A cDNA is a double stranded DNA made from a mRNA

							BIMM110        Human Molecular Genetics                                            Spring 2004
I.E. Scheffler

                                      MIDTERM EXAM
                                         ANSWERS

There are 11 questions. All answers should be written in the blue book. Leave the first inside page
blank for scoring and write a column of numbers from 1 to 11
If you want your blue book returned in the hallway near my office, please sign the waiver on the back
of the blue book.

                                  *****************************

QUESTION 1 (12 points)

a) What is the distinction between a cDNA and a piece of genomic DNA containing some of the same
sequences?

A cDNA is a double-stranded DNA made from a mRNA; it contains all the exons (5’UTR, ORF,
3’UTR). The mRNA is the product of a transcription and splicing reactions
A genomic DNA with some of the same sequences represents the gene from which the mRNA was
transcribed; it contains exons and introns.

b) What is the distinction between a reverse transcriptase and a DNA polymerase?

Reverse transcriptase uses an RNA as a template to make the complementary (deoxy) strand; DNA
polymerase uses DNA as a template to make the complementary deoxy strand.

c) What is the distinction between an endonuclease and a restriction enzyme?

A general endonuclease cleaves a DNA (or RNA) randomly and in the interior of the strand. A
restriction enzyme makes highly selective breaks (blunt end or staggered) at a specific recognition site
that is unique for each restriction enzyme.

QUESTION 2 (12 points)

a) A circular plasmid vector with a poly-linker is used to clone and amplify a particular human cDNA.
Draw a highly schematic diagram of the construct, labeling some of the essential elements of the
vector.

The drawing should show a circular DNA with the following elements: the cDNA segment, flanked by
short segments from the poly-linker, an origin of DNA replication on the plasmid and a selectable
(drug-resistance) marker on the plasmid.

b) What is a cDNA library?

A cDNA library is a collection of plasmids each containing a different cDNA made from all the
mRNAs of a particular tissue or cell type.
QUESTION 3 (12 points)

a) What are two methods by which a radioactive probe can be produced corresponding to the human
cDNA cloned in Question 2?

1) nick translation: a limited number of single strand breaks are made by an endonuclease, and then
DNA polymerase is used with deoxynucleotide triphosphates (one of which is labeled) to move the
nick by digesting the 5’ end and extending the 3’ end

2) Use of random primers: the probe is denatured and annealed with a large excess of small (~ 6 nt)
oligonucleotides of random sequences. Some of these primers will find a short complementary DNA
segment and anneal, and then a polymerase and dXTPs can extent the primers

3) End-labeling is also acceptable, although it is less common, because relatively little label is
incorporated

b) This probe is to be used in conjunction with a Northern blot to study gene expression in skeletal
muscle. What are we looking for in such an experiment?

A Northern blot is used to detect the presence of a specific mRNA in a population of mRNAs from a
specific tissue or cell population. The mRNAs are separated according to size by electrophoresis,
blotted to paper and the specific probe is used to see if an mRNA with a sequence complementary to
the probe is present.

c) The same probe can also be used on a Southern blot with a restriction digest of genomic DNA from
lymphocytes. Several bands are detected. Explain.

In a Southernblot we are looking at genomic DNA fragments obtained by a digestion of genomic DNA
with a specific restriction enzyme. There may be several bands, if the gene has several exons and if
there are restriction sites present in either the exons or in the introns.

QUESTION 4 (12 points)

A mouse gene is known to have three exons, as shown in the following diagram. A knock-out vector is
constructed to be used with ES cells to inactivate the gene. The HSV-TK gene is included in the
vector, but we select against its retention with a drug called gancyclovir.
a)(4) Draw this diagram into your blue book and indicate where homologous recombination would
have to take place to achieve our goal of inactivating the gene.

Recombination has to occur on the left between the homologous segments of exon 1 and on the right
between the homologous segments of exon 3.
Exon 2 would be lost and replaced with the selectable marker; the HSV-TK gene should also be lost.


b)(8) A potential ES cell line with the desired modification is to be verified by the use of a Southern
blot. Sites B1 - B5 are restriction sites for the enzyme BamHI. Indicate in your drawing what you
would use as a probe. Also show a very schematic appearance of the Southern blot with two lanes; the
first with DNA from the ES cells before the modification and the second lane with DNA from the
modified ES cells.

Various probes could be used, but one such probe would be made from exon 3. In the original ES cells
we would see two bands corresponding to the fragments B3-B4 and B4-B5.
If a successful homologous recombination has been achieved with one chromosome, a Southern blot
with DNA from the modified ES cells would show the same two bands from the unmodified
chromosome, and an additional much larger band from cuts at B2 and B4 in the modified
chromosome; B3 is no longer present, because exon 2 has been replaced by the selectable marker


QUESTION 5 (12 points)

Outline the steps and describe the major reagents you would need to characterize a pedigree using
VNTRs ("variable number of tandem repeat polymorphisms") using PCR. Illustrate with a two-
generation pedigree involving three children. The paternity of one of these children is being
challenged. Show the pedigree and the corresponding "raw data" used in the analysis. What kind
of data would establish that the father of the third child is not the real (biological) father

Select oligonucleotide primers from unique regions flanking the tandem repeat region.
Obtain DNA samples from the mother, the biological father, the “other man” (if available), and
from each of the children
Perform a PCR reaction with the pair of primers with each DNA sample. In the most favoarable
case (illustrated in the figure) one would get two bands from each sample (two alleles).The alleles
of the father, mother and “suspect” should all be different, i.e. the bands have different sizes
because the number of repeats between the two primer sis variable. In the figure the numbers
above the individuals in generation I represent the number of repeats in the VNTR
If one of the children has an allele that does not come from the husband, then its paternity could
be in question, i.e. any of the combinations of the left column would suggest legitimacy, any
combination on the right would suggest a liaison with the individual I-3

QUESTION 6 (10 points)

a)(5) Draw a hypothetical pedigree for at least three generations that would be consistent with an
autosomal dominant disease . State briefly what you would be looking for to be reasonably sure of
your conclusion.

Individuals in each generation should be affected (if a statistically sufficient number is available);
Both male and female siblings should be affected.
If the number of siblings is sufficiently large there should be about 50% affected siblings.




b)(5) What would be the most critical criteria for establishing that a disease is due to a recessive X-
linked mutation? Drawing a hypothetical three-generation pedigree would be helpful.




Only male siblings would be affected
The sisters of such an affected male would not be affected, but they may be carriers (heterozygotes)
and have affected sons.
\
QUESTION 7 (6 points)

List and describe at least three well-known structural chromosomal abnormalities; for each, give a
hypothetical example in the short-hand notation used by cytologists.

1) an interstitial deletion in the long arm of chromosome 1: del(1) (q23q31)
2) a balanced translocation between two autosomes, eg. chromosomes 2 and 9:
        t(2;9) (q24; q31)
chromosome 2 is broken in the long arm in region 2, band 4; chromosome 9 is broken in the long arm
in region 3, band 1
3) a pericentric inversion in chromosome 4: inv(4) (p15q27)
chromosome is broken in the short arm at p15 and in the long arm at q27, the piece in the middle (with
the centromere) is rotated 180 degrees and inserted again


QUESTION 8 (10 points)

Cystic fibrosis is a relatively common genetic disease due to mutations in the CFTR locus. The
                                                    Key:

                                                               Cy st ic Fibro sis


                                                                Normal




gene is on an autosome, and the mutations are recessive. Consider the following pedigree.
Individual II-2 is concerned about being a carrier.
    a) Using only her pedigree data, what is the probability that she carries an allele for cystic
        fibrosis?

Since individual II-2 has an affected brother, it is likely that both her parents are heterozygous
carriers. The probability that she is also a carrier is 2/3, but she is not affected.

    b) What is the probability that the child is NOT affected?

(the unborn child, III-1, should have been represented by a diamond symbol)
If the father, II-3, is from the general population, the chance that he is also a carrier is less than
one in a thousand: he is most likely (+/+), and the chance of an affected child III-1 is very low.
QUESTION 9 (12 points)

In the following pedigree, a shaded symbol represents an individual affected with Down
syndrome, and an open symbol represents a phenotypically normal individual. Individuals II-4
and II-5 present for genetic counseling. He is 26 years old and she is 24
years old and pregnant with their first child. They are concerned about the possibility that the
child might have Down syndrome. II-4 has two siblings who are affected. His mother was 25
years old when she gave birth to II-2, and 28 years old when she gave birth to II-3. (Hint: multiple
affected births and the young age of the mother (I-2) suggest a chromosomal abnormality)
Gen
 I




II


III



      a. II-4 and II-5 elect to have amniocentesis performed. A karyotype is determined from cells
         obtained from the amniocentesis, and an abnormal genetic constitution is seen. What is a
         likely problem?

The occurrence of two Down syndrome children in a young family is quite abnormal and suggests
a genetic predisposition in one of the parents I-1 or I-2.
One likely possibility is the existence of a Robertsonian fusion involving chromosome 21 and
another acrocentric chromosome, e.g. 14.
The father, II-4 is normal because he carries a normal 14, anormal 21 and a 14/21 fusion.
However, from meiosis I and subsequent gametogenesis the gamete involved in the fertilization
had a normal 21 and a 14/21 fusion. A normal oocyte from II-5 was fertilized and the resulting
zygote III-1 had three copies of the long arm of 21.


      b. Draw and explain the segregation event during Meiosis I in the father that is the cause of
         the fetus’s abnormal karyotype.
QUESTION 10 (6 points)

Turner's syndrome (45, X) can arise:
       A. only from nondisjunction in the mother
       B. only from nondisjunction in the father
       C. only from nondisjunction in the first meiotic division
       D. only from nondisjunction in the second meiotic division
       E. from nondisjunction in either division in either sex

Klinefelter’s' syndrome (47, XXY) can arise:
       A. only from nondisjunction in the mother
       B. only from nondisjunction in the father
       C. only from nondisjunction in the first meiotic division
       D. only from nondisjunction in the second meiotic division
       E. never from MII nondisjunction in the father
QUESTION 11 (16 points)

   a) Explain the difference between a cloned animal and a transgenic animal expressing a novel
      gene (e.g encoding the green fluorescent protein)

A cloned animal is obtained by introducing a somatic nucleus from a diploid adult cell into an
enucleated oocyte, followed by mitotic divisions first in vitro and then after implantation in a foster
mother.

A transgenic animal is obtained by introducing a foreign gene either into a pronucleus of a fertilized
egg or by transfection of embryonic stem cells. The transfected oocyte can be introduced into a foster
mother and lead to an offspring with the foreign gene integrated into a chromosome at a random
location. The transfected ES cells have to be introduced into a blastocyst and then a chimeric offspring
has to be brought to term and subsequently bred to obtain an animal with the transgene in every
somatic cell.

   b) How could one restrict the expression of the transgene to a particular chosen tissue, e.g. heart
      muscle?

There are well-characterized promoters from genes that are expressed in a highly tissue-specific
manner. Such promoters can be attached to other genes, and such genes then are also subject to the
same regulatory signals for tissue-specific transcription.

A more complicated scheme might be to place a transgenic gene in all cells with a promoter element
that is disrupted by an insert which in turn is flanked by to loxP sites. Mating such a mouse to another
expressing the Cre recombinase in a specific tissue only would lead to the excision of the insert in the
promoter and restoration of promoter activity in only this tissue.

   c) When a gene is knocked out in an ES cell, we are just at the beginning of a procedure towards
      the production of a homozygous (-/-) animal in which the gene product is completely missing.
      Outline the steps that have to be followed to make such an animal, starting with the (-/+) ES
      cell.
   - introduce the modified ES cell into the blastocyst obtained from another fertilized egg
   - implant into a pseudopregnant female
   - live birth of a chimeric animal
   - verify by DNA analysis from a tail snip
   - with luck the modified ES cells will also be used for the development of the gonads, so that the
      chimeric animal will produce gametes with modified chromosomes
   - use gametes/spermatozoa from the chimeric animal to fertilize a normal gamete/oocyte to
      obtain heterozygous animals
   - cross heterozygous animals to obtain homozygous animals in which both chromosomes carry
      the mutation.