BIOLOGY 283 Lecture Notes - Lecture 26: Zygote, Northern Blot

20 views4 pages

viridianhornet922

4 Dec 2018

School

Department

Course

Professor

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

Document Summary

Use of three point test cross an mapping using. * parental types are always the most numerous step two. Use of three point test cross on step 4 = determine the gene order . step 5- Calculate distances between the middle gene and each end gene utu der . 16 on outline: mapping in humans markers in pedigrees, lo d= log of odds what constitutes proof that the correct has been identified ? gene am. Northern blotting and steady state rna levels causative. Mutations constructing genetic maps in humans using pedigrees and molecular markers one pedigree is not enough ! Not present likelihood present i likelihood not present ) Lod will be positive it it linkage linkage is is very unlikely likely and will be higher the more likely linkage is . Mrna from the gene is expressed in the right place and at the right time an.

Related Questions

ANSWER ALL the PARTS you can

PART1 Alleles are considered linked if recombination is less than 50% in:

a) a Px P Fl cross

b) an F1 x F1F2 cross

c) an F2 x F2F3 cross

d) an Fl x homozygous recessive cross

e) any F2 progeny x homozygous recessive cross 1.

PART2 If two loci A/a B/b are 9 map units apart, and an Ab/aB individual is testcrossed,

a) There will be 9% AB individuals

b)There will be 91% Ab individuals

c) There will be 4.5% ab individuals

d) There will be 91% recombination individuals

If two genes a and b show 50% recombination genes a and c show 35% recombination, and genes b and show 32% recombination, then

a) Genes a and b are on the same chromosome, and 50 m.u. apart

b) Genes a and b are on different chromosomes

c)Genes a and b are on the same chromosome, and 67 map units apart

d) One cannot tell if genes a and b are on the same or different chromosomes from given data.

PART4 If two genes a and b are 8.3 m.u. a part, one expects to find

a) 8.3% recombinant gametes from a double heterozygous parent

b) 91.7% recombinant gametes from a double heterozygous parent

c) A chiasma between the a and b loci in 8.3% of the meiosis

d) A chiasma between the a and b loci in 91.7% of the meiosis.

PART5 Map distances are most accurate when genes are closely linked because as genes become further apart,

a) They always show independent assortment

b)The chance of multiple crossovers increases

c) They no longer recombine

d) There is chiasma interference

PART6 In a three-point testcross, the most frequent classes of progeny were ABc and abC. The less frequent classes of progeny were AbC and aBc. Which gene is in the middle?

a) Aa

b) B/b

c)C/c

d) It is not possible to tell from just these data

PART7 In a three-point testcross with 100 progeny, there were 16 single crossovers between genes A and B, 29 single crossovers between genes B and C, and 4 double crossovers. The correct gene order was determined to be A-B-C. What is the map distance between A and B?

a) 20%

b) 0.2 m.u

c) 20 m.u.

d) 16 m.u.

e) 29 m.u.

f) 16%

Chromosome Mapping (determining the relative positions of genes on a chromosome) - This is a two point cross since it involves two loci on a chromosome.

Follow the directions for the crosses below involving 2 traits located together on the same chromosome (Drosophila Chromosome IV): eyeless and shaven bristles. Please only typed answer, no hand-written work or screen shots please.

1. a. Perform a monohybrid cross between a female fly with the eyeless mutation and a male fly with shaven bristles. Show results for the F₁ offspring. (NOTE: You will use an F₁ female in the next cross.)
ENTER DATA HERE:

Question: Are the phenotypes of the F₁ offspring what you would have predicted for this cross: why or why not?

Question: After looking at your F₁ offspring, which allele is dominant: wild-type eyes or eyeless? How can you tell?

Question: After looking at your F₁ offspring, which allele is dominant: wild-type bristles or shaven bristles? How can you tell?

Draw the Chromosome IV pair of an F₁ offspring: (HINT: One chromosome was inherited from the mother, with the eyeless allele at the eye shape location (locus) and the wild-type allele at the bristles locus. The second chromosome was inherited from the father, with the wild-type allele at the eye shape locus and the shaven bristle allele at the bristles locus.)

1. b. Perform a testcross between an F₁ female offspring from the previous cross and a new male with both the eyeless and the shaven bristle traits to produce an F₂ generation.
ENTER DATA HERE:

Question: What phenotypes are found in the F₂ offspring?

Question: Are the phenotypes of the F₂ offspring what you would have predicted for this cross: why or why not?

Draw the 2 different Chromosome IV pairings you expected in an F₂ offspring: (HINT: The F₁ female will donate one or the other of the chromosomes you drew above to each offspring. The new male with both the eyeless and the shaven bristle traits can only donate one type of chromosome, with both mutations.)

Question: What are the 2 phenotypes that would result from the 2 different chromosome pairings you just drew?

Question: What are the other 2 phenotypes that you obtained in your F₂ offspring? What process, which occurred during meiosis, was responsible for the creation of these recombinant phenotypes?

Draw a map that shows the map distance (in map units or centimorgans) between the locus for the shaven bristle allele and the locus for the eyeless allele. (NOTE: You need to add the proportion of SV;EY flies to the proportion of wild flies OR add the number of SV;EY flies and wild flies and then divide by the total number of flies. Then multiply either number times 100.)âSHOW ALL CALCULATIONS BELOW

CALCULATIONS:

CHROMOSOME MAP:

There are several different ways to draw maps of the chromosomes. You can draw a simple line graph using a free online graphing tool, such as:

http://www.onlinecharttool.com/

http://nces.ed.gov/nceskids/createagraph/

Or you can simply use your keyboard to create a line with the dash or underline, with the abbreviations for the genes inserted, and then put the map units below, such as:

___________A_____________________B_________

50 map units

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

BIOLOGY 283 Lecture Notes - Lecture 26: Zygote, Northern Blot

Document Summary

Get access

Related Documents

BIOLOGY 283 Chapter Notes - Chapter 6.1: Euchromatin, Heterochromatin, Zygosity

BIOL239 Chapter Notes - Chapter 5: Gene Mapping, Chromosome, Verb

BIO 325 Chapter Notes - Chapter 5: Vulgate, Product Rule, Chromosome

Related Questions