Genetics Chapter 8

Origins of Genetics Heredity Character Trait The passing of characters from parents to offspring Character Inherited characteristic (ex. flower color) Trait Single form of a characteristic (ex. purple flower)

Genes Control the expression of traits (i.e. flower color) Have two parts - alleles One allele is from Mom other from Dad.

Alleles Dominant - the expressed allele of the gene Capital letter Ex. P Recessive - the allele that was not expressed Lower case letter Ex. p Homozygous (Homo = same) Two alleles are the same Ex. PP or pp Heterozygous (Hetero = different) The alleles are different Ex. Pp

Modern Terms Genotype Phenotype The set of alleles for a Gene Ex. PP, Pp or pp Phenotype Physical appearance Ex. Purple or White Flowers The dominant allele determines phenotype Ex. Pp  Purple Flower or PP  Purple Flower If both alleles are recessive then the individual will express the recessive phenotype. Ex. pp  White Flower

Homozygous- genes are the same HH hh Heterozygous- genes are different Hh

Probability The likelihood that a specific event will occur Expressed in multiples ways: Fractions ½ Percentages 50% Decimals .5 Ratios 3:1 or 3 to 1 Probability = # of one kind of possible outcome Total # of all possible outcomes

Y Y y y Monohybrid Crosses Cross 1 character – Seed Color Yellow Seeds are dominant – Y Green Seeds are recessive – y Cross a Homozygous Dominant Yellow with a Green. Cross: TT X tt Genotype Ratio: 4Yy Phenotype Ratio: 4 Yellow Probabilities: Being Yellow: 4 out of 4 Being Green: 0 out of 4 Y Y y Y y Y y y Y y Y y

P p p p Monohybrid Crosses Cross 1 character – Flower Color Purple Flowers are dominant – P White Flowers are recessive – p Cross a Heterozygous with a homozygous recessive Cross: Pp X pp Genotype Ratio: 2 Pp; 2 pp Phenotype Ratio: 2 Purple 2 White Probabilities: Being Purple: 2 out of 4 Being White: P p p P p p p p P p p p

Incomplete Dominance Inheritance that is intermediate between two parents Ex. Snapdragon with red flowers are crossed with a snapdragon with white flowers  a snap dragon with pink flowers is produced. Neither the allele for red flowers nor the allele for white flowers is dominant.

R R W W Incomplete Dominance Cross: RR X WW R W R W R W R W Genotype for red flowers: RR Genotype for white flowers: WW, Genotype for pink flowers: RW. Cross a red with a white. Genotype Ratio: 4 RW Phenotype Ratio: 4 Pink Flowers Probabilities: Red Flower: 0 out of 4 White Flower Pink Flower: 4 out of 4 Cross: RR X WW R R W R W R W W R W R W

R W W W Incomplete Dominance Cross: RW X WW R W W W R W W W Genotype for red flowers: RR Genotype for white flowers: WW, Genotype for pink flowers: RW. Cross a pink with a white. Genotype Ratio: 2 RW; 2WW Phenotype Ratio: 2 Pink; 2 White Probabilities: Red Flower: 0 out of 4 White Flower 2 out of 4 Pink Flower: Cross: RW X WW R W W R W W W W R W W W

Codominance Case in which both forms of the character are displayed. Both traits are displayed Ex. The ABO blood groups Neither IA nor IB are dominant is dominant over the other. Both are dominant over i. When both IA and IB are present they are codominant and the individual is type AB

Blood Types Blood Typing IA IB i IA IB i O B AB A ii IBi IAi IBIB IAIB IAIA IA IB i

Blood Typing IAIA X ii IA IA i i IAi Cross a homozygous type A with a type O Homozygous type A: IAIA Type O: ii Genotype Ratio: 4 IAi Phenotype Ratio: 4 Type A Probabilities: Type A: 4 out of 4 Type O: 0 out of 4 IAIA X ii IA IA IAi i i

Blood Typing IAIB X ii IA IB i i IBi IAi Cross a type AB with a type O Homozygous type A: IAIB Type O: ii Genotype Ratio: 2 IAi; 2 IBi Phenotype Ratio: 2 Type A 2 Type B Probabilities: Type A: 2 out of 4 Type B: 2 out of 4 IAIB X ii IA IB IBi IAi i i

Sex-Linked Traits Body cell has 46 chromosomes - 44 autosomes, 2 sex chromosomes. Gamete (egg/sperm) has 23 chromosomes – 22 autosomes, 1 sex chromosome Sex-Linked Gene A gene located only on the X or Y Most are carried on the X chromosome Males have only one X chromosome Male who carries a recessive allele on the X chromosome will exhibit the sex-linked condition

Hemophilia (Heme: blood, philia: liking of) Condition that impairs the blood’s ability to clot A sex-linked trait Carriers and victims contain the recessive allele on their X-chromosomes Only females can be carriers  XHXh Because males only have one X chromosome Males CAN NOT be carriers only victims.

Sex Linked Traits Xh XH XH Y XHY XhY XHXH XHXh The normal gene: XH The hemophilia gene: Xh Cross a normal male with a carrier female Genotype Ratio: 1 XHXh: 1 XHXH: 1 XhY: 1XHY Phenotype Ratio: 1 norm. female: 1 carrier female: 1 norm. male: 1 hemophilic male Probabilities Normal Female: 1 out of 2 Carrier Female: Normal Male: Hemophilic male: XHY X XHXh Xh XH XHY XhY XHXH XHXh XH Y

Sex Linked Traits XH Xh Xh Y XhY XHY XhXh XHXh The normal gene: XH The hemophilia gene: Xh Cross a hemophilic male with a carrier female Genotype Ratio: 1 XHXh: 1 XhXh: 1 XHY: 1XhY Phenotype Ratio: 1 carrier female: 1 hemophilic female: 1 norm. male: 1 hemophilic male Probabilities Hemophilic Female: 1 out of 2 Carrier Female: Normal Male: Hemophilic male: XhY X XHXh XH Xh XhY XHY XhXh XHXh Xh Y

Genetic Disorders Sickle Cell Anemia Recessive disorder among African Americans Caused by a mutated allele that produces a defective hemoglobin protein. Hemoglobin in RBC’s bind and transport oxygen through the body

Genetic Disorders Cystic Fibrosis Fatal hereditary recessive disorder among Caucasians. Clogging of the airways and lungs with thick mucus and the blockage of ducts of the liver and pancreas. There is no known cure.

Genetic Disorders Albinism Recessive Body is unable to produce an enzyme necessary to make melanin. Melanin is a pigment that gives color to hair, skin and eyes

Pedigrees A family history that shows how a trait is inherited over several generations Helps to track down the carriers (heterozygotes) of recessive disorders. Reading Pedigrees A shaded box – affected male ………… A shaded circle - affected female………. A clear box - normal male………….. A clear circle - normal female………...

Pedigree Practice ? ? I II III IV A man and woman marry. They have five children, 2 girls and 3 boys. The mother is a carrier of hemophilia, an X-linked disorder. She passes the gene on to two of the boys who died in childhood and one of the daughters is also a carrier. Both daughters marry men without hemophilia and have 3 children (2 boys and a girl). The carrier daughter has one son with hemophilia. One of the non-carrier daughter’s sons marries a woman who is a carrier and they have twin daughters. What is the percent chance that each daughter will also be a carrier? I II III IV Legend: Hemophilic Male: ……… Hemophilic Female: …... Normal Female: ……… Normal Male: ……… ? ?

Pedigree Practice The non-carrier normal daughter in the 2nd generation must have a genotype of XHXH her normal husband must have the genotype XHY; therefore, it is not possible for any of their children to have hemophilia or be a carrier of the trait. XHXh XHY XHY XHXh XhY XhY XHY XHXH XHY XHXh XHY XHY XHXH XhY XHY XHXH Probability of XHXh? Probability of XHXh?

Pedigree Practice However, when their normal son (genotype XHY) marries a woman who is a carrier of the trait (genotype XHXh), then the probability that each of his twins will be a carrier is 1 out of 2 (50%). XH Xh XhY XHY XHXh XHXH XH Y XHXh XHY XHY XHXh XhY XhY XhY XHXH XHY XHXh XHY XHY XHXH XhY XHY XHXH Probability of XHXh? Probability of XHXh?