BIOL 102 Lecture 18: Chromosomal Basis of Inheritance

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16 Nov 2016

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Chromosomal theory of inheritance: genes located on chromosomes; signi cance of inheritance of genes are located on chromosomes. Krogh principle: whatever biological trait you are looking at, there is an organism that you can most conveniently study. Wild type (normal and more common) vs mutant type. Sex-linked genes; x-linked and y-linked (aren"t many one y gene) The genotypes of the eggs of a female aren not all the same. X and y chromosomes behave as homologous structures during meiosis. One of 2 x chromosomes in each cell randomly inactivates during embryonic development. Get random model (mosaic) pattern (reason why cats have patches of ex. orange and black fur) When cross over is between the genes, hybrid will be switched of phenotype. Farther apart 2 genes are, the higher the probability that the crossover will occur between them; therefore higher recombination frequency. Max % that recombination frequency can have is 50% Nondisjunction: homologous structures don"t separate in meiosis due to spindle.

Related Questions

The white eye mutation studied by Thomas Hunt Morgan is X-linkedand recessive to wild-type red eyes. When true-breeding white-eyedmales carrying the white mutation were crossed with true-breedingpurple-eyed females, all the F1 progeny had wild-type (red) eyes.When the F1 progeny were intercrossed, the F2 progeny emerged inthe following ratio:

3/8 wild-type females
1/4 white-eyed males
3/16 wild-type males
1/8 purple-eyed females
1/16 purple-eyed males.

Formulate a hypothesis to explain the inheritance of these eyecolors and match the correct genotype with each individual listedbelow.

P Generation Male:

P Generation Female:
F1 Generation Male

F1 Generation Female:

F2 Generation Purple-eyed Female

F2 Generation White-eyed Male

F2 Generation Wild-type Male:

F2 Generation Purple-eyed Male:

F2 Generation Wild-type Female:

Options:

w+/Y; p+/_

w/Y;_ _

w+/_; p/p

w+/Y;p+/p

w+/_;p+_

w/Y; p+/p+

w+/W+; p/p

w+/Y; p/p

emeraldswan267

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 102 Lecture 18: Chromosomal Basis of Inheritance

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