Problem 1
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The haploid human genome is ~3x109 base pairs. If we assume the final estimate of human
genes will be 100,000 (It is currently ~20687 protein coding for Homo sapians and — 17,387 for
C. elegans. Of those genes the function for ~5898 are unknown. Note: humans have between
10-100 trillion cells and C. elegans have 959 cells.) and that the average human gene coding
sequence is 2kb, what can be concluded about the average spacing between genes measured
from the central point of one gene to the central point of the next. What can be concluded about
the proportion of the genome that codes for protein? What occupies the noncoding regions?
The E. coli DNA polymerase III makes a mistake ~1 in 108 nucleotide. After DNA replication
approximately how many cells out of 100,000 cells would E. coli DNA polymerase III make a
nucleotide mistake in the gene coding sequence of the haploid human genome. The genome of
the bacterium nicenumberas is 1000 kb in size, sequencing has shown it has 1000 genes and it
has an average gene size of 1kb. After DNA replication approximately how many cells out of
100,000 would E. coli DNA polymerase III make a mistake in a gene of the nicenumberas
bacterium genome.
Answer Problem 1
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The average spacing between central points of adjacent genes must be
3,000,000kb/100,000genes which equal 30 kb/gene
Since the protein-coding segment is only 2kb, it is ~2kb/30kb or 6% of the
total sequence
The remaining sequence must be composed of regulatory regions adjacent
to genes, as well as repetitive DNA and nonrepetive spacer DNA between
adjacent protein-coding regions and introns within the protein-coding
regions
Since only 6.667% of the genome is protein-coding, only 2x108 nucleotides
code for protein (0.0667x3x109 or 2000*100,000). Because E. coli DNA
polymerase makes a mistake approximately every 1x108,~2 genes would
have a mistake in each cell (so of the 100,000 cell approximately every cell
would have roughly two genes with a mistake in their sequence).
For nicenumberas 100% of genome is genes
(1000kb/1000gene=1000bp/gene so a 1000 bp
gene/1000bp/gene*100=100%) In 100,000 approximately 1000 cells would
have an error in one of its genes. (100,000 cellx1,000,000 bpx1/1x108
errors/bp)
Problem 2
Draw a pedigree from the words of the song I’m own grandpaw and indicate the degrees of separation from the proband
Many, Many years ago when I was twenty-three
I was married to a widow who was pretty as could be.
This widow had a grown up daughter who had hair of red
My father fell in love with her and soon, they too, were wed
Chorus
This made my dad my son-in-law, and changed my very life.
For my daughter was my mother cause she was my father’s wife.
To complicate the matter even though it brought me joy,
I soon became the father of a bouncing baby boy.
Chorus
My little baby then became a brother-in-law to Dad,
And so became my uncle, though it made me very sad.
For if he was my uncle then that also made him brother
Of the widow’s grown-up daughter who, of course, was my stepmother.
Chorus
Father’s wife then had a son who kept them on the run.
And he became my grandchild for he was my daughter’s son.
My wife is now my mother’s mother and it makes be blue
Because, although she is my wife, she’s my grandmother too.
Chorus
Oh, if my wife is my grandmother, then I am her grandchild
And every time I think of it it nearly drives me wild
For now I have become the strangest case you ever saw;
As husband of my grandmother I am my own grandpaw.
Chorus
I’m my own grandpaw
I’m my own grandpaw
It sounds funny, I know,
But it really is so;
I’m my own grandpaw
Answer to problem 2
Degrees of separation based on number of different meiosis from proband
1o
1o
I was married to a widow who was pretty as could be
My father fell in love with her and soon, they too, were wed
Proband
I soon became the father of a bouncing baby boy
1o
2o
This widow had a grown up daughter who had hair of red
Father’s wife then had a son who kept them on the run
Problem 3
The accompanying pedigree was obtained for rare kidney disease. From the
pedigree deduce what the mode of inheritance is, state your reasons. If
individuals 1 and 2 marry, what is the probability that their first child will have the
kidney disease.
1
2
Answer Problem 3
From the pedigree deduce what the mode of inheritance is, state your reasons
Likely to be autosomal recessive. Affected individuals
inherited from unaffected parents. It affects both males and
females inherited the disease
1
what is the probability that their first child will have the kidney disease.
2
Answer Problem 3
If individuals 1 and 2 marry, what is the probability that their first child will have
the kidney disease.
Both parents must be heterozygous for them
to have a ¼ chance of passing the disease
For individual 2, the parent is heterozygous
because her father in homozygous
For individual 1, has ½ chance because her
father is heterozygous
1
So the chance that any offspring will be
affected is 1*1/2*1/4=1/8
2
Problem 4
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Considering the accompanying pedigree of a rare autosomal recessive disease, PKU.
A. List genotypes of many of the family members as possible
B. If individuals A and B marry, what is the probability that their first child will have PKU
C. If their first child is normal, what is the probability that their second child will have PKU?
D. If their first child has the disease, what is the probability that their second child will be affected?
I
II
III
IV
A
B
Answer Problem 4
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A. List genotypes of many of the family members as possible
Because the disease PKU is a rare autosomal recessive disorder the affected males must be homozygous.
Because the disease is rare it is a good assumption that the individuals marring into the family are homozygous for
the wildtype
P/P
p/p p/p P/P
I
P/P
II
P/p
P/p
P/P
III P/P
For II because the mothers are
homozygous for the wildtype
and the fathers are
homozygous for the disease
gene they must be
heterozygous
IV
p
P/?
P
P/p
p A
P/p
P
P/p
P/p
B
The remaining family member could be
homozygous for the wildtype or heterozygous
Answer Problem 4
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B. If individuals A and B marry, what is the probability that their first child will have PKU
We know that for the grandmother of individual A is heterozygous and the grandfather is homozygous. Therefore the
chance that individual A’s father is heterozygous is ½. We know that individuals mother is homozygous. If individual A
father (1/2 chance) has the mutant allele he has ½ chance of passing the allele on. Therefore individual a has
½*1/2=1/4 of being a carrier of the mutant allele.
P
P
P/P
p
P/p
P
P/P
P/p
I
II
P/P
1/2
P/p
III
1/2 P/p
¼ P/p
IV
A
B
Answer Problem 4
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B. If individuals A and B marry, what is the probability that their first child will have PKU
We know the grandmother and grandfather of individual B are heterozygous. Therefore the chance that individual B’s
mother is heterozygous is 2/3. We know that individual B’s father is homozygous. If individual B mother (2/3 chance)
has the mutant allele she has ½ chance of passing the allele on. Therefore individual B has 2/3*1/2=1/3 of being a
carrier of the mutant allele.
P
P
P/P
p
P/p
p
P/p
p/p
X
I
¼ chance to be homozygous
P/p
II
2/3
P
P
P/P
p
P/p
p
P/p
p/p
2/3 P/p
III
1/3 P/p
¼ P/p
IV
A
B
1/48
So if individual A is heterozygous ¼ and individual B is heterozygous 1/3. The chance of having a homozygous
child is ¼*1/3*1/4=1/48
Answer Problem 4
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C. If their first child is normal, what is the probability that their second child will have PKU? If the
first child is normal we do not gain any additional information, so the probability of having an
affected child is still 1/48
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D. If their first child has the disease, what is the probability that their second child will be affected?
If the first child has the disease we know that both parents must be heterozygous, therefore the
chance for the second child is affected is ¼.
I
II
III
IV
A
B