1. Mating two organisms produces a 3:1 ratio of the phenotype in progeny. The parental genotypes are

A. Aa x Aa
C. AA x aa
B. Aa x aa
D. AA x AA

2. An observant gardener finds that some of his bean plants have pubescent (Hairy of fuzzy) leaves and others have glabrous (no hair) leaves. He crosses different plants and obtains the results shown.

1. Pubescent x Glaborous 56 61 Pp x pp
2. Pubescent x Pubescent 63 0 PP x PP or PP x Pp
3. glabrous x glabrous 0 44 pp x pp
4. Pubescent x glaborous 59 0 PP x pp
5 Pubescent x Pubescent 122 41 Pp x Pp

(a) Explain these results

Single gene trait with two alleles; complete dominance.

(b) Using your own gene symbols, give the genotypes of the parents of each cross.

P = pubescent and p = Glabrous. P is completely dominant to p therefor Pp individuals would be pubescent.

See above for genotypes

(c) How many of the pubescent progeny in crosses 2, 4, and 5 would you expect to produce glabrous progeny when self-fertilized?

The frequency of pubescent progeny resulting from a selfing of the F1 in cross 2 is undeterminable. We are not sure of the genotypes of both parents. Here one parent could be PP while the other is P_ (either Pp or PP) These would both give different results in the progeny of selfing. The F2 progeny of the selfed F1 from cross 4 should result in a 3:1 ratio of pubescent to glabrous as the phenotypic ratio of the F1 can only be explained by a PP x pp cross in the parental generation. The 3:1 ration of Pubescent to glabrous suggest a cross between two heterozygous individuals.

3. (2 points): A pure-breeding red flower line is crossed with a pure-breeding white flower line. All the F1 individuals are green, while the F2 shows red, white and green flowered individuals. Crossing red and green individuals gives equal numbers of red and green-flowered offspring. Crossing white and green individuals gives equal numbers of white and green-flowered offspring. What genetic model accounts for these observations (i.e., give the genotypes and their corresponding phenotypes)?

This trait can be explained by a single gene with two alleles. Red flowers have the genotype RR, White floweres have the genotype rr and green flowers have the genotype Rr.

4. [5 points] The American bulfrog has 13 pairs of chromosome (2n=26). Considering only three of these pairs and desginating the chromosomes from the male parent with solid lines and those from the female parent with broken ones, show diagrmmatically all the possible alignments of these pairs at metaphase I. Assuming no crossing-over, how many different kinds of meiotic products (spermatids, ootids) are possible from such a 26n individual?


213 possible combinations = 8192 possible chromosome combinations.

5 You wish to generate a complementation problem for the upcoming genetics midterm.You decide on using 8 mutations (a1 -a8) and generate the following complementation>group data diagrammed below.

Fill in the table below with the appropriate pluses and minuses according to thecomplementation group data above.

 

 

a1

a2

a3

a4

a5

a6

a7

a8

a1

-

+

+

+

+

+

+

-

a2

 

-

+

-

+

-

+

+

a3

 

 

-

+

+

+

+

+

a4

 

 

 

-

+

-

+

a5

 

 

 

 

-

+

-

+

a6

 

 

 

 

 

-

+

+

a7

 

 

 

 

 

 

-

+

a8

 

 

 

 

 

 

 

-

 

 

 

 

 

 

 

 

 

6. The dominant allele Cy (Curly) in Drosophila results in curly wings. The
cross Cy/+ x Cy/+ (where + represents the wildtype allele of Cy) results in a ratio of 2
curly : 1 wildtype F1 progeny. The cross between curly F1's also gives a ratio of 2 curly :
1 wildtype F2 progeny. How can this result be explained?

We expect a 3:1 ratio of the dominant to the recessive phenotype when
two heterozygotes are crossed. In this case, this would correspond to a
genotypic ratio of 1 Cy/Cy : 2 Cy/+ : 1 +/+. Since we only see 2
curly-winged flies for every wildtype one, the simplest explanation is
that the Cy/Cy homozygotes die. Thus, Cy is dominant with respect to
wing phenotype and recessive with respect to lethality.

7 . In plants of the genus Primula, the K locus controls synthesis of a compound called malvidin. Two plants heterozygous at the K locus were crossed, producing the following distribution of progeny: 1010 make malvidin 345 do not make malvidin You wish to determine if the D locus (located on a different chromosome than K), affects production of malvidin as well. Pure-breeding plants that make malvidin (KKdd) were crossed to pure-breeding plants that do not make malvidin (kkDD). All F1 plants fail to produce malvidin. F1 heterozygotes were self-fertilized and the F2 progeny were assayed for malvidin synthesis, with the following distribution: 522 make malvidin 2270 do not make malvidin

(a) Write the genotypes and the corresponding phenotypes of all the F2 progeny obtained.
The 3:1 ratio in the first cross tells us that making malvidin (K) is dominant to not making malvidin (k). The modified dihybrid ratio in the second cross is 3:13. Twelve out of sixteen F2 progeny will be K-, but only three make malvidin, so the D locus must be homozygous dd in order for malvidin to be made:

9 K-D- does not make
3 K-dd makes
3 kkD- does not make
1 kkdd does not make

(b) Does the D locus affect malvidin synthesis?
(circle one) Yes No

(c) Briefly explain your answer to (b).

D is a dominant suppressor of K. In other words, malvidin is only made when the plant is not kk, but is dd.

8. You and your friend Cartman work together in a genetics lab in the prestigious South Park University. Your advisor gives you a plant showing the dominant phenotype for a simple, monogeneic Mendelian trait, and asks you to determine if the plant is homozygous or heterozygous. Cartman lets the plant self-fertilize and records 11 progeny showing the dominant phenotype and 3 progeny showing the recessive phenotype. Cartman then asks you to calculate the probability of obtaining, out of 14 progeny from a selfed heterozygote, 11 with the dominant phenotype and 3 with the recessive phenotype. What is the probability?

The probability of obtaining 11 dominant-phenotype and 3 recessive-phenotype plants from a selfed heterozygote is given by the binomial formula:
When you tell Cartman, he gets upset and says, "Dude, that's way to low! You must have done something wrong in your calculation, because if that were really the right probability then we wouldn't have much confidence that the plant we selfed was a heterozygote." In one sentence, tell Cartman why his logic is wrong.

Cartman is wrong because getting any recessive-phenotype progeny means that the selfed plant must have been heterozygous. (He probably is also confused about a more subtle point that doesn't really apply to this question: when calculating probabilities to do statistical testing, we must consider not just the probability of the event that actually occurred but the sum of the probabilities of the event we observed and of all other possible events that are as deviant from the null expectation.) Anyway, tell Cartman the following: "Cartman, you *@# *@$!, the fact that we got any recessive-phenotype progeny at all tells us that the selfed plant was heterozygous." (14!) /(11!)(3!) (3/4) 11 (1/4) 3 = 0.24

9 For each pedigree (A, B and C), state, by answering Yes or No in the appropriate blank space, whether transmission of the trait can be accounted for on the basis of each of the listed simple modes of inheritance.


Pedigree A Pedigree B Pedigree C
Autosomal Recessive Yes Yes Yes
Autosomal Dominant Yes Yes No
X-linked Recessive Yes Yes No
X-linked Dominant No No No


10. An experiment to determine the inheritance opf flower color in snap dragons crossed two planbts with pink flowers (Ii). The following progeny were recovered from the cross. 84 plants whad red flowers (II), 172 had pink flowers (li) and 78 had white flowers (ii). Use a Chi-square test to determine whether those numbers fit a mendelian pattern of inheritance.

Answer: 0.26 with 2 degrees of freedom

X2 = Sum{84-83.5/83.5) + (172 - 167/ 167) + (78-83.5/83.5) = 0.26