Genetics Lecture 8, 9/17: Epistasis and Sex linked traits

• • (13:00ish) When we do epistasis problems we are going to work with a set of assumptions.

• • 1st assumption: All phenotype classes will be unique from each other. This allows you to clearly identify one from the other.
• • 2nd assumption: Assume that our traits are not linked and assort independently of each other. For simplicity we will label our alleles A, B, C.
• • 3rd assumption: When complete dominance occurs we will designate AA and Aa as A_
• • 4th assumption: Our crosses will be parental AABBxaabb. Therefore our F1 crosses are AaBbxAaBb
• • 5th assumption: We will focus our analysis on the F2 generation. How do our relationships between the genes yield the final F2 ratio.
• • Mice. Agouti (gray) is dominant A_ and black is recessive aa. Second locus in which the presence of the dominant allele gives coat color. B_ → gives coat color, bb → albino.
• • Based on our assumptions . . .
• o 9/16 A_B_ → Agouti
• o 3/16 A_bb → Albino
• o 3/16 aaB_ → Black
• o 1/16 aabb → albino
• o Final ratio is 9 agouti, 4 albino and 3 black.
• • A dominant allele at one locus masks the genotypes at the 2nd locus
• • EX. Fruit color in summer squash
• o A dominant A allele = white fruit color
• o If the A allele is recessive = aaB_ Yellow and aabb green
• o 9/16 A_B_ → 9 white
• o 3/16 A_bb → 3 white
• o 3/16 aaB_ → 3 yellow
• o 1/16 aabb → 1 green
• o final ratio is 12:3:1
• • EX. Fruit color in sweat peas. (27:45)
• o Cross two white pea plants and produce 9/16 purple and 7/16 white. Proposed that pea color is controlled by two genes. Proposed two parent plants: AAbbxaaBB
• • 9/16 A_B_ → purple .. 9/16
• • 3/16 A_bb → white
• • 3/16 aaB_ → white
• • 1/16 aabb → white .. 7/16
• • Propose that one dominant allele for each gene is necessary for purple color.
• • Sex Linked Traits (34:10)
• o Chromosomes → XX in female and XY in male. Autosomal chromosomes are everything else.
• o In males the X and Y chromosomes synapse. But the Y chromosome has very little genetic information. The male phenotype is a direct reflection of the X chromosome from Mom. For a female there are two copies of the X chromosome and as a result of more normal dominant/recessive relationships. The end result is that you have unique patterns of expression for these traits.
• o EX. (39:00) Eye Color in fruit flies
• • Wildtype eye color is red and mutant color is white. Gene for this trait is on the X chromosome.
• • Cross between wild type female x white eyed male. XwXw ¬x XwY
• • Females are all XwXw
• • Males are all XwY
• • All offspring in the F1 gen are wildtype
• • Cross the F1
•  XWXwx XWY
•  Females are all WT
•  Males ½ Wildtype and ½ white-eyed
• o One hallmark of a sexlinked trait is that there are different ratios for different sexes.
• o Another hallmark of an X-linked trait is that reversing the parental phenotypes alters the ratio.
• o White eyed female x red eyed male
• o XwXw ¬x XWY → cross it to get the F1
• • Females are all wildtype and males are all white
• • F2 XWXw ¬x XwY →
• • ½ females are red eyes.
• • ½ xwxw white eyes
• • Males ½ xWY redyes
• • 1/2XWY white eyes