GENETIC DISORDERS 1: Single Gene Disorders * GENETIC DISORDERS 1: Single Gene Disorders * Take notes from all pages with an asterisk in the corner – you can just read the other pages without taking notes.

Autosomes vs. Sex Chromosomes * Autosomes vs. Sex Chromosomes All chromosomes have hundreds of different genes on them Some alleles cause severe disorders The sex chromosomes are the final pair of chromosomes (pair #23 in humans) Autosomes are the other 22 pairs A karyotype is a photo of the chromosomes in a cell

Human Karyotype All others are Autosomes!! Sex Chromo-somes

* Autosomal Disorders Genes for these disorders are on the autosomes They follow Mendel’s Laws Examples- Achondroplasia (dwarfism) Huntington’s Disease

Achondroplasia AKA, Dwarfism

“A” = without “chondro” = cartilage “plasia” = growth * Achondroplasia “A” = without “chondro” = cartilage “plasia” = growth Caused by a dominant allele Heterozygous individuals receive the dwarfism phenotype (Aa) Homozygous dominant (dwarf) individuals die (lethal allele) (AA) Homozygous recessive individuals grow to full height as adults (aa)

Achondroplasia Dorothy meets and marries a dwarf from the “Lollipop Guild” Can they have a normal child? If they can, what are the odds that they will?

Achondroplasia A a Aa aa A a Dorothy = Standard Adult = aa Lollipop = Dwarf = Aa A a Aa aa A a Phenotype ratio= 1:1 Genotype ratio= 1:1 Final Answer: Yes, they have a 50% chance of having a child that grows to standard height, and a 50% chance of having a dwarf.

Achondroplasia A member of the “Lullaby League” marries a member of the “Lollipop Guild” Can they have a child of normal height? If they can, what are the odds that they will?

Achondroplasia A a AA Aa aa A a Lullaby = Dwarf = Aa Lollipop = Dwarf = Aa A a AA Aa aa A a Phenotype ratio= 1:2:1 Genotype ratio= 1:2:1 Final Answer: Yes, they have a 25% chance of having a child that grows to standard height, a 50% chance of having a dwarf, and a 25% chance of having a child that will not survie.

* Huntington’s Disease Lethal autosomal disorder Dominant allele on the fourth chromosome – so if you get the gene, you get the disease  Degenerative disease of the central nervous system; causes involuntary dancelike movements Strikes at age 35- 45 Normal Brain Brains with Huntington’s Disease

Huntington’s Disease, aka Huntington’s Chorea (Why is it similar to the word choreography?) Normal Brain has small Ventricles (fluid-filled cavities) and large Striatum Diseased Brain has large Ventricles (fluid-filled cavities) and small Striatum

Huntington’s Disease H = Huntington’s h = normal Cross a heterozygous male with an unafflicted female. What are the chances their children will develop Huntington’s disease?

Huntington’s Disease H h Hh hh Heterozygous male (Hh) crossed with Unafflicted female (hh) H h Hh hh Their children have a 50/50 chance of developing Huntington’s Disease If this is a dominant lethal allele, why is it still in existence? Because the trait does not kill people until after they have lived long enough to have children! CAT scan of brain with Huntington’s Disease

* Sex- Linked Disorders These traits are controlled by alleles on the sex chromosomes The traits they control are NOT related to gender Both males and females can inherit these disorders, with modifications to Mendel’s Laws Not related to gender, but often more common in males – why? Examples- Hemophilia (bleeding disorder – blood won’t clot) Muscular dystrophy (muscles wear down over time) Color blindness (can’t tell red from green)

* Hemophilia Recessive allele on the X chromosome This disorder prevents blood from clotting – so they can easily bleed to death from a cut or a bruise 1: 10,000 ♂ versus 1 : 100,000,000 ♀

Hemophilia Genotypes & Phenotypes * Hemophilia Genotypes & Phenotypes X H = normal blood clotting X h = hemophilia X H X H = Normal Female X H X h = Carrier Female X h X h = Female Hemophiliac X HY = Normal Male X hY = Male Hemophiliac Notice that both the X and the Y chromosomes are shown in the genotypes! But why don’t the Ys have the h subscript like the Xs? Because the gene is only on the X chromosome, and is missing from the Y! Carrier = Does not have the trait, but can give it to their children (must be heterozygous)

Sex-Linked Punnett: Hemophilia * Sex-Linked Punnett: Hemophilia A heterozygous female marries a normal male. Describe the percent of sons and daughters with hemophilia. Mom X H X H X H X h X H X HY X hY X H X h Y Dad Final Answer 50% of the daughters are normal = 50% of the daughters are carriers = 50% of the sons are normal = 50% of the sons are hemophiliacs =

Muscular Dystrophy Recessive allele on the X chromosome 1: 10,000 ♂ versus 1 : 100,000,000 ♀ X D = normal X d = abnormal

* Muscular Dystrophy Affected individuals lack gene coding for muscle protein dystrophin Results in a progressive wasting of muscle mass. Always fatal. Sex-Linked Recessive Trait, so more common in males than females X D = normal X d = abnormal

Muscular Dystrophy X D = normal X d = abnormal A carrier female and a normal male have a child. What are the child’s chances for being normal?

* Color Blindness Recessive disorder on the X chromosome Affected individuals lack pigment needed for color vision Red/ green is the most common form 8% ♂ vs. 1 %♀ population XB= normal vision Xb= color blind

Color Blindness Test Patterns If you can see the number within each pattern, you are not colorblind.

Color Blindness Practice Punnett! * Color Blindness Practice Punnett! A carrier female marries a colorblind male. Describe the percents of sons and daughters with each kind of vision. Hints: XB= normal vision Xb= colorblind Y = male Use a punnett square showing both sex chromosomes of both parents, since this is a sex-linked trait. Give your final answers separately for sons & daughters. Show all work for full credit!

Color Blindness Practice Punnett! * Color Blindness Practice Punnett! A carrier female marries a colorblind male. Describe the percents of sons and daughters with each kind of vision. X B = X chromosome with Color Vision allele X b = X chromosome with Color Blindness allele Y = Y chromosome without either allele for color vision

Color Blindness Practice Punnett! * Color Blindness Practice Punnett! A carrier female marries a colorblind male. Describe the percents of sons and daughters with each kind of vision. X B = X chromosome with Color Vision allele X b = X chromosome with Color Blindness allele Y = Y chromosome without either allele for color vision Mother Father Phenotype Carrier Colorblind Genotype X B X b X bY Gametes X B or X b X b or Y

Color Blindness Practice Punnett! * Color Blindness Practice Punnett! A carrier female marries a colorblind male. Describe the percents of sons and daughters with each kind of vision. Mother Father Phenotype Carrier Colorblind Genotype X B X b X bY Gametes X B or X b X b or Y X B X b Y

Color Blindness Practice Punnett! * Color Blindness Practice Punnett! A carrier female marries a colorblind male. Describe the percents of sons and daughters with each kind of vision. X B X b X BX b XbX b Y X BY X bY

Color Blindness Practice Punnett! * Color Blindness Practice Punnett! A carrier female marries a colorblind male. Describe the percents of sons and daughters with each kind of vision. Final answer: 50% of males are normal, 50% of males are colorblind 50% of females are carriers, 50% of females are colorblind X B X b X BX b XbX b Y X BY X bY

So WHY are Sex-Linked Recessive Disorders more common in Males? * So WHY are Sex-Linked Recessive Disorders more common in Males? Since males only have one X chromosome, they only get one gene for sex-linked traits – this means that a recessive allele is ALWAYS expressed since there is no dominant allele to hide it. Since females have two X chromosomes, they get two alleles for these traits. In order for the trait to show up in their phenotype, they must inherit two of the disorder genes, which is much less likely.

YAY!! You are now a master of autosomal and sex- linked disorders!!!!