Download Chapter 6 - University of Hawaii System

WEEK OF 10/16:
Chapter 6:
Question #7
A pure-breeding strain of squash that produced disk shaped fruits was crossed with a pure
breeding strain that produced long fruits. The F1 had DISK fruits, but the F2 showed a new
phenotype, sphere, and was composed of the following proportions:
Disk
Sphere 178
Long
270
32
Propose an explanation for these results, and show the genotypes of the P, F1, and F2
generations.
1-Ask yourself whether this deviation in the ratio could be due to linkage! Hmm…Nope!
Because in linkage, you don’t ever see a NEW phenotype introduced in the F2 generation. Also,
the F2 progeny were generated by selfing the F1 individuals, and a cross to test linkage is
always a testcross, right? 
2-So, then maybe this is a deviation of a Mendelian ratio, namely the 9:3:3:1 phenotypic ratio.
How can you figure out what the deviation is?
a. Sum all the individuals and determine the number of progeny you would EXPECT to
see if this were a 9:3:3:1 ratioTotal: 480480/16=30
So, the expected 9:3:3:1 would be 270:90:90:30 individuals. Do we see that? Well, what the
data tells you is that it appears the two-3/16th classes have been grouped together so that what
you are actually seeing is a 9:6:1 phenotypic ratio!
3- Okay, so what genotypes are associated with this ratio?!
Go back and think about the genotypes that are associated with the 9:3:3:1 phenotypic ratio!
ALWAYS remind yourself of this when answering the problems from Chapter 6!
9 A_B_  disk
3 aaB_
3 A_bb
 sphere
1 aabb  long
So, by recalling what genotypes are associated with the 9:3:3:1 phenotypic ratio, you have just
SIMPLIFIED the problem solving! Now you know which genotypes correspond with the
phenotypes in the F2.
So, Pgen disk X long (remember that they’re PURE BREEDING)
AABB X aabb
F1
all AaBb (disk)self to obtain the F2selfing 2 dihybrids will usually give you a
9:3:3:1 phenotypic ratio, but as we see from the data given in the problem, this ratio is
presented in a 9:6:1 phenotypic ratio
Chapter 6 (contd.)
Question #14:
The petals of the plant Collinsia parviflora are normally blue, giving the species its common
name “blue-eyed Mary”. Two pure-breeding lines were obtained from color variants found in
nature; the first line had pink petals and the second line had white petals. The following crosses
were made between pure lines with the results shown:
Parents
Blue X White
Blue X Pink
Pink X White
F1
Blue
Blue
Blue
F2
101 blue, 33 white
192 blue, 63 pink
272 blue, 121 white, 89 pink
a.) Explain these results genetically. Define allele symbols you use and show the genetic
constitution of the parents, F1, and F2.
This question is generally an extension of Question 7!
Look at the data from the first 2 crosses. Blue in the F1 for both of these implies that blue is
dominant to BOTH pink and white. Also, notice that the phenotypic ratio in the F2 for both
crosses rings bells of a MONOHYBRID phenotypic ratio (where we’re dealing with ONE gene).
BUT, now look at the third cross!! Not only are the F1 all BLUE, but the ratio of the F2 does not
imply that there is only ONE gene at play here. Here is another case of a deviation of the
9:3:3:1 ratio!
So, calculate the EXPECTED number of progeny for each phenotype if this were a 9:3:3:1
270:90:90:30. Now, compare with the data and identify the modified ratio! The data appear to
fit a 9:4:3 ratio.
Recall the genotypes associated with the 9:3:3:1 phenotypic ratio:
9 A_B_  blue
3 aaB_  pink
3 A_bb
1 aabb
 white
NOTE: It doesn’t matter which of the 3/16th ratios you choose to include in the “4” part of the
ratioIt all works out in the end…Try it out! See what happens when you choose to group
aaB_ and aabb, then see what happens when you choose to group A_bb and aabb. You’ll see
that it doesn’t make a difference.
So, now we can go back to each of the crosses and decide what the genotypes of the
individuals are (it might also help explain WHY there was the 3:1 ratio)
Cross 1—blue X white (PURE BREEDING)
Blue = AABB ; white = either AAbb or aabb
Pgen AABB X AAbb
F1
all AABb (blue)
F2
¾ B_AA (blue)
¼ bbAA (white)
OR
AABB X aabb
all AaBb (all blue)
9:4:3
So, the genotypes of the parents in Cross 1 MUST be AABB X AAbb
Cross 2—blue X pink (PURE BREEDING)
Pgen AABB X aaBB
F1
all AaBB (blue)
F2
¾ A_BB (blue)
¼ aaBB (white)
Cross 3—pink X white (PURE BREEDING)
Pink = aaBB ; white = either aabb or AAbb
Pgen aaBB X aabb
OR
aaBB X AAbb
F1
all aaBb (pink)
all AaBb (all blue)
F2
¾ B_aa (pink)
9:4:3
¼ bbaa (white)
So, the genotypes of the parents in Cross 1 MUST be aaBB X AAbb
b.) A cross between a certain blue F2 plant and a certain white F2 plant gave the following
progeny :
3/8 blue
1/8 pink
½ white
What must the genotypes of these 2 F2 plants have been?
Since the question does NOT state that the individuals are PURE BREEDING, you cannot
assume that they are…So,
Blue= AABB or AABb or AaBB or AaBb
White = AAbb or Aabb or aabb
▪white will always contribute “bb”
If White:
If Blue:
AA
Aa
AA
all AA
all A_
Aa
all A_
¾ A_, ¼ aa
REMEMBER! In order to get PINK, there
must be “aa”
So, we know that the blue individual MUST be Aa and the white individual MUST be Aa in
order to get pink progeny…
If White:
If Blue:
bb
BB
Bb
all Bb
REMEMBER! In order to get a WHITE, there must be “bb”
¾ B_, ¼ bb
So, we know that blue individual MUST be Bb and the white individual can only be bb in
order to get white progeny…
Put all of this information together!! Blue X White AaBb X Aabb
¾ A_ 3/8 A_Bb (blue)
½ Bb
¼ aa  1/8 aaBb (pink)
¾ A_  3/8 A_bb (white)
½ bb
½ white
¼ aa  1/8 aabb (white)
Chapter 7
Question #9
Describe the Streisinger model for frameshift formation. Show how this model can explain
mutational hotspots in the lacI gene of E. coli.
DON’T PANIC!!  Look up Streisinger in your glossary…It should direct you to page 206 of
your text along with pages 210-212. This model describes slipped mispairing!! Pay special
attention to Figure 7-9 (it explains HOW the model works to create additions and deletions) and
the figures on page 211 (these explain how this mechanism creates addition and deletion
mutations in a specific bacterial gene).
But this model does not only apply to bacterial genes, right?? Think about some human
disorders about which you’ve read and Dr Cann lectured! See pages 213-215 for a description
of how slipped mispairing has been proposed as the mechanism for trinucleotide repeat
disorders!!
Question #14
Describe the repair systems that operate AFTER replication.
There’s a very nice table provided in this chapter (Table 7-3, page 220) that tackles the repair
systems. There’s also reading about post-replication repair mechanisms on pages 217-219.
Question #21
A certain compound that is an analog of the base cytosine can become incorporated into DNA.
It normally hydrogen bonds just like cytosine, but it quite often isomerizes to a form that
hydrogen bonds LIKE thymine. Do you expect this compound to be mutagenic and, if so, what
types of changes might it induce at the DNA level?
Remember that a base analog is a chemical whose structure MIMICS that of a DNA base (but it
is NOT a proper base).
So CG, but if the analog isomerizes and starts bonding LIKE thymine analogA
What once had base pairing properties of a C, now has properties of a T. This is what is known
as a CGT=A TRANSITION mutation. It’s a transition because C and T are both pyrimidines;
a transversion mutation would (in this case) alter CG A=T. In this case, C and A are not the
same type of base.