Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 14 – Mendelian Genetics. Blending vs. Particulate Theory of Inheritance Traits of offspring were a “blend” of the parental traits. Infer  over.

Similar presentations


Presentation on theme: "Chapter 14 – Mendelian Genetics. Blending vs. Particulate Theory of Inheritance Traits of offspring were a “blend” of the parental traits. Infer  over."— Presentation transcript:

1 Chapter 14 – Mendelian Genetics

2 Blending vs. Particulate Theory of Inheritance Traits of offspring were a “blend” of the parental traits. Infer  over time, populations begin to look uniform and look alike Mendel observed that genes retain their separate identities Over time, populations do not become uniform Often traits that seemed to have “disappeared” would reappear in subsequent generations His observations lead to what’s now accepted as the Particulate Theory of inheritance

3 Vocabulary Review Trait: Characteristic of an organism Ex. Plant Height Allele: Different forms of a gene responsible for different traits Ex. T = Tall t = short Genotype = Genetic make-up of characteristics represented by 2 alleles in diploid organisms Phenotype = expressed trait, based on the genotype TT or Tt tt TallShort Homozygous = When both alleles of the genotype are the same Heterozygous = When both alleles of the genotype are different (Hybrid) TTtt Tt

4 P = Parental Generation F 1 = 1 st Filial Generation F 2 = 2 nd Filial Generation XTT tTt t XTt TTTTt t tt Genotypic Ratio of monohybrid cross 1:2:1 Phenoytypic Ratio of monohybrid cross: 3:1 Single-Factor Crosses Memorize these ratios!

5 Two Factor Crosses (2 traits) X Y = yellow, y = greenR = Round, r = wrinkled RrYy X Dihybrid Cross Memorize these ratios!

6 Test Crosses If a plant has a dominant phenotype, (for example yellow seeds) and we are unsure of its genotype (YY, or Yy), you can determine it’s genotype by crossing it with another with a recessive phenotype (green seeds) with the genotype yy. YY or Yy? X yy If F 1 = 50% Yellow, 50% green Then P = Yy (hybrid) Yy If F 1 = 100% Yellow Then P must be = YY Yyyy

7 Pedigree male female Mating couple Children/Siblings Shaded = trait being followed aa A = tongue roller a = can not roll tongue ? ? ? ?? aa Aa AA Can you figure out the rest of the genotypes on your own?

8 Other Pedigree Symbols Fraternal twins Identical twins Examples of connected symbols:

9 Unique Degrees of Dominance (exceptions to the rule) Incomplete Dominance: Dominant trait “blends” when combined with a recessive allele Notice how the genotypes are written… X CRCRCRCR CWCWCWCW CRCWCRCW

10 Unique Degrees of Dominance Co-Dominance: When there are multiple alleles that are dominant and are of equal strength - then both dominant alleles will be expressed when combined - a dominant allele will always mask a recessive allele. Notice how the genotypes are written… Example: Blood Type I A I B i I’m the dominant allele for Type A blood! I’m the dominant allele for Type B blood ! Don’t count me out just ‘cuz I’m recesssive. I’m type O allele! What’s expressed when we’re together? I A I A or I A i = type A bloodI A I B = type AB blood I B I B or I B i = type B blood i i = type O blood

11 Unique Gene Interactions Pleiotrophy: The ability of a gene to affect an organism in many phenoytypic ways Ex. Sickle Cell Anemia Blood Clumping Physical Weakness or Brain Damage

12 Unique Gene Interactions Polygenic Inheritance: when multiple genes have an added effect on a single phenotype (Opposite of Pleiotrophy) – ex. Skin color, height Notice the range in genotypes… aabbcc AABBCC

13 Unique Gene Interactions Epistasis: when a gene at one locus alters the expression of a gene at another locus Alleles for Fur Color: B = Black Fur b = brown fur C = Color c = albino BBCC, BBCc, BbCC, BbCc  bbCC, bbCc  BBcc, Bbcc, bbcc  Since cc genotype is albino, the alleles for fur color (B or b) are not expressed

14 Genetics is cool! But wait…sample probability problems to come!

15 How do you calculate probability? (Alternatives to Punnett Squares) Sample Problem Brown eyes are dominant over blue eyes. Parent A has blue eyes, while Parent B is heterozygous for brown eyes. What is the probability that they will have a child with blue eyes? Parent A (blue eyes) = bb Parent B (brown eyes) = Bb Process: Blue-eyed Child has to be bb Probability of parent A donating one “b” allele= 1 Probability of parent B donating the other “b” allele = ½ 1 X ½ = ½ (50% probability) Xbb BBb bbb Check your work! 50% probability for blue eyes Rule of Multiplication

16 Rule of Multiplication and Addition Sample Problem #2: In a cross between AaBbCc x Aabbcc, what is the probability that at least two of the three recessive traits is present in the offspring? Steps for solving: 1.Write out the genotypic possibilities 2.Use rule of X (multiply probabilities of each genotypic combination) 3.Use rule of + AAbbcc Aabbcc aaBbcc aabbCc aabbcc  AA (1/2) x bb (1/2) x cc (1/2) = 1/8  Aa (1/2) x bb (1/2) x cc (1/2) = 1/8  Aa (1/4) x Bb (1/2) x cc (1/2) = 1/16  Aa (1/4) x bb (1/2) x Cc (1/2) = 1/16  Aa (1/4) x bb (1/2) x cc (1/2) = 1/16 Sum of the fractions: 6/16 = 3/8


Download ppt "Chapter 14 – Mendelian Genetics. Blending vs. Particulate Theory of Inheritance Traits of offspring were a “blend” of the parental traits. Infer  over."

Similar presentations


Ads by Google