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 F1 offspring: All had red eyes. This suggests that white eyes is a _________? Recessive. F1= w + w What is the predicted phenotypic ratio for the F2 generation?

F1 X F1 = F2 Expected F2 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 Male XX 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, sex- linked, 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 - X h Y Daughter - must get the dad’s X chromosome X H X h (normal phenotype, so she’s a carrier) Daughter’s husband X H Y (normal phenotype) A. daughter must get X H 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 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)

F1: YyRr x YyRr 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 (non- parental 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 X B = black fur X O = orange fur Calico is heterozygous, X B X O.

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: 2N - 1 Trisomy: 2N + 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

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.