Bio 97 Midterm 2 Review Session October 26 th 2010 MSTB 118 5:30PM-7:30PM

Bio 97 Midterm 2 Review SessionOctober 26th 2010

MSTB 118 5:30PM-7:30PM

Emily LingTalar Tfnakjian

Sample Problem– L7-1.

• George, Gregor Mendel’s friend and next-door neighbor, tries to repeat Mendel’s experiments. He obtains a true-breeding line of the wrinkled pea plants, but his line of round pea plants is not true-breeding. Some of his round pea plants when self-crossed give rise to wrinkled pea plants. He crosses a large number of his round pea plants with a large number of his wrinkled pea plants. Which genotype(s) would he expect to see in the F1 progeny assuming that only one locus is determining this trait? (Remember that round is the dominant trait and that the alleles for this trait are denoted by W and w).

• A) WW only • B) Ww only • C) WW mainly and some Ww • D) Ww mainly and some WW • E) Ww mainly and some ww

Details Regarding Meiosis

Linked Genes Violate Mendel’s Second Law

• Mendel’s second law states that each gene segregates independently of others.

• Problem is we now know there are multiple genes that are on the same chromosome!

Non-disjunctionNormal Case:

Non-disjunctionMeiosis I

Non-disjunctionMeiosis II

Compare the two types of non-disjunction

Things That Alter Mendelian Phenotypic and Genotypic Ratios

Sample Problem• 21. A heterozygous round pea plant is crossed to a

wrinkled pea plant. Among the progeny, there is the following phenotypic ratio: 2 very round : 2 slightly round : 1 wrinkled.

• Which of the following best explains this unexpected phenotypic ratio?

• A) Incomplete penetrance • B) Variable expressivity • C) Both incomplete penetrance and variable expressivity • D) Incomplete dominance • E) Complementation

Epistasis and Complementation

Epistasis

• More than one gene is affecting the phenotype

•In this example, two types of genes that are coding for different enzymes are responsible for the final color of the organism

• If you only had enzyme B (either as Bb or BB) but enzyme A’s genotype is aa, then what is the final phenotype?

Complementation

• Wildtype (specified, so read the question carefully) phenotype is restored

Sample ProblemMidterm 2009C 6. Wild-type Martian flies have yellow eyes; mutants have white, red, or green eyes. Eye-color is determined by the production of red pigment and green pigment by two separate pathways shown below. If both pi gments are made, the eyes are yellow; if neither pigment is made, the eyes are white. If only the red pigment is made, the eyes are red; if only the green pigment is made, the eyes are green.

The dominant alleles (A, B, C, D) represent the wild-type allele that makes each enzyme; the recessive alleles (a, b, c, d ) represents the null mutation allele that does not make each enzyme. The following cross is made:

AA bb CC DD X AA bb cc DD

What is the phenotypic ratio of this cross? A) All green B) All red C) All yellow D) All white E) None of the above

Mutant Screens

• Mutant screens are followed by a complimentation test to determine where the gene mutations are located relative to each other (on the same chromosome or another one)

• Two types can occur:– Loss-of-function– Gain-of-function

Sample Problem• L11-2. • Wild-type yeast cells are red. Professor Yi isolates 6 mutant white cells: 1, 2, 3, 4, 5, 6. He performs a

complementation test by crossing the mutants to each other and observes the following results: • 1 X 3 = White • 1 X 4 = Red • 2 X 3 = Red • 2 X 6 = White • 3 X 5 = White • 4 X 6 = Red • (i) What is the phenotype of the progeny of a 1 X 5 cross? • (ii)What is the phenotype of the progeny of a 2 X 5 cross?

• A) (i) White (ii) White • B) (i) White (ii) Red • C) (i) Red (ii) White • D) (i) Red (ii) Red