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Vesuvian volocano voles are usually black, but some are agouticolor. Agouti is the result of a dominant allele at an autosomallocus. Crosses were made between volcano voles known to behomozygous for the agouti allele and normal black individuals knownto be homozygous for the recessive allele. Out of 232 offspring,218 were agouti in color, but 14 were black.

a. What is the most likely genetic explanation for thisresult?

b. Is it possible that a cross between two black volcano voleschosen at random from the wild could yield three offspring, all ofwhich were agouti in color? Briefly explain your answer.

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Beverley Smith
Beverley SmithLv2
29 Sep 2019
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Related questions

Problem A. You cross two homozygous parents: RRYY (round yellow) and rryy (wrinkled green) to produce an F1 generation. Round seeds are dominant, yellow seeds are dominant. You cross two F1s to produce the F2 generation. In the F2 generation, you find an individual with round yellow seeds. You do a testcross to find out the genotype of this “mystery” dominant/dominant phenotype. In the offspring of the cross, you find 1?2 have round yellow seeds and 1?2 have wrinkled yellow seeds.
1. Based on these data what was the genotype of the mystery F2?

Problem B. You carry out a dihybrid cross between a red-flowered rose with long pollen grains (A1A1 LL) and a white-flowered rose with round pollen (A2A2 l l). The flower color gene (A) shows incomplete dominance, and long pollen is completely dominant to round pollen. The genes for these traits are on different chromosomes so assort independently
2. What would be the phenotypes of the F1 generation?
3. What fraction of the F2 generation is expected to have red flowers and long pollen? (As always when traits are independent, what rule can you apply to save time?)

Problem D. In the example of epistasis presented in class, the agouti allele in mice (A) codes for striped hairs and is dominant to the allele for all black hairs (a). A loss-of-function allele for the pigment gene (b) produces no pigment and is recessive to the normal, dominant allele (B). The recessive allele at b masks the effect of the A gene. A dihybrid cross of AABB and aabb is carried out.
5. Name all of the genotypes in the F2 generation that can produce an albino (no pigment) phenotype.

Problem E. A cross is made between homozygous fruit flies that differ for eye color (the st gene) and
body color (the e gene). The F1 double heterozygote is shown below using chromosome notation.
st+ e
--------------------------------------
st e+
st = the recessive allele for scarlet eyes; st+ = the dominant wildtype allele for red eyes e = the recessive allele for ebony body; e+ is the dominant wildtype allele for gray body

6. Are the chromosomes shown in the genotype above in cis or trans configuration?
7. Based on this notation, what were the genotypes and phenotypes of the two parents of the F1? (Think carefully here...)
8. With what phenotype would you cross this F1 to measure recombination frequency?

(tautomerization) The most common form of guanine is the ketoform, not the enol form.

True

False


What term describes the allowed degree of base pair matchingduring annealing, varying from lots of base-pair mismatches allowedto allowing only perfect base-pair matching.

stringency

aneuploidy

polyploidy

quantitative PCR

restriction fragment length polymorphism


I have an individual with a disease we will call SSI-itis,brought on by extreme exposure to cannabis. I do a southern andfind the gene affected. Another of his classmates exhibits theexact same symptoms, but that gene in hiw is wt. A southern in thesecond student finds a mutation in another gene. This is an exampleof

pleitropy

allelic heterogeneity

locus heterogeneity

dominant traits

recessive traits

penetrance


If a person carrying a dominant allele does not demonstrate thephenotype strongly, but still differs from wt, thisdemonstrates?

haploinsufficiency

penetrance

expressivity

monogenic traits

polygenic traits


I find three alleles for a gene in a fruit fly. One allelecreates the ability for the fly to eat graphite. Another allelecreates the ability for the fly to eat coal. The third alleleallows the fly to eat diamond. These alleles are all dominant, andwt cannot eat any of these substances. Which statement below is themost true?

Upon mating a bunch of the flies carrying these alleles, Ishould not find a fly that can eat graphite, coal, and diamond.

I would consider the ability to eat these substances an exampleof pleitropy because they are all carbon.

I would expect the ability to eat diamond to be dominant to theability to eat graphite.

These flies are demonstrating locusheterogeneity.


Expression vectors in prokaryotes do not make functionaleukaryotic gene products in bacteria very well because

the codon sequence for prokaryotes is different than the codonsequence in eukaryotes

there are no disulfide bridges formed in proteins normally madein prokaryotes

prokaryotic expression vectors cannot translate eukaryticsequences

eukaryotic genes have introns, and prokaryotes don't

eukaryotic genes have exons and prokaryotes don't


Which of the following is not used in PCR amplification ofDNA?

Double-stranded DNA template

Oligonucleotide primers

DNA polymerase

DNA ligase

ddNTPs


I create a knockout mouse using the agouti/black fur systemdiscussed in the online lecture and in the book. When I cross theagouti offspring of the originally engineered mouse, I find a ratioof 2 agouti mice to 1 black furred mouse. What is the bestexplanation?

The gene knocked out was recessive.

The gene knocked out was dominant.

The gene knocked out was a lethal gene.

The knockput was integrated into a random spot, and did notknockput the original gene.


A loss of function mutation is usually recessive, but canactually be a dominant mutation because of what effect?

haploinsufficiency

dominant negative mutations

allelic heterogeneity

locus heterogeneity

pleitropy


A genetic carrier of a disease can have a form of the diseasethat only shows intermittent symptoms, while the full homozygousrecessiuve individual would show all of the symptoms of thedisease.

True

False


Wild type, or wt, refers to the genotype that is the typicalgenotype found in nature. Now read that again. Once more. Nowanswer.

Answer

True

False



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 “B” 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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