The document discusses various genetic experiments involving diploid organisms, including inheritance patterns, gene mapping via linkage analysis, and mutation effects. It describes crossing plants and insects, as well as experiments with E. coli to explore genetic recombination and mutations. Additionally, it covers concepts like interference in genetic crossing, the mechanics of conjugation, and examples of Mendelian traits impacted by nondisjunction.

1. Problems:
2. This is a map for a diploid plant:
R--------35--------E--------17---------F
One parental plant is R e f/r E F. It is mated with a homozygous recessive plant: r e f/r e f. Give
the proportion of the offspring assuming there is both no interference, and complete interference.
3. In a newly discovered insect, the three genes R, A, and P located on the same chromosome
(R= red body, A= spotted coloring, P=purple eyes, r=white body, a=solid coloring, p=white
eyes). A heterozygous parent(R A P/r a p) is mated with a homozygous recessive parent (r a p/ r
a p).
The following offspring are shown below:
(210) dominant for all traits
(16) red body, solid coloring, purple eyes
(70) red body, spotted coloring, white eyes
(50) red body, solid coloring, white eyes
(200) recessive for all traits
(55) white body, spotted coloring, purple eyes
(13) white body, spotted coloring, white eyes
(72) white body, solid coloring, purple eyes
Perform a three-point linkage to determine the gene order and relative distance between the
genes. How did the distance between the 2 genes affect the frequency of crossing-over events?
4. Here are four recombinant sequences that came from four Hfr matings during an interrupted
mating experiment. Determine the order and relative distance by setting A to 0/100 minutes.
Hfr1) A T1 H T2
8 26 33 53
Hfr2) E1 T1 E2 E3
4 12 24 54
Hfr3) T2 S T3 A
7 37 52 62
Hfr4) T3 C E1 E3
2 17 22 72
5. In a u-tube experiment, a researcher placed a kanamycin resistant strain of E. coli on one side
of a membrane impervious to the cells, and a wild-type strain on the other side. After a day, he
then plated a sample from both sides of the tube on plates containing kanamycin and observed
growth on both plates. He then repeated this experiment, but this time he added DNase to the
tube and observed the same results when he plated both cell types. What happened inside the

2. tube?
6. Pretend the following is a chromosome that underwent some changes. What were they? (the
dot is the centromere).
A B C D E F . G H I J A B C G . F E H I J
a. pericentric inversion, deletion
b. paracentric inversion, deletion
c. deletion, translocation
d. Robertsonian translocation,
pericntric inversion
7. Given:
Type of mutation Cistron A or B
1. point A
2. point A
3. point B
4. point B
5. deletion A
Results from 5 crosses:
1x2 (plaque) 2x3 (plaque) 3x4 (plaque) 4x5 (plaque)
1x3 (plaque) 2x4 (plaque) 3x5 (plaque)
1x4 (plaque) 2x5 (no plaque)
1x5 (plaque)
8. The following nucleotide sequence is found in a piece of DNA
5’-TTAACA-3’
3’-AATTGT-5’
if the cytosince undergoes a tautameric shift for one replication cycle draw two round of
replication and explain what type of mutation occurred (Transition or Transversion)?
Bonus:
Explain the relationship between mosaicism and x-inactivation using an example.
9. In a diploid fly, the three loci R E and D are linked as follows
R--------------22cm------------E----------16cm-------------D
One fly is available to you and its genotype is R e D /r E d
If you cross the parent with a homozygous recessive fly, what is the proportion of the progeny
given
No Interference
Complete Interference
Genotype

3. Proportion
Genotype
Proportion
What is Interference?
10. Explain conjugation and what 3 cell types are involved in this process. Draw pictures.
11. The ability to roll your tongue into a TACO shape is an X-linked recessive trait. A young boy
with Klinefelter syndrome has a TACO tongue. His normal (XY) brother also has a TACO
tongue and no non disjunction. The boy’s father has a TACO tongue but his mother sadly does
not have a TACO tougue. Where could have non-disjunction occurred?
Draw the meiotic event that caused the non-disjunction.
No Interference
Complete Interference
Genotype
Proportion
Genotype
Proportion
Solution
2a) Complete Interference = No double crossover
Total map units = 35 + 17 = 52 m.u
52 m.u = 52% recombination
Remaining 48% parental - type offspring
48 /2 = 24/100 = 0.24
Parental type offspring (female) - 0.24
Parental type offspring (female) - 0.24
Recombinant between R and E = 35/2 = 17.5/100 = 0.175 - Single Crossover (SCO)
Recombinant between R and E = 35/2 = 17.5/100 = 0.175 - SCO
Recombinant between E and F = 17/2 = 8.5/100 = 0.085 - SCO
Recombinant between E and F = 17/2 = 8.5/100 = 0.085 - SCO
2b) No interference = Two double crossover
The expected double crossover is simply the chance of recombination in one interval multiplied
by the chance of recombination in another interval
0.17 x 0.35 = 0.0595 x 100 = 5.95 % DCO
Total recombination = SCO + DCO = 46.05 + 5.95
Parental offspring will be 48 (P) + 5.95 (DCO) = 53.95

4. Corrected P = 53.95/2 = 26.975/100 = 0.26975
Parental type offspring (male) = 0.26975
Parental type offspring (female) = 0.26975
Recombinant between E and F = (17- 5.95)/2 = 5.525/100 = 0.05525 - SCO
Recombinant between E and F = (17-5.95)/2 = 5.525/100 = 0.05525 - SCO
Recombinant between R and E = (35-5.95)/2 = 14.525/100 = 0.14525 - SCO
Recombinant between R and E = (35-5.95)/2 = 14.525/100 = 0.14525 - SCO
Double cross over between R and E = 5.95/2 = 2.975/100 = 0.02975
Double cross over between R and E = 5.95/2 = 2.975/100 = 0.02975