BIOL 2107 Lecture Notes - Lecture 6: Wild Type, Siamese Cat, Genetic Predisposition

69 views11 pages

amethysthamster869

30 Jan 2017

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

In all of these cases, f2 phenotypes from dihybrid crosses are in a variation of the 9:3:3:1 ratio expected for independently assorting genes. We had been saying two genes had two different traits. Here, we"re talking about two genes interacting together to do something. These are all deviations from the 9:3:3:1 ratio, because that was two genes and two different traits. Novel phenotypes resulting from gene interactions, i. e. seed coat in lentils. Dihybrid cross of lentils, tan x gray (two genes, one phenotype). F2 progeny: 9/16 brown, 3/16 tan, 3/16 gray, 1/16 green. 9:3:3:1 ratio in f2 suggests two independently assorting genes for seed coat color. F1 will be all heterozygous all brown. Brown, tan, grey, and a new colour green. We hadn"t seen that in either of the parents. First time we"ve seen a new phenotype that wasn"t in the parents our old example was wrinkled, round, green, or yellow in the parents.

Related Questions

Canaries: Green (wild-type) Pied phenotype Yellow type

Canaries are popular pets and easy to breed in captivity. Additionally, they display great phenotypic diversity, which allows you to do some interesting genetic investigations with them.

When you take a canary from a true-breeding green strain and cross it with a canary from a true-breeding yellow strain, all offspring display the pied phenotype. Based on this you suspect that this color variation is inherited by a gene with a co-dominant inheritance pattern (showing both phenotypic traits in hybrid phenotypes).

You decide to test this hypothesis. To do so, you take pairs of your hybrid F1 individuals (of pied phenotype), and cross them to produce an F2 generation.

After many dozens of crosses and hundreds of offspring, you arrive at the peculiar ratio of:

3/16 green : 6/16 pied : 7/16 yellow

Which of the following hypotheses are consistent with these data? (Note that not all of these hypotheses are mutually exclusive. Assume that the two genes are independently assorting)

Hint: This problem is very hard to work from the given ratios to what is happening genetically. It is easier to "work backwards" and start with each choice below, and see if you can obtain all or important aspects of the results of your cross when making the assumption that the answer choice is correct.

Select all that apply:

Select one or more:

a. At least one of the genes involved is sex linked.

b. One gene controls production of darker pigment (melanin which appears green when mixed with yellow). A second gene acts as a patterning gene with one allele causing expression and the other non-expression of melanin and when in the heterozygous condition each allele controls melanin expression on different parts of the body so long as the bird's genotype of the first gene allows for the production of melanin.

c. this is a single-gene trait with co-dominance

d. one of the two genes involved displays complete dominance, while the other displays co-dominance.

e. the genes involved are epistatically interacting

f. more than one gene is involved

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.

fuchsiasnake5

BIOL 2107 Lecture Notes - Lecture 6: Wild Type, Siamese Cat, Genetic Predisposition

Document Summary

Get access

Related Documents

BIOL 2107 Lecture Notes - Lecture 3: Heterogametic Sex, Cleavage Furrow, S Phase

BIOL 2107 Lecture Notes - Lecture 5: Xeroderma Pigmentosum, Environmental Factor, Wild Type

BIOL 2107 Lecture Notes - Lecture 5: Zygosity, Wild Type, Anemia

Related Questions