (1) Page 183 All purple-flowered, tall Dihybrid Cross: F2 Results Dihybrid AaBb X AaBb 1/4 AB 1/4 Ab 1/4 aB 1/4 AB 1/4 ab 1/16 AABB 1/16 AABb 1/16 AaBB 1/16 AABb 1/16 AAbb 1/16 AaBb 1/16 Aabb 1/4 aB 1/16 AaBB 1/16 AaBb 1/16 aaBB 1/16 aaBb 1/4 ab 1/16 AaBb 1/16 Aabb 1/16 aaBb 1/16 aabb 1/4 Ab 1/16 AaBb 9/16 purple-flowered, tall 3/16 purple-flowered, dwarf 3/16 white-flowered, tall 1/16 white-flowered, dwarf Figure 11.9(2) Page 183 Dihybrid Problem #1 Dihybrid A person is heterozygous for tonguerolling and homozygous recessive for rolling freckles. This person marries an individual who is heterozygous for both tonguewho rolling and freckles. Show all the possible rolling genotypes for their children. Tongue roller – Dominant and Non-tongue Tongue roller Recessive roller Freckles- Dominant and No Freckles Frecklesand Recessive Recessive Dihybrid Problem #1 Dihybrid (Tongue roller – no-freckles) x (Tongue (Tongue roller- freckles) rollerT = tongue rolling dominant tt = non-tongue rolling F = freckles dominant ff = no freckles Ttff x TtFf (the cross) Ttff produces Tf and tf gametes Tf tf TtFf produces TF, Tf, tF, and tf gametes TF Tf tF and gametes Dihybrid Problem #1 Dihybrid Ttff x TtFf (symbolizes the cross) TF Tf Tf TTFf TTff TtFf tf sperm Ttff tF tf TtfF Ova Ttff ttFf ttff zygotes Dihybrid Problem #1 Dihybrid Ttff x TtFf (symbolizes the cross) TF Tf Tf TTFf TTff TtFf Ttff tF tf TtfF Ova Ttff ttFf ttff zygotes tf sperm Rollerfrecked Rollerno freckles Phenotypic ratios: 3:3:1:1 Nonroller freckles Nonroller No freckles Independent Assortment Independent Mendel concluded that the two “units” for Mendel the first trait were to be assorted into gametes independently of the two “units” for the other trait for Members of each pair of homologous Members chromosomes are sorted into gametes at random during meiosis Independent Assortment Independent Metaphase I: A Aa B Bb OR a A b Aa a b bB B Metaphase II: A a a A A a a B Gametes: A B b b b b B B B A B A 1/4 AB b a b a 1/4 ab b A b A 1/4 Ab B a B a 1/4 aB Dihybrid Problem Set Dihybrid Form cooperative groups of 2 to 4 and Solve the problems in the Dihybrid Solve Problem Set. Problem We’ll check the answers in class. Tremendous Variation Tremendous Number of genotypes possible in Number offspring as a result of independent assortment and hybrid crossing is 2n (n is the number of gene loci at which the parents differ) at Impact of Mendel’s Work Impact Mendel presented his results in 1865 Paper received little notice Mendel discontinued his experiments in Mendel 1871 1871 Paper rediscovered in 1900 Paper Dominance Relations Complete dominance Incomplete dominance Incomplete Codominance Incomplete Dominance X Incomplete Homozygous Homozygous parent parent Dominance Dominance All F1 are heterozygous X Figure 11.10 Page 184 F2 shows three phenotypes in 1:2:1 ratio Codominance: ABO Blood Types Gene that controls ABO type codes for Gene enzyme that dictates structure of a glycolipid on blood cells glycolipid Two alleles (IA and IB) are codominant are when paired when Third allele (i) is recessive to others ABO Blood Type: ABO Allele Combinations Range of genotypes: I A IA IB IB or or IA i Blood types: I A IB IB i ii A AB B O Figure 11.11 Page 184 ABO and Transfusions ABO Recipient’s immune system will attack Recipient’s blood cells that have an unfamiliar glycolipid on surface glycolipid Type O is universal donor because it has Type neither type A nor type B glycolipid neither Pleiotropy Alleles at a single locus may have effects Alleles on two or more traits on Marfan syndrome - Mutation in gene for Marfan fibrillin affects skeleton, cardiovascular system, lungs, eyes, and skin Marfan Syndrome Marfan Campodactyly: Unexpected Phenotypes Effect of allele varies: Bent fingers on both hands Bent fingers on one hand No effect Many factors affect gene expression Continuous Variation Continuous A more or less continuous range of small more differences in a given trait among individuals individuals The greater the number of genes and The environmental factors that affect a trait, the more continuous the variation in versions of that trait versions Human Variation Human Some human traits occur as a few discrete Some types types Attached or detached earlobes Attached Many genetic disorders Other traits show continuous variation Height Weight Eye color IQ Polygenic Inheritance Polygenic Suppose height in humans is controlled by Suppose three sets of genes each on a different chromosome (independently assorted). Assume that each dominant allele contributes “1 unit of height” and a recessive allele produces only “½ unit of height”. height”. Two average heigth parents have the Two genotypes: genotypes: AaBbCc x AaBbCc Polygenic Inheritance Polygenic Each parent can produce 8 gametic gene Each combinations: combinations: ABC, ABc, AbC, aBC, Abc, aBc, abC, abc What size Punnett Square would we What need? need? ABC AB c Ab C aB C ab C aB c A bc ab c ABC AAB BCC AABB Cc AABb CC AaBB CC AaBb CC AaBB Cc AABb Cc AaBb Cc ABc AABB Cc AABB cc AABb Cc AaBB Cc AaBb Cc AaBB cc AABb cc AaBb cc AbC AABb CC AABb Cc AAbb CC AaBbC C AabbC C AaBb Cc AAbb Cc Aabb Cc AaBb CC aaBBC C aaBb CC aaBB Cc AaBb Cc aaBb Cc AaBb Cc Aabb CC aaBb CC aabb CC aaBb Cc AaBb cc aabb Cc AaBB Cc AaBB cc AaBb Cc aaBB Cc aaBb Cc aaBB cc AaBb cc aaBbc c AABb Cc AABb cc AAbb Cc AaBb Cc Aabb Cc AaBb cc AAbb cc Aabb cc AaBb Cc AaBb cc Aabb Cc aaBb Cc aabb Cc aaBbc c Aabb cc aabbcc aBC abC aBc Abc `abc AaBB CC AaBb CC AaBB Cc 6 units 5.5 units 5 units 4.5 units Polygenic inheritance produces a more or less continuous distribution of phenotypes. The more genes involved, the smoother the distribution. Parents AaBbCc 4.5 units each 4 units 3.5 units 3 units Roughly bell shaped (line of bell-shaped curve indicates continuous variation in population) Range of values for the trait Number of individuals with some value of the trait Number of individuals with some value of the trait Describing Continuous Variation Range of values for the trait Temperature Effects on Phenotype Rabbit is homozygous for Rabbit an allele that specifies a heat-sensitive version of an enzyme in melaninenzyme producing pathway Melanin is produced in Melanin cooler areas of body cooler Figure 11.18 Page 190 Environmental Effects on Plant Phenotype Phenotype Hydrangea macrophylla Action of gene responsible for floral Action color is influenced by soil acidity color Flower color ranges from pink to blue ...
View Full Document

This note was uploaded on 01/13/2012 for the course IS 3020 taught by Professor Staff during the Spring '08 term at Kennesaw.

Ask a homework question - tutors are online