Chapter 5:
Question #21
The Neurospora hist-1 gene, which is on the right arm of chromosome 1, is involved in
the biochemical pathway for the synthesis of histidine.
Two mutations of the hist-1 gene were obtained in separate experiments—
hist-1a and hist-1b. Both left the strains auxotrophic—that is, they would not grow
unless histidine was added to the growth medium. These mutant strains were stable
and never reverted to the wild type.
1
2
3
4
The 2 mutants were crossed together (hist-1a X hist-1b) and thousands of ascospores
were spread onto medium lacking histidine. Most ascospores did not grow, but THREE
colonies were observed.
Draw a diagram to account for the origin of these rare prototrophic colonies!!
What you have to remember when solving this problem is that when recombination
occurs in Neurospora, the ORDER of the ascospores in the sac is what changes (from a
4:4 arrangement to 5:3, 3:5, 3:1:1:3, 6:2, 2:6). See page 152 of your text for an
explanation of this. So, general recombination does not cause a mutant allele to
become wild type.
BUT!! There is a RARE recombination event that COULD result in creation of a wild
type spore. It is called INTERGENIC RECOMBINATION; it results from recombination
WITHIN a gene. These occur at a frequency of 1/1000 to 1/1000000 and, where this
problem is concerned, would result in the following:
Double Mutant
WILD TYPE
Question #25: Pattern Recognition Problems:
These are fairly simple questions!! You should be able to assign genotypes based on
the pattern of phenotypes observed!!
Question #31:
In beans:
-tall (T) is dominant to short (t),
-red flowers (R) are dominant to white flowers (r), and
-wide leaves (W) are dominant to narrow leaves (w)
The following cross is made and progeny obtained are shown:
Cross: tall, red, wide X short, white, narrow
478 tall, white, wide
21 tall, red, wide
19 short, white, wide
482 short, red, wide
a.) Explain why these progeny phenotypes were obtained and in the proportions
observed (list all genotypes and show chromosomal positions gene map)
1.) Three genes, so EXPECT 2n phenotypes8 phenotypes, but only see FOUR. Four
phenotypes is reminiscent of how many genes?? 2 genes
2.) In all of the progeny, only the WIDE phenotype is seen despite one of the parents
being narrow! Does this scenario ring bells? Wide and narrow parents, but only WIDE
progeny?? Should tell you that wide must be dominant to narrow.
This information (all progeny WIDE) should also lead you to realize that 2 of the 3 genes
being observed here are LINKED! The gene for HEIGHT and FLOWER COLOR appear
to be linked, while the gene for leaf shape appears to be assorting independently.
3.) So, we realize that 2 of the 3 genes are linked, so we can attempt to assign
GENOTYPES to the parents.
▪We can deduce that the wide parent must be WW and the narrow parent must be ww,
so that all of their progeny are Ww (wide). So, we can think about the 2 linked genes
separately…
▪The short, wide parent is obviously t r/t r since both genes are recessive. But, the
genotype of the tall, red parent could be TR/TR, TR/tr, or Tr/tR. So which one is it??
Question #31 (contd)…
Tall, red
X
short, wide
t
r
If TR/TR, then -------------------- all progeny would be TR/tr (tall, red)
NOPE!!
If TR/tr (cis), then----------------
t
r
T
R
T
t
R
r
Parental: Tall,Red
t
r
t
t
r
r
Parental: short, white
t
R
t
R
r
T
t
r
r
t
r
T
r
t
If Tr/tR (trans), then----------------
T
r
T
t
r
r
Parental: Tall,white
t
R
t
t
R
r
Parental: short, red
T
t
R
r
T R
T
r
t
r
t
r
Looking back at the data, the progeny that represent non-recombinant/parental types
are: 478 tall, white, wide
 tall, white and short, red
21 tall, red, wide
So, the genotype of the tall, red
19 short, white, wide
individual is T r/t R !! 
482 short, red, wide
NOTE!! It makes a difference if the arrangement of linked genes is cis or trans!!
So, now that we know the genotype of the parents, we can assemble a GENE MAP:
What information will you need to be able to assemble the map?? Recall that
Recombination Frequency (RF) = # of recombinants / total
So, recall the data:
478 tall, white, wide
21 tall, red, wide
19 short, white, wide
482 short, red, wide
Which classes are the recombinant classes?? You should have guessed tall, red and
short, white. So, RF = 21+19/100040/1000=4% 4 map units or 4cM
T
t
r
R
4 map units
b.) Under your hypothesis, if the tall, red, wide parent is SELFED, what will be the
proportion of short, white, wide progeny?
T r▪ W X
t R W
T r▪ W
t R W
XXXXXXXX
XXXXXXXX
T
r
t
R
T
R
t
R
T
T
T
r
r
t R
T r
T
T
R
r
t
T
R
r
T r
t R
t
t
R
R
T R
t R
t
t
R
R
T
r
t
R
T
t
R
T
R
T
R
T
R
T
R
T
t
r
R
t
t
R
R
T
t
R
R
t
t
R
R
t
T
t
r
R
R
R
R
NOTICE!!
------- : recombinant
-------: non-recombinant
So, there are a set of
individuals who have
inherited
▪---/--- (both nonrecombinant
chromosomes)
▪---/--- (one nonrecombinant and one
recombinant)
▪---/--- (both recombinant
chromosomes)
So, the short, white progeny is created from two recombinant chromosomes.
Remember that the percent recombination is 4%! Since only 2 of the 4 chromatids
recombine at a single chiasma, the actual frequency would be 2% for every
recombinant chromosome.
So, the probability of getting t R/t R is p(tR) * p(tR) (2%)(2%) 0.0004=0.04%
Since all the progeny will be WIDE from this cross, the
p(W_) = 1, so (0.04%)(1) = 0.04%