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Solutions to 7.012 Problem Set 2

Question 1



In unicorns, coat color (brown or white) is controlled by a single gene with two alleles,

A and a. The brown phenotype is dominant over the white phenotype. Height (tall or

short) is controlled by a single gene with two alleles, H and h. The tall phenotype is

dominant over the short phenotype. These two loci are on different chromosomes.



a) What is the phenotype of an HhAa unicorn?

The phenotype of an HhAa unicorn is tall and brown





b) What is the phenotype of an hhAa unicorn?

The phenotype of an hhAa unicorn is short and brown



c) If an HhAa unicorn mates with an hhAa unicorn, what fraction of the progeny will be

short and brown?

i) Show how you would answer this question using a Punnett square.









ii) Show how you would use the laws of probability to answer the same question.



p(short, brown) = p(hhAA) X p(hhAa)

= 1/2 X 1/2 X 1/2 X 1/4

= 3/8

2





Question 2

For each of the following pedigrees, give the likely mode of inheritance (autosomal

dominant, autosomal recessive, sex-linked dominant, or sex-linked recessive). Assume

that individuals marrying into a family do not carry the mutant allele.





KEY



unaffected male

affected male

unaffected female

affected female



a) Mode of inheritance: Autosomal dominant









b) Mode of inheritance: X-linked recessive









c) Mode of inheritance: Autosomal recessive

3





Question 3

A rare and delicious fish lives in the shallows surrounding a remote desert island soon to

be famous as the site for a SURVIVOR sequel. This fish has two phenotypes of interest:

shininess and speed. The shiny phenotype is dominant over the dull phenotype, and the

fast phenotype is dominant over the slow phenotype. The producers of SURVIVOR hired

you to breed large numbers of shiny slow fish to increase the chance that the new

contestants can catch a meal.



Phenotype alleles

shininess D or d

speed E or e



You cross a true breeding shiny slow fish to a true-breeding dull fast fish to get an F1

class of progeny that is entirely shiny and fast.



a) What is the genotype of the true-breeding shiny slow fish?

DDee





b) What is the genotype of the true-breeding dull fast fish?

ddEE





c) What is the genotype of the F1 shiny fast fish?

DdEe





d) You then perform a test cross with an F1 fish. What are the genotypes of the

individuals in this test cross?

A test cross is a cross with the double recessive parent, in this case, DdEe X ddee



e) If the genes controlling these two traits are unlinked, what phenotypes and in what

ratios would you expect as a result of this test cross?

You would expect:

1:1:1:1, shiny fast fish : shiny slow fish : dull fast fish : dull slow fish





f) If the traits were completely linked, what phenotypes and in what ratios would you

expect as a result of this test cross?

1:1, shiny slow fish : dull fast fish

4





Question 3, continued

f) What you actually see when you complete the test cross is:



48 shiny fast fish



199 shiny slow fish



208 dull fast fish



45 dull slow fish







i) Circle the recombinant phenotypes.



ii) What is the recombination frequency between the shininess and speed loci?

RF = # of recombinants/ total number X 100

RF = (48 + 45/48 + 199 + 208 + 45) X 100

RF = 93/500 X 100 = 18.6%





g) Interestingly, you find that these same fish can be either sharp-eye or blind, where the

sharp-eyed phenotype is dominant over the blind phenotype. Use B or b to designate the

alleles of the eyesight locus.



You determine that the loci for shininess and for eyesight are linked with a recombination

frequency of 4.5%. Draw the two potential maps for the D, B, and E loci. On each map,

express the distance between the loci as the recombination frequency.









h) What would you have to do to determine which of these maps is the correct one?

You would determine the RF between the speed and the eyesight loci. One way this could

be done is to cross a true breeding sharp-eyed fast fish to a true breeding blind slow fish

to get an F1 class of progeny that is entirely sharp-eyed and fast. Then set up a test cross

between an F1 individual and the double recessive parent, in this case, BbEe X bbee.

Count the number of blind fast fish and the number of sharp-eyed slow fish to get the

number of recombinant progeny. RF = # of recombinants/ total number X 100

5





Question 4

On the same island is a seaweed (a diploid organism) that exists in three different colors

(this seaweed will soon be woven into bikini tops and breech cloths and will become

THE fashion statement in Paris). This seaweed can be gold, green or purple. The color is

controlled by a pathway of two distinct enzymes encoded by the F and G genes. For

questions a) and b) below, use F and G to designate the wild-type alleles. Use f and g to

designate the loss-offunction alleles. See the partial pathway below.





You cross two true-breeding plants, one that is purple and one that is green. The F1 plants

are all gold.



a) Given only the information above, what are two possible genotypes for the purple

parent?

FFgg or ffGG



b) Given only the information above, what are two possible genotypes for the green

parent?

ffGG or FFgg



c) What is the genotype of the F1 progeny?

FfGg



You cross two F1 progeny and expect to see a 9:3:3:1 ratio of phenotypes in the F2

generation. Surprisingly, you notice that the F2 progeny have the following ratio of

phenotypes:

9 gold : 4 purple : 3 green

d) Given this data, circle the correct pathway.









*Note, for parts a) and b) above....

The two enzymes are part of the same pathway, and the purple phenotype is the first step in the pathway.





This means that if the enzyme that carries out the transformation from purple to green is missing, it doesn't

matter what the alleles of the second enzyme are. So in this particular case ffGG, ffGg, and ffgg are all

possible phenotypes for the true-breeding purple parent. The true-breeding green parent could only be FFgg.

CHECK IT OUT FOR YOURSELF!



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