Pedigree Chart Symbols Male Female Person with trait

Sample Pedigree

Dominant Trait Recessive Trait

Chapter 15 The Chromosomal Basis of Inheritance

First Experimental Evidence to connect Mendelism to the chromosome • Thomas Morgan (1910) • Used fruit flies as model organism • Allowed the first tracing of traits to specific chromosomes.

Fruit Fly • Drosophila melanogaster • 3 pairs of Autosomes • 1 pair of sex chromosomes

Morgan Observed: • A male fly with a mutation for white eyes.

Morgan crossed • The white eye male with a wild type (red eyed) female. • Wild type is most common – NOT always DOMINANT • Male ww x Female w+w+

The F 1 offspring: • All had red eyes. • This suggests that white eyes is a _____? • Recessive. • F 1= w+w • What is the predicted phenotypic ratio for the F 2 generation?

F 1 X F 1 = F 2 • Expected F 2 ratio - 3: 1 of red: white • He got this ratio, however, all of the white eyed flies were MALE. • Therefore, the eye color trait appeared to be linked to sex.

Morgan discovered: • Sex linked traits. • Genetic traits whose gene are located on a sex chromosome

Fruit Fly Chromosomes • Female • XX Male XY • Presence of Y chromosome determines the sex • Just like in humans!

Morgan Discovered • There are many genes, but only a few chromosomes. • Therefore, each chromosome must carry a number of genes together as a “package”.

Sex-Linked Problem • A man with hemophilia (a recessive, sexlinked, x-chromosome condition) has a daughter of normal phenotype. She marries a man who is normal for the trait. • A. What is the probability that a daughter of this mating will be a hemophiliac? • B. That a son will be a hemophiliac? • C. If the couple has four sons, what is the probability that all four will be born with hemophilia?

• Original Man - Xh. Y • Daughter - must get the dad’s X chromosome XHXh (normal phenotype, so she’s a carrier) • Daughter’s husband XHY (normal phenotype) • A. daughter must get XH from the dad. 0% (50% carrier, 50% homo dom. ) • B. son must get Y from dad. 50% chance to be hemophiliac • C. ½ x ½ x ½ = 1/16

Multiple Genes • Parents are two true-breeding pea plants • Parent 1 Yellow, round Seeds (YYRR) • Parent 2 Green, wrinkled seeds (yyrr) These 2 genes are on different chromosomes (all problems so far have assumed this)

• F 1: Yy. Rr x Yy. Rr • What are the predicted phenotypic ratios of the offspring? • ¾ yellow ¾ round • ¼ green ¼ wrinkled ¼ (green) x ¼ (wrinkled) = 1/16 green, wrinkled 9: 3: 3: 1 phenotypic ratio

Linked Genes • Traits that are located on the same chromosome. • Result: • Failure of Mendel's Law of Independent Assortment. • Ratios are different from the expected

Example: • Body Color - gray dominant/wild • b+ - Gray • b - black • Wing Type - normal dominant/wild • vg+ - normal • vg – vestigial (short)

Example b+b vg+vg X bb vgvg Predict the phenotypic ratio of the offspring

Show at board b+b x bb vg+vg x vgvg ½ gray ½ black ½ normal ½ vestigial --------------------------¼ gray normal, ¼ gray vestigial, ¼ black normal, ¼ black vestigial 1: 1: 1: 1 phenotypic ratio

Conclusion • Most offspring had the parental phenotype. Both genes are on the same chromosome. • bbvgvg parent can only pass on b vg • b+b vg+vg can pass on b+ vg+ or b vg

b+b vg+vg - Chromosomes (linked genes)

Crossing-Over • Breaks up linkages and creates new ones. • Recombinant offspring formed that doesn't match the parental types. • Higher recombinant frequency (nonparental types) = genes further apart on chromosome

If Genes are Linked: • Independent Assortment of traits fails. • Linkage may be “strong” or “weak”. • Strong Linkage means that 2 alleles are often inherited together.

• Degree of strength related to how close the traits are on the chromosome.

Genetic Maps • Constructed from crossing-over frequencies. • 1 map unit = 1% recombination frequency. • Can use recombination rates to ‘map’ chromosomes.

• Comment - only good for genes that are within 50 map units of each other. Why? • Over 50% gives the same phenotypic ratios as genes on separate chromosomes

Genetic Maps • Have been constructed for many traits in fruit flies, humans and other organisms.

Barr Body • Inactive X chromosome observed in the nucleus. • Way of determining genetic sex without doing a karyotype.

Lyon Hypothesis • Which X inactivated is random. • Inactivation happens early in embryo development by adding CH 3 groups to the DNA. • Result - body cells are a mosaic of X types.

Examples • Calico Cats. • Human examples are known such as a sweat gland disorder.

Calico Cats • XB = black fur • XO = orange fur • Calico is heterozygous, XB XO.

Chromosomal Alterations • Changes in number. • Changes in structure.

Number Alterations • Aneuploidy - too many or too few chromosomes, but not a whole “set” change. • Polyploidy - changes in whole “sets” of chromosomes.

Nondisjunction • When chromosomes fail to separate during meiosis • Result – cells have too many or too few chromosomes which is known as aneuploidy

Meiosis I vs Meiosis II • Meiosis I – all 4 cells are abnormal • Meiosis II – only 2 cells are abnormal

Aneuploidy • Caused by nondisjunction, the failure of a pair of chromosomes to separate during meiosis.

Types • Monosomy: 2 N - 1 • Trisomy: 2 N + 1

Question? • Why is trisomy more common than monosomy? • Fetus can survive an extra copy of a chromosome, but being hemizygous is usually fatal.

Structure Alterations • • Deletions Duplications Inversions Translocations

Translocations

Result • Loss of genetic information. • Position effects: a gene's expression is influenced by its location to other genes.

Summary • Know about linkage and crossing-over. • Sex chromosomes and their pattern of inheritance.

Summary • Be able to work genetics problems for this chapter.