BIOL 2000 Lecture Notes - Epistasis, Penetrance, Wild Type

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1 Mar 2013

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Understand how the phenotype of a double mutant can be used to infer the relative positions of gene sin a pathway. No gene interaction is indicated in the 9:3:3:1 ratio. Analyzing double mutants: variations on the 9:3:3:1 ratio. Getting albino snake (ratio of 1) from orange and black cross (recessive for both means the snake can not make either pigment. What if we self the dihybrid resulting from the second cross: same as both single mutants (9:7) Act in the same pathway, no distinguishable intermediates. Mutations in genes for a target protein and its regular give the same phenotype, as does the double mutant. Same as one single mutant (9:3:4)- genes not alleles. Similar in idea to dominance (alleles), but is used for interactions between genes. Cross white (w/w; m+m+) with magenta (w+/w+ ; mm)= blue (w+/w; m+/m) W is epistatic to m: w m blue.

Related Questions

Problem A. You cross two homozygous parents: RRYY (round yellow) and rryy (wrinkled green) to produce an F1 generation. Round seeds are dominant, yellow seeds are dominant. You cross two F1s to produce the F2 generation. In the F2 generation, you find an individual with round yellow seeds. You do a testcross to find out the genotype of this âmysteryâ dominant/dominant phenotype. In the offspring of the cross, you find 1?2 have round yellow seeds and 1?2 have wrinkled yellow seeds.

1. Based on these data what was the genotype of the mystery F2?

Problem B. You carry out a dihybrid cross between a red-flowered rose with long pollen grains (A1A1 LL) and a white-flowered rose with round pollen (A2A2 l l). The flower color gene (A) shows incomplete dominance, and long pollen is completely dominant to round pollen. The genes for these traits are on different chromosomes so assort independently

2. What would be the phenotypes of the F1 generation?

3. What fraction of the F2 generation is expected to have red flowers and long pollen? (As always when traits are independent, what rule can you apply to save time?)

Problem D. In the example of epistasis presented in class, the agouti allele in mice (A) codes for striped hairs and is dominant to the allele for all black hairs (a). A loss-of-function allele for the pigment gene (b) produces no pigment and is recessive to the normal, dominant allele (B). The recessive allele at b masks the effect of the A gene. A dihybrid cross of AABB and aabb is carried out.

5. Name all of the genotypes in the F2 generation that can produce an albino (no pigment) phenotype.

Problem E. A cross is made between homozygous fruit flies that differ for eye color (the st gene) and

body color (the e gene). The F1 double heterozygote is shown below using chromosome notation.

st+ e

--------------------------------------

st e+

st = the recessive allele for scarlet eyes; st+ = the dominant wildtype allele for red eyes e = the recessive allele for ebony body; e+ is the dominant wildtype allele for gray body

6. Are the chromosomes shown in the genotype above in cis or trans configuration?

7. Based on this notation, what were the genotypes and phenotypes of the two parents of the F1? (Think carefully here...)

8. With what phenotype would you cross this F1 to measure recombination frequency?

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11. You suspect that two genes interact with one another to determine the product of a pathway you are studying. You cross two pure breeding plantsâone with white flowers and one with pinkâto make F1 individuals that have blue flowers. You cross the F1 generation and observe the following numbers of individuals with different phenotypes: Number of individuals Phenotype 91 Blue 30 Pink 39 White Which pathway best explains these results? (Note that arrows indicate enzymatic reactions carried out by products of genes 1 and 2. Assume that buildup of metabolites will determine flower color.)

Question 12 and 13: Consider the biochemical pathway presented below. This pathway determines pigment production in pea flowers. (Note: gene products are required for each reaction to proceed.)

12. Do you expect that genes A and B will interact with one another?

A. Yes B. No C. Insufficient information to determine

13. What phenotype ratio would you expect in offspring produced by crossing AaBb parents?

A. 9:3:3:1 (blue : green : yellow : white) B. 9:3:4 (blue : green : white) C. 9:7 (blue : white) D. 15:1 (blue : white)

14. How can haplosufficiency explain dominance of an allele that codes for a functional protein over a null (loss of function) allele?

A) a single functional copy of the gene produces sufficient gene product for normal biological function B) a single functional copy of the gene produces insufficient gene product for normal biological function C) the null allele produces a product that interferes with proper function of the normal allele D) the null allele produces a product that does not interfere with proper function of the normal allele E) the null allele rapidly mutates to reacquire normal functionality

15. In a plant species with genes A and B, you have isolated recessive mutant alleles at each locus, which you call a and b. You would like to know whether these two genes are linked. To test this hypothesis, you obtain an individual that is heterozygous at both genes (AB/ab) and perform a test cross. Which results would be most consistent with these loci being tightly linked? (Results are counts of the numbers of offspring from the testcross.)

A. Ab/ab 390 aB/ab 410 AB/ab 350 ab/ab 420

B. Ab/ab 80 aB/ab 73 AB/ab 760 ab/ab 754

C. Ab/ab 700 aB/ab 695 AB/ab 56 ab/ab 59

D. Ab/ab 400 aB/ab 65 AB/ab 50 ab/ab 659

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In this study you will learn about a modification of Mendelian inheritance which is also an important source of genetic variation.

You are about to perform what looks like a classic dihybrid testcross ("A" is the gene for eye color and "B" is the gene for wing color.)

A allele for red eyes

a allele for white eyes

B allele for clear wings

b allele for spotted wings

(AaBb) X (aabb) Note the genotypes and phenotypes.

Question #1: Predict the expected phenotypic ratio of this P X P cross.

Observed results:

500 red eyes clear wings

500 white eyes spotted wings

Question #2: How do the expected results compare with the observed results?

Question #3: Considering Mendel's second law (regarding dihybrid crosses), suggest an explanation for the discrepancy between the observed results and those predicted.

Genes are located at specific positions on a chromosome. The distance between genes is measured in terms of "mapping units". In this exercise the gene for wing phenotype âBâ will be moved from its original position of zero to different positions on the chromosome. You will observe the effect of this gene movement.

First, gene "B" (wing color) is moved from position zero to position 10 on the same chromosome. Gene "A" (eye color) stays at position 0.

Observed Results: (position 10)

450 red eyes clear wings

50 red eyes spotted wings

50 white eyes clear wings

450 white eyes spotted wings

Question #4: How do these results differ from those predicted by Mendel's second law?

Question #5: From those in the first cross?

Question #6: Can you suggest a hypothesis to explain these new results?

Gene "B" (wing color) is now moved to position 20 on the same chromosome. Gene "A" (eye color) stays at position 0.

Observed Results: (position 20)

400 red eyes clear wings

100 red eyes spotted wings

100 white eyes clear wings

400 white eyes spotted wings

Question #7: How do the proportions of the phenotypic classes in this cross differ from the proportions you obtained in the earlier crosses?

The wing color gene "B" is now moved farther away (positions 30, 40, 50, and 60).

Observed Results: (position 30)

350 red eyes clear wings

150 red eyes spotted wings

150 white eyes clear wings

350 white eyes spotted wings

Observed Results: (position 40)

292 red eyes clear wings

220 red eyes spotted wings

195 white eyes clear wings

293 white eyes spotted wings

Observed Results: (position 50)

250 red eyes clear wings

250 red eyes spotted wings

250 white eyes clear wings

250 white eyes spotted wings

Observed Results: (position 60)

250 red eyes clear wings

250 red eyes spotted wings

250 white eyes clear wings

250 white eyes spotted wings

Question #8: Prepare a line graph plotting the % of recombinant offspring (Y-axis) vs. difference in position between the two genes (X-axis). (Hint: your graph should be one line with labeled points at map positions 0, 10, 20, 30, 40, 50, & 60).

Question #9: A genetic definition of "1 map unit" is the distance between 2 genes that gives 1% recombinants out of the total number of progeny. Using this definition, how many map units separate the "A" and "B" genes when the eye color gene "A" is at position 0 and the wing color gene "B" is at position 40? (Show how you determined the results.)

Question #10: How does the distance between two genes on the same chromosome affect the proportion of recombinants? Why?

Question #11: Maximum crossover frequency can be defined as the point at which no more recombination can be detected no matter how far apart the genes are located. Do your results indicate that you are approaching the maximum crossover frequency?

Question #12: Show (using a Punnett Square) the expected phenotypic rations if the two genes were located on different chromosomes.

Question #13: Which of your simulated cross or crosses do the data from the Punnett Square most closely fit? Why?

Question #14: A wild-type fly (heterozygous for gray body color and normal wings - b⁺ vg⁺/b vg) was mated to a black fly with vestigial wings (b vg/b vg). The F1 had the following phenotypic distribution: wild type, 850; black body - vestigial wings, 899; black body - normal wings, 79; gray body - vestigial wings, 85. What is the recombination frequency (RF) between these genes for body color and wing type? (show work)

Question #15: A cross produced 915 offspring with normal pigment and 310 with albinism. Conclusion?

a. One of the parents was homozygous for albinism.

b. Both parents were heterozygous.

c. One parent was homozygous for normal pigmentation.

d. Both parents were albinos. e. 605 albino zygotes must have failed to develop.

NOTE: Please show all work to support your answers.

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BIOL 2000 Lecture Notes - Epistasis, Penetrance, Wild Type

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