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, redXshort, 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 R

t r
t r /

t r

t r
t R /

t R

t r
T r /

T r

t r

Parental: Tall,Red

Parental: short, white

If Tr/tR (trans), then------

t r
T r /

T r

t r
t R /

t R

t r
T R /

T R

T r
t r /

t r

t r

Parental: Tall,white

Parental: short, red

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, wideSo, the genotype of the tall, red

19 short, white, wideindividual 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 r

t 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 r /
T r
T r / t R
T r /

T R

T r / t R
T r
t R / T r
t R
/ t R
t R /

T R

t R / t R
t R

T R

/
T r
T R / t R
T R /

T R

T R / t R
T R
t R / T r
t R / t R
t R /

T R

t R / t R
t R

NOTICE!!

------: recombinant

------: non-recombinant

So, there are a set of individuals who have inherited

▪---/--- (both non-recombinant chromosomes)

▪---/--- (one non-recombinant 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%