BICD 100 Lecture Notes - Genetic Linkage, Zygosity, Wild Type

In Drosophila, the autosomal recessive brown eye color mutation (b) displays interactions with both the X-linked recessive vermilion mutation (v) and the autosomal recessive scarlet (s) mutation. Flies homozygous for brown and simultaneously hemizygous or homozygous for vermilion have white eyes. Flies simultaneously homozygous for both the brown and scarlet mutations also have white eyes. Flies that are wildtype at all 3 loci have wildtype eye color. Flies that are homozygous or hemizygous for the recessive mutant at only one locus show that mutations color. (i.e. brown, vermillion or scarlet).
Use this information to answer the next 4 questions.

Predict the F1 progeny of a cross between a true breeding scarlet female X brown male.

A three point cross was performed using fruit flies. The three loci are physically linked on the same autosome. The d allele, in the homozygous state, results in the recessive ‘double dorsal’ phenotype. Similarly, the r allele results in the recessive ‘round ears’ phenotype and the e allele in the recessive ‘endless eye’ phenotype. In contrast, the D, R, and E alleles encode for the dominant wild type phenotypes at each of these loci.

True breeding (homozygous) flies were crossed to produce F1 offspring. Due to poor book-keeping, the genotypes of the parent flies were not recorded – but one parent fly had one mutant phenotype, while the other parent had the other two mutant phenotypes.

Phenotype Number of F2 Flies

Ears Dorsal Eyes

1) Round ears Double Dorsal Endless Eyes 230

2) Wild-type Wild-type Wild-type 260

3) Round ears Wild-type Wild-type 4,390

4) Wild-type Double Dorsal Endless Eyes 4,100

5) Round ears Double Dorsal Wild-type 15

6) Wild-type Wild-type Endless Eyes 15

7) Round ears Wild-type Endless Eyes 510

8) Wild-type Double Dorsal Wild-type 480

10,000 total

a) What is the phenotype of the F1 flies?

b) What is the genotype of the F1 flies? Your answer should indicate the phase of linkage.

c) What are the genotypes and phenotypes of the original parent flies (previously not recorded)? Your answer should indicate the phase of linkage.

d) Calculate the map distance in cM between each of the three loci and determine the gene order. Show all work for full credit. (6 points)

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 “B” 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.