Download Biology 207 Workshop 9

Biology 207 Workshop 9
1. A true-breeding red snapdragon was crossed to a true-breeding white snapdragon. The
F1 progeny were red. When F1 x F1 matings were made, the following F2 progeny were
observed:
1850
670
831
red
pink
white
a) P: RR; WW and rr; ww
F1: Rr; Ww
F2: R-; W-, and rr; W- and -/-; ww
b) 1red: 1 pink: 2white
2 . In sweet peas the allele C is needed for color expression (c results in no color: white).
The flower colour that is expressed is determined by the alleles R (red) and r (blue). A
cross between certain red and blue plants resulted in progeny as follows:
3/8 red
a) What were the genotypes of the plants crossed?
Rr;Cc x rr;Cc
3/8 blue
1/4 white
3. In wheat, a cross between true-breeding red-kernel and true-breeding white-kernel
strains yielded F1 offspring with red kernels. When the F1 were intercrossed, the F2
plants had a ratio of 15 red-kernel : 1 white-kernel. A testcross of the red-kernel plants
yielded 3 red-kernel : 1 white-kernel.
a. What are the parental genotypes?
b. What are the F2 genotypes and phenotypes?
c. What conclusions can be made about the allelic and gene interactions?
a. Their F1 parents are R/R; C/C (red) x r/r;c/c (white).
b. R/_:C_ (red), r/r;C/_ (red), R_; c/c (red), r/r; c/c (white)
c. The relationship between alleles of R and C could be either dominant suppression
or a duplication.
4. In mice, C (colored) vs. c (albino) and B (black) vs. b (brown) are two autosomal pairs
of alleles and show dominance as indicated (capitalized alleles are dominant). Mice from
the true-breeding brown strain are crossed with albinos of genotype ccBB; when the F1’s
were crossed with albinos of genotype ccbb, three phenotypes were produced:
black
102; brown 198; albino 300
a. Explain why one can conclude that the two genes are linked.
b. Calculate the percentage recombination between the two genes.
c. If each of the 102 black offspring is used as a parent in a testcross, what
phenotypes would you expect to appear in the progeny? Explain your
answer.
d. In what proportion would you expect the different phenotypes?
a. Based on information in the question, there is a 2:1:1 albino: brown: black ratio
expected from the testcross. The results does not match the expected ratios,
therefore the alleles of B and C are not segregating independently and are linked.
b. The recombinant genotypes in the progeny are CB/cb (black) and cb/cb (albino).
The albino phenotype is produced by the parental combination cB/cb. Usually, the
rf is determined by dividing the total recombinant phenotypes by the total progeny.
In this case, parent and recombinant genotypes are giving the same phenotype.
Since the number of black progeny should equal the number of recombinant cb,
then doubling the number of black progeny should equal the total recombinant
progeny expected. Therefore the answer is 2*102/600*100=34 m.u..
c+d. The black (CB/cb) parents are heterozygous because they are crossed to a
tester. The expected ratio is is 33% black, 50% white, and 17% brown.
5. A DNA segment, that is radioactively labeled at one end, is partially digested with
the restriction enzyme EcoRI. The resulting fragments are then separated on an
agarose gene. This gel is exposed to X-ray film and you then obtained the
following autoradiogram: a) Draw the restriction map of this DNA segment. Note:
Numbers along the side are sizes in kilobases (kb).
origin
13.5
12
9.5
6.5
0.5
0.5 kb
6 kb
3 kb
2.5 kb
1.5 kb
b) Draw the restriction digest you would observe if instead
of being radioactively labeled, the restriction digest was
stained with ethidium bromide.
You would get the bands of length:
0.5, 6, 3, 2.5, 1.5
6.5, 9, 5.5, 4
9.5, 11.5, 7
12, 13
13.5
c) When ethidium bromide is used to stain the DNA, the bands with larger DNA
fragments appear darker or more intense than bands containing smaller sized DNA
fragments. In the autoradiogram, all the bands occur with equal intensity. Assuming
that all the bands have the same number of DNA molecules can you explain this
result?
Ethidum bromide intercalates between the paired nucleotides in a double stranded DNA molecule.
The longer the DNA molecule the more ethidum bromide is bound and the brighter the band will
appear. With the autoradiogram each DNA molecule is labelled once on each end. Consequently, all
the bands appear with the same intensity if they have the same number of molecules.
6. Locus A has four different alleles with restriction sites for BamH1 as shown in the
following diagram. Sizes between each restriction site is given in kilobases (kb).
2.5
Allele A1
Allele A2
1
Allele A3
1
Allele A4
1
1.5
2
0.5
2
0.5
3.5
1.5
0.5
2.5
Probe
You design a probe for the locus of interest (locus A) and perform a southern blot on
restriction digested DNA. Can you determine the genotypes of individuals X, Y and Z
based on the gel you obtained? (Gel is shown below, numbers along the side are sizes in
kb).
Ind X
Ind X(A2/A2)
Ind Y(A2/A3)
Ind Z(A1/A3)
Ind Y
Ind Z
3.5
3
2.5
2
1.5
1