Overview of meiosis: Two consecutive cell divisions *meiosis I *meiosis II. AKA: Reduction-Division

Transcription

1 Overview of meiosis: Two consecutive cell divisions *meiosis I *meiosis II AKA: Reduction-Division

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3 Life Cycle

4 Mistakes in Meiosis Aneuploidy an abnormal set of chromosomes. Down syndrome, which involves an extra copy of chromosome 21.

5 How Do Mistakes Occur? The failure of homologous chromosomes to separate is called nondisjunction *trisomy & monosomy

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7 Genetics Heredity transmission of traits from one generation to the next Variation asexual vs. sexual (offspring differ from parents) Genetics study of heredity and variation Gene a discrete unit of hereditary info consisting of a specific nucleotide sequence unique combo of genes

8 Two main hypotheses on how traits were transmitted: *blending inheritance *particulate inheritance The father of transmission genetics: Gregor Johann Mendel *Made new flowers *Noticed patterns of inheritance. *Deduced that some particle is transmitted from parent to offspring.

9 Mendel tracked heritable characters for three generations -Example: *P 1 -parental generation *F 1 first filial generation *F 2 second filial generation F 2 P 1 X Tall Dwarf F 1 all Tall Tall

10 Mendel s Hypotheses allele for tall plants 1. Alternative versions of genes (alleles) account for variation in inherited characters *gene unit of inheritance which occupies A specific chromosomal location (locus) *allele alternative forms of a single gene homologous pair locus for stem length allele for dwarf plants 2. For each character, an organism inherits two alleles, one from each parent

11 3. If two alleles differ, one is dominant, the other recessive 4. The two alleles for each character segregate (separate) during gamete production. P 1 F 1 all Tall X Tall Dwarf DD dd Tall Dd Mendel s Law of Segregation

12 Punnett Square predicts the results of a genetic cross between individuals of known genotype P: Gamete formation: Tall DD X Dwarf dd D D d d *genotype *phenotype *Homozygous *Heterozygous

13 Mendel s Law of Independent Assortment *What happened when he looked at two characters? If they segregate together: If they segregate independently:

14 Dihybrid cross- A genetic cross between two individuals involving two characters Example: Punnett square and the law of independent assortment: yellow, round GGWW P 1 X green, wrinkled ggww gw gw GW GW GW GW gw F 1 gw All yellow, round

15 Punnett square and the law of independent assortment: GW Gw gw gw F 1 All yellow, round X F 1 All yellow, round GW Gw gw gw GGWW GGWw GgWW GGWw GGww Ggww GgWW ggww ggww Ggww ggww ggww F 2 9/16 yellow, round 3/16 yellow, wrinkled 9:3:3:1 Phenotypic ratio; Genotypic ratio as follows: 1/16 GGWW, 2/16 GGWw, 2/16 GgWW, 4/16 1/16 GGww, 2/16 Ggww 3/16 green, round 1/16 ggww, 2/16 ggww 1/16 green, wrinkled 1/16 ggww

16 Mendelian inheritance is based on probability Example- coin toss *1/2 chance landing heads *Each toss is an independent event *Coin toss, just like the distribution of alleles into gametes *The rule of multiplication determines the chance that two or more independent events will occur together ½x ½= ¼

17 Sample problem Albinism in humans is inherited as a simple recessive trait. Determine the genotypes of the parents and offspring for the following families. When two alternative genotypes are possible, list both. (A) Two non albino (normal) parents have five children, four normal and one albino. (B) A normal male and an albino female have six children, all normal. 1) establish gene symbols: 2) Establish: genotype phenotype AA normal A=normal (not albino) Aa normal a=albino aa albino Move on to part (A): Parents are both phenotypically normal, genotypes could be EITHER AA or Aa, an albino phenotype could only result from an aa genotype. *One a had to come from the mother and one a had to come from the father, so, the parents must be genotypically Aa. A a A AA Aa a Aa aa Answer to (A): Genotype of the parents = Aa Genotype of normal children = AA or Aa Genotype of albino child = aa

18 2. (B) A normal male and an albino female have six children, all normal. 1) The female is phenotypically albino; genotype can only be aa 2) The male is phenotypically normal; genotype can be AA or Aa 3) Since all children are normal one might assume the male to be AA a a A Aa Aa F1: A Aa Aa Genotype: all Aa Phenotype: all normal 4) BUT male COULD also be Aa! a a A Aa Aa F1: a aa aa Genotype: ½ Aa, ½ aa or a 1:1 ratio Phenotype: ½ normal, ½ albino or a 1:1 ratio *IF the father was genotypically Aa, then what is the likelihood (chance or probability) of this couple having 6 normal children? Recall the product law! ½x ½x ½x ½x ½x ½ = 1/64