Semester 1, Day 8 Mendelian Genetics

Agenda  Review for Quiz  Turn in HW  Quiz on Meiosis  Lecture on Mendelian Genetics  Work/Reading Time

Review  Purpose of Meiosis  Meiosis I  How many cells do you start with? How many chromosomes do those cells have?  How many cells do you end with? How many chromosomes do those cells have?  Meiosis II  How many cells do you start with? How many chromosomes do those cells have?  How many cells do you end with? How many chromosomes do those cells have?  Spermatogenesis Products and Oogenesis Products  Draw crossing over. What phase does it occur in?  Label a diagram of Meiosis  What is a gamete? Is it haploid or diploid?  What is a zygote? Is it haploid or diploid?  How much DNA do you get from each parent?  Describe fertilization

Review

Homework Due  Cornell Notes on Section 10.2 (Meiosis)  10.2 Section Assessment:  #1-4  Chapter 10 Assessment:  #1, 4-6, 9, 10, 13, 15, 18, 25, 26-28

Quiz  Meiosis

Sex Chromosomes  Karyotype: Chromosomes of an organism organized into homologous pairs from largest to smallest.  In humans, first 22 pairs always have homologous pairs of chromosomes with identical centromere position, lengths, and traits (may have different alleles).  23 rd pair may or may not have two similar chromosomes  Sex Chromosomes Image Source:

Sex Chromosomes  Probability that child will be male or female:  50:50 chance the child will be male:female Mom 100% chance of passing on an X = 1.0 Dad 50% chance of passing on a Y = % chance of passing on an X = 0.5 Child Probability #1 X Y 1.0 x 0.5 = 0.5 (50%) Child Probability #2 X 1.0 x 0.5 = 0.5 (50%)

Sex Chromosomes  Easier Method: Punnett Square X X Y XX Female X Y Male Child Female: X X = 2 /4 = 50% Male: X Y = 2 / 4 = 50%

Mendelian Genetics  Gregor Mendel: Austrian monk & plant breeder. Considered the father of genetics due to his findings in breeding pea plants in Image Source: commons.wikimedia.org

Mendelian Genetics  Inheritance / Heredity: Passing of traits to the next generation.  Genetics: Study of heredity. Image Source:

Mendelian Genetics  Mendel’s Work  Mendel determined there must be 2 forms of a trait in pea plants  alleles!  Same Trait: Color  Different Alleles: Yellow or Green  Also said the trait seen in F 1 = dominant, while masked (hidden) trait in F 1 = recessive.  Dominant Allele: Yellow  Recessive Allele: Green  Dominant allele is labeled with a capital letter and the recessive allele is labeled with the corresponding lower case letter.  Yellow (Dominant): Y  Green (Recessive): y Generation Parental (P) (Pure-Breeding) First Filial Generation (F 1 ) Second Filial Generation (F 2 ) x YellowGreen All Yellow 6022 Yellow: 2001 Green = 3:1

Mendelian Genetics  Genotype: the organism’s allele pair  Zygosity: the similarity of alleles for a trait  Phenotype: observable characteristic of allele pair  Homozygous Dominant Genotype  Phenotype  Heterozygous Genotype  Phenotype  Homozygous Recessive Genotype  Phenotype Example Y = yellow pea color y = green pea color GenotypeZygosityPhenotype Y Homozygous Dominant Yellow Y y Heterozygous Yellow yy Homozygous Recessive Green “Same”“Dominant Alleles” “Same”“Recessive Alleles” “Different Alleles” Y Y = “Dominant” “Dominant” = Yellow “Yellow” “Yellow” Y y = “Dominant” “Recessive” = Yellow “Yellow” “Green” (Dominant allele masks recessive allele) y y = “Recessive” “Recessive” = Green “Green” “Green” (No dominant allele to mask recessive)

Mendel’s Laws  1. Mendel’s Law of Segregation  The two alleles for each trait separate during meiosis. Y Y Y y y y Grows to Plant Gamete Formation Yellow Pea (2n) 100% = 1.0 Gametes (n) x 2 Grows to Plant Gamete Formation Green Pea (2n) 100% = 1.0 Gametes (n) x 2 Y y Fertilization Y y F 1 Generation Zygote Heterozygous Yellow (2n) (n) 1.0 x 1.0 = 1.0 = 100%

Mendel’s Laws  1. Mendel’s Law of Segregation (cont.) Y y Y y Y y Grows to Plant Gamete Formation Yellow Pea (2n) Gametes (n) x 2 Grows to Plant Gamete Formation Green Pea (2n) 50% = 0.5 Gametes (n) x 2 y y Fertilization y F 2 Generation Zygote Homozygous Green (2n) (n) 50% = x 0.5 = 0.25 = 25%

 Monohybrid Punnett Square: Genotype Probabilities Y Y = 1 / 4 = 25% (homozygous dominant) Y y = 2 / 4 = 50% (heterozygous) y y = 1 / 4 = 25% (homozygous recessive) Phenotype Probabilities Yellow = 3 / 4 = 75% Green = 1 / 4 = 25% Y Y y y Yy Female Y y Male Child Y y Law of Segregation!!! Y y Law of Segregation!!! Mendel’s Laws “one trait”

Mendel’s Laws  2. Mendel’s Law of Independent Assortment  Alleles are randomly distributed during gamete formation Example 2 Traits: Color & Shape Different Alleles for Color: Yellow & Green Different Alleles for Shape: Round & Wrinkled Yellow = Y (dom.) Green = y (rec.) Round = R (dom.) Wrinkled = r (rec.) Same Trait (Color) Same Trait (Shape) Mom’s Body Cell (2n): YyRr YyRr Heterozygous Yellow Heterozygous Round YyRr Gamete Formation (FOIL): Y y R r YR yR Yr yr Each gamete should ALWAYS have one allele for EACH trait!

Mendel’s Laws  Dihybrid Punnett Square “two traits” YyRr YR yR Yr yr Mom’s Genotype yyRR yR Dad’s Genotype yR YRYryRyrMom yR Dad YyRRYyRryyRRyyRr Law of Independent Assortment!!! Genotype Probabilities YyRR = 1 / 4 = 25% (heterozygous color, homozygous dominant shape) YyRr = 1 / 4 = 25% (heterozygous color, heterozygous shape) yyRR = 1 / 4 = 25% (homozygous recessive color, homozygous dominant shape) yyRr = 1 / 4 = 25% (homozygous recessive color, heterozygous shape) Phenotype Probabilities Yellow & Round = 2 / 4 = 50% Green & Round = 2 / 4 = 50% Yellow & Wrinkled = 0 / 4 = 0% Green & Wrinkled = 0 / 4 = 0%

Reading/Work Time  Section 10.1  Cornell Notes  Section Assessment: #1-6  Chapter 10 Assessment  2, 3, 7, 8, 11, 12, 14, 16, 17, 20-24