6. A research group has selected three independent Trp−haploid strains of Neurospora, each of which cannotgrow in the absence of the amino acid tryptophan.They first mated these three strains with a wild-typestrain of opposite mating type, and then they analyzedthe resultant octads. For all three matings, two of thefour spore pairs in every octad could grow on minimalmedium (that is, in the absence of tryptophan), whilethe other two spore pairs were unable to grow on thisminimal medium.a. What can you conclude from this result?In the matings of mutant strains 1 and 2 with wild type,one of the two topmost pairs in some octads had sporesthat could grow on minimal medium while the other ofthe two topmost pairs in the same octads had spores thatcould not grow on minimal medium. In the mating ofmutant strain 3 with wild type, either all the spores inthe two topmost pairs could grow on minimal mediumor all could not grow on minimal medium.b. What can you conclude from this result?The researchers next prepared two separate cultures ofeach mutant strain; one of these cultures was of matingtype A and the other of mating type a. They matedthese strains in pairwise fashion, dissected the resultantoctads, and determined how many of the individualspores could grow on minimal medium. The resultsare shown here.% of octads with x number of spores viableMating on minimal mediumx = 0 2 4 6 81 × 2 78 22 0 0 01 × 3 46 6 48 0 02 × 3 42 16 42 0 0c. For each of the three matings in the table, howmany of the 100 octads are PD? NPD? T?d. Draw a genetic map explaining all of the precedingdata. Assume that the sample sizes are sufficientlysmall that none of the octads are the result ofdouble crossovers.e. Although this problem describes crosses inNeurospora, it does not help in this case to presentthe matings in the table as ordered octads. Why not?
Genetic Recombination
Recombination is crucial to this process because it allows genes to be reassorted into diverse combinations. Genetic recombination is the process of combining genetic components from two different origins into a single unit. In prokaryotes, genetic recombination takes place by the unilateral transfer of deoxyribonucleic acid. It includes transduction, transformation, and conjugation. The genetic exchange occurring between homologous deoxyribonucleic acid sequences (DNA) from two different sources is termed general recombination. For this to happen, an identical sequence of the two recombining molecules is required. The process of genetic exchange which occurs in eukaryotes during sexual reproduction such as meiosis is an example of this type of genetic recombination.
Microbial Genetics
Genes are the functional units of heredity. They transfer characteristic information from parents to the offspring.
6. A research group has selected three independent Trp−
haploid strains of Neurospora, each of which cannot
grow in the absence of the amino acid tryptophan.
They first mated these three strains with a wild-type
strain of opposite mating type, and then they analyzed
the resultant octads. For all three matings, two of the
four spore pairs in every octad could grow on minimal
medium (that is, in the absence of tryptophan), while
the other two spore pairs were unable to grow on this
minimal medium.
a. What can you conclude from this result?
In the matings of mutant strains 1 and 2 with wild type,
one of the two topmost pairs in some octads had spores
that could grow on minimal medium while the other of
the two topmost pairs in the same octads had spores that
could not grow on minimal medium. In the mating of
mutant strain 3 with wild type, either all the spores in
the two topmost pairs could grow on minimal medium
or all could not grow on minimal medium.
b. What can you conclude from this result?
The researchers next prepared two separate cultures of
each mutant strain; one of these cultures was of mating
type A and the other of mating type a. They mated
these strains in pairwise fashion, dissected the resultant
octads, and determined how many of the individual
spores could grow on minimal medium. The results
are shown here.
% of octads with x number of spores viable
Mating on minimal medium
x = 0 2 4 6 8
1 × 2 78 22 0 0 0
1 × 3 46 6 48 0 0
2 × 3 42 16 42 0 0
c. For each of the three matings in the table, how
many of the 100 octads are PD? NPD? T?
d. Draw a genetic map explaining all of the preceding
data. Assume that the sample sizes are sufficiently
small that none of the octads are the result of
double crossovers.
e. Although this problem describes crosses in
Neurospora, it does not help in this case to present
the matings in the table as ordered octads. Why not?
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