Fall SectionSpring Section 1Spring Section 2
Solutions  Problems for Chapter 1 and 2

Bacteria Division Problem
Overall Cell Generation Time Problem
Transgenic Mice
X-linked Inheritance
Gene Product Interaction
Hin Recombinase
Fly Lab
Bacteria problems (Solutions2 page)

Solutions to DNA Problems

1.  5 HCN --> 1 Adenine
Well, it takes a lot of rearranging, but the atoms are all there!  Remember, where adenine would link to the sugar, there is a hydrogen atom:

 2.  The charge on proteins in chromatin (the structural material of chromosomes) would be largely positive, because positively charged proteins can bind to negatively charged DNA.  The DNA has negative charges on its phosphate groups, whose OH groups ionize and give off H+ in water solution at near-neutral pH (the conditions found inside living cells.)

3.  If a DNA sample has 23% guanine, then it has to have 23% cytosine.  The remaining 54% must contain equal amounts of adenine and thymine, or 27% each.

4.  Sequences of DNA which need to come apart for regulatory function, such as initiation of transcription, will tend to have A=T base pairs because they have only two hydrogen bonds.  Guanine-cytosine pairs have three hydrogen bonds, and hold  together more strongly.

5.  A molecule that slides inbetween two base pairs and stays they can cause the sequence to be misread by enzymes trying to replicate the DNA.  An extra base might get inserted into newly synthesized sequence, resulting in a mutation.  (More on this, next week.)

6.  Some species of microbes living at extremely high temperature have evolved to maintain positively supercoiled DNA.  This helps to increase stability of their DNA, which ordinarily tends to melt open more easily at high temperature.

Solution -- Bacteria Division Problem

Bacterial DNA is circular.  In bacteria, as in eukaryotes, replication occurs bidirectionally.  In bacteria, this replication starts at one point attached to the cell wall, and continues in both directions around the cirle.

If the cell is growing very fast, it will start to replicate each daughter chromosome BEFORE COMPLETION of the first replication.  This allows the chromosome to be replicated in one half the time that it would take if it waited till the first round of replication was complete.

Solution -- Overall Cell Generation Time Problem

Assume that all of the cells are evenly distributed throughout the different stages in the cell cycle (G1, S, G2, Mitosis).  If only five percent of the cells are in mitosis, then mitosis must only take five percent of the total time that is required for overall generation.  If five percent of the time is five minutes, then one hundred percent of the time is one hundred minutes.

Solution -- Fly genes

The brown-eyed fly has to be missing a functional allele for a gene B that makes scarlet pigment.  The scarlet-eyed fly has to be missing a functional allele for a DIFFERENT gene locus S that makes BROWN pigment.

Solution -- Transgenic Mice Problem

To simplify the notation, let's convert this to Mendel's peas.
First, which alleles of each gene locus are dominant?

The  transgene containing human globins is dominant, because it provides a product not expressed by the standard mouse, certainly not at the location where the transgene ended up.

The normal mouse alpha and beta genes, however, are dominant to the null alleles created by transgenic allele replacement  (designated by hyphens.)

So let's redraw the genotypes:

TT AA BB  X  tt Aa Bb

What progeny are desired?  Purebreeding for transgene (T);
and for mouse null alleles (a, b):

 TT aa bb

But -- because the human-transgenic parent is pure-breeding, the closest we can get to what we want is this:

 Tt Aa Bb

The proportion of these progeny would be:

 T t    A a    B b

(1)(1)  (1)(1/2)  (1)(1/2)  =  ?

These mice are hybrid for all three genes.
We need to intercross them--a trihybrid cross.
The proportion of desired progeny is:

  T  T     a   a      b  b

(1/2)(1/2)  (1/2)(1/2)  (1/2)(1/2)  =  1/64

Better breed plenty of mice!

Actually, we would be helped out by the fact that all the tt aa bb, tt A- bb, and tt aa B- mice would die before birth (Why?  What proportion?).  But some of the sickle-cell  TT aa bb mice would also die, either shortly after birth, or later before producing offspring.

The moral is, if you want to get into mammalian transgenics—plan to write a big grant to grow all those mice.

Solution --  X-Linked Inheritance

          A   A            a                     A   a       A
        X   X   with  X    Y   ---->     X   X  or  X    Y

          a   a            A                     A   a       a
        X   X   with  X    Y   ---->     X   X  or  X    Y

          A   a            a                     A   a     a   a      A               a
        X   X   with  X    Y   ---->     X   X,    X   X,    X    Y,  or   X    Y

          A   a            A                     A   A     A   a     A               a
        X   X   with  X    Y   ---->     X   X,    X   X,    X    Y,  or  X    Y

Solution --  Gene Product Interaction

1.      The phenotypes of the sweet peas are determined by the interaction between two gene loci.  The presence of two recessive white alleles at one locus of the white flowers that were crossed made useless the presence of a dominant purple allele at the other locus (recessive epistasis).  The ratio of organisms in the self-cross that have dominant purple alleles at both gene loci, to those that have a dominant allele at the first locus, to those that have a dominant allele at the second gene locus,  to those that have no dominant alleles is 9 : 3 : 3 : 1.  The 3 and 3 and 1 all have white flowers because they have at least one double recessive gene locus.  This makes the ratio of purple to white 9:7.

2.    Again, the phenotypes of the fancy fowl are determined by the interaction between two gene loci.  In this case, however, the presence of a dominant feathered allele at one locus compensates for the lack of dominant feathered alleles at the other locus (dominant epistasis).  The ratio of the offspring that had a dominant feathered allele at both, to those that had a dominant allele in the first locus, to those that had a dominant allele in the second locus, to those that had no dominant alleles was 9 : 3 : 3 : 1.  The 9 and 3 and 3 all have at least one dominant allele and are therefore feathered.  This makes the ratio of feathered to non-feathered 15:1.

3.         9 : 4 : 3 --->   AaBbcc
      12 : 3 : 1 --->   AABbCc
           13 : 3 --->   AaBBCc

Solution -- Hin Recombinase

Hin recombinase is involved in site-specific recombination because it catalyzes a DNA inversion in the Salmonella chromosome.  This recombination involves no homology, but involves enzymatic recognition of a particular short DNA sequence.

Solution -- Pedigrees

1.  The trait is recessive, because two parents  who don't show the trait do pass it on to their children.

2.  Dominant, because two parents who show the trait have one child without it; this one child must have inherited "normal" recessive alleles from each parent.

3.  More likely dominant, because if it were recessive, two many people by coinicidence would have to have married to pass on the trait.

4.  Recessive, because two  two parents  who don't show the trait do pass it on to their children.

5.  Autosomal dominant--but a trait that only affects males, such as beard hair color.  It cannot be X-linked because fathers pass it on to sons.

Solution: Fly Lab, Stubble Aristapedia

To determine the map distance between Stubble and Aristapedia, we cross a Stubble Aristapedia (SB AR) female with a wild-type male.  Remember that SB and AR are each lethal when homozygous; so the genotype of the female has to be:

SB    AR                SB     +
------------    or        ------------
 +        +                    +    AR

The male is of course:

+        +
+        +

The actual results of one cross are shown:

Summary of Results

(Female: SB;AR) x (Male: +)
Phenotype Number Proportion Ratio
Female: + 255 0.02515 1.186
Male: + 252 0.02486 1.172
Female: SB 2225 0.21947 10.349
Male: SB 2296 0.22647 10.679
Female: AR 2329 0.22973 10.833
Male: AR 2329 0.22973 10.833
Female: SB;AR 237 0.02338 1.102
Male: SB;AR 215 0.02121 1.000
TOTAL 10138
Recombinants/Total =  959/10138 = 9.5%
(Note: Your numbers may differ in detail, but the same statistical results should apply.)

Note that the female genotype has to be:

SB     +
  +    AR

because the recombinant phenotypes are either all wild type, or both SB and AR.