P. 280

Linkage MappingProblem 1: Mating: AaBb x aabbA = Long antennae a = Short antennae B = Green eyebrows, and b = Blue eyebrows.

 How do you tell your genes are linked?Prediction for 2000 offspring?

Problem 1 ResultsLong Green - 850 Long Blue - 150 Short Green - 150 Short Blue - 850

(Total # crossovers/total # offspring) x 100 = LMU (or cM).

Problem 2 – Autosomal MappingDrosophila:

Wing shape – wild/arcBody color – wild/black

Progeny:660 wild530 black490 arc712 arc black

Problem 3: Sex-linked crossSex linked:

eye color – wild type/whiteBody color – wild type/yellow

Male Progeny:Yellow white – 43White – 2146Yellow – 2302Wild - 22

Problems for additional practice:Example: p. 280p. 291, #4p. 291, #5Online textbook quiz questions for Ch. 15

Problem 4: Three point test crossCorn:

Growth habit – normal/lazy

Leaf appearance – normal/glossy

Endosperm – normal/sugary

Progeny:286 wild272 lazy glossy sugary40 glossy sugary33 lazy44 lazy sugary59 glossy2 lazy glossy4 sugary

Problem 5 - Imaginary CrittersD = Calm personality d = Dithery personality F = Five toed f = Four toed G = Smooth fur, and g = Grizzled fur.

Progeny:Calm, Five, Smooth – 616 Dithery, Five, Smooth - 5 Calm, Five, Grizzled – 104 Dithery, Five, Grizzled - 75 Calm, Four, Smooth – 78 Dithery, Four, Smooth - 100 Calm, Four, Grizzled – 2Dithery, Four, Grizzled - 620

Problem 6 - GroodiesGroodies are useful (but fictional) haploid

organisms that are pure genetic tools. A wild-type groody has a fat body, a long tail, and flagella. Mutant lines are known that have thin bodies, or are tailless, or do not have flagella. Groodies can mate with each other (although they are so shy that we do not know how) and produce recombinants. A wild-type groody mates with a thin-bodied groody lacking both tail and flagella. The 1000 baby groodies produced are classified as shown in the following illustration. Assign genotypes, and map the three genes. (Problem 18 from Burton S. Guttman.)

Assign genotypes, and map the three genes.

Problems for additional practice:p. 292, #10 & 11Online textbook quiz questions for Ch. 15

Constructing a Linkage Map

Problem 7 – Chromosomal Linkage MappingFour genes on chromosome: ABCD

Linkage data:A & B = 7 LMUA & C = 6 LMUA & D = 4 LMUB & C = 1 LMUB & D = 11 LMUC & D = 10 LMU

Problem 8 – Chromosomal Linkage MappingFour genes on chromosome: ABCD

Linkage data:A & B = 17 LMUA & C = 7 LMUA & D = 13 LMUB & C = 10 LMUB & D = 4 LMUC & D = 6 LMU

Problem 9 – Chromosomal Linkage MappingFive genes on chromosome: ABCDELinkage data:

A & B = 3 LMUA & C = 11 LMUA & D = 9 LMUA & E = 9 LMUB & C = 8 LMUB & D = 6 LMUB & E = 12 LMUC & D = 2 LMUC & E = 20 LMUD & E = 18 LMU

Problems for additional practice:p. 292, #8Online textbook quiz questions for Ch. 15http://www.tinyurl.com/linkagemap