BIO 2306 Lecture Notes - Lecture 29: Mendelian Inheritance, Elephantiasis, Drosophila Melanogaster

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corallouse434

28 Aug 2016

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Non- mendelian inheritance: maternal effects, larvae of some kind of bug. Larvae are normally dark brown with eyes becoming progressively darker. Adults are dark brown and dark brown eyes. The mutant forms do not get dark and red eyes(adult) See book: fruit fly grandchild less mutant. Females who are homo fro this mutation produce viable offspring but those offspring are sterile. This gene directs the development of an embryonic region responsible for eventual formation of gonads. This is still nuclear and mendelian: cytoplasmic traits, traits which are inherited via dna not associated with the nucleus. Intracellular symbionts: small prokaryotic cells that exist inside eukaryotes, they have their own dna and metabolism, mosquito susceptibility to filarial parasites, elephantitis, crosses. Susceptible female with a normal male: all susceptible. All cytoplasm genes come from the mother: streptomycin resistance in chlamydomonas - green algae, crosses: streptomycin sensitive (ss) streptomycin resistant (sr) Cross b/w ss female and sr: all ss.

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1. The eye color of wild-type Drosophila flies is red. Differentmutations in a single or in multiple genes lead to flies with whiteeyes. Mutations in the w gene on the X chromosome lead to whiteeyes instead of the normal red. You have isolated both a white-eyedmutation (designated W-1) that gives a dominant phenotype, and awhite-eyed mutation (designated w-2) that gives a recessivephenotype.

(a) A white-eyed male from the W-1 line is crossed to awild-type female. What color eyes will the female progeny from thiscross have?

(b) What color eyes will the male progeny from the cross in part(a) have?

(c) One of the female progeny from the cross in part (a) ismated to a white-eyed male from the w- 2 line. What fraction of thewhite-eyed progeny from this cross will be female?

2. The body color of Drosophila adult flies is regulated bymultiple genes, including the autosomal ebony gene. Ebony body (e)is recessive to the wild type (yellowish) body (e+). A second gene,which is X-linked, controls bristle shape. Normal bristle shape(f+) is dominant to the mutant forked bristles (f). A female withforked bristles that is heterozygous for the body color gene ismated with an ebony body male (his bristles are normal).

(a) What are the genotypes of the parents?

(b) Using a Punnett square, determine the expected proportionsof each possible progeny phenotype. Since there is a sex-linkedgene here, be sure to include consider sex as part of thephenotype.

3. Describe in your own words what a balancer chromosome is andwhy it is important in fly genetics (you will have to do someresearch for this one!)

mauvepug893

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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BIO 2306 Lecture Notes - Lecture 29: Mendelian Inheritance, Elephantiasis, Drosophila Melanogaster

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