BIOL 110 Chapter Notes - Chapter 17: Allosome, Y Chromosome, Wild Type

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14 Jan 2017

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The relationship between chromosomes behavior and mendel"s laws: dihybrid individual has two different alleles for each gene, each gene is located at the same position on each member of the homologous pair. Linking genes and chromosomes: mutant phenotype - anything other than the wild-type. Nomenclature used by drosophila geneticists: wild-type (+, allele for red eyes is w, white eye allele is w, lowercase/uppercase. Sex-linked genes: to investigate the mutant white-eye allele: Mate white-eyed male fly with red-eyes (wild type) female. All of the females and half of the males had the wild-type trait. This led morgan to believe that the gene coded for eye color must be related to sex determination: fruit flies also have x and y chromosomes. The gene for eye color is located on the x chromosome, then this would explain why the trait rarely appears in females. The female would have to be homozygous for the recessive mutant.

Related Questions

In Drosophila, the map positions of genes are given in map units numbering from one end of a chromosome to the other. The X chromosome of Drosophila is 66 m.u.. long. The X-linked gene for body color-with two alleles, y ⁺for gray body and y for yellow body-resides at one end of the chromosome at map position 0.0. A nearby locus for eye color, with alleles w ⁺for red eye and w for white eye, is located at map position 1.5. A third X-linked gene, controlling bristle form, with f ⁺ for normal bristles and f for forked bristles, is located at map position 56.7. Each gene resides on the Xchromosome, and at each locus the wild-type allele is dominant over the mutant allele.

Predict the frequency of gray with white eyes and forked bristles progeny type produced by this mating.

Enter your answer as a percentage to three decimal places (example 0.235 or 0.230).

Frequency=_________

Phenotype	Number
Pale	648
Pale, oval	64
Pale, short	10
Pale, oval, short	102
Oval	6
Oval, short	618
Short	84
Wild type	98
	1630

A wild-type trihybrid soybean plant is crossed to a pure-breeding soybean plant with the recessive phenotypes pale leaf l, oval seed (r), and short height (t). The results of the three-point test cross are shown in the table. Traits not listed are wild type.

Calculate the interference value for these data.

Enter your answer to three decimal places (example 0.201).

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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!)

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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 110 Chapter Notes - Chapter 17: Allosome, Y Chromosome, Wild Type

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