GENETICS: EXERCISES
CHAPTER 1
FROM GENE TO PROTEIN
1. Example
Gene (DNA):
5’ TCGCC ATG GAA GCA TCA CCC TAG 3’ (positive strand or forward
strand)
3’ AGCGG TAC CTT CGT AGT GGG ATC 5’ (negative strand or reverse
strand)
Alone, the negative strand should be written:
5' CTA GGG TGA TGC TTC CAT GGCGA 3’
messenger RNA:
5’ UCGCC AUG GAA GCA UCA CCC UAG 3’
Protein:
Met–Glu–Ala–Ser–Pro–(stop) (= MGASP)
In the exercises on gene, if there is no any mention, please consider the provided sequence as a
positive strand, direction 5’- 3’.
Example on Chargaff’s rule in a double strand DNA:
5’ T T A T A T G C C C A T G G C C G C A G C C G G 3’
│ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │
3’ A A T A T A C G G G T A C C G G C G T C G G C C 5’
2. Problem
1. This is the sequence of the first exon of a gene (negative strand):
5’ –TCCGCACAGGGTTCCCCAATGCATTTTCCT -3’
Write the complementary DNA sequence (positive strand).
Determine the mRNA and peptide sequences encoded by this gene exon.
1
Solution:
Positive strand
5’- AGGAAAATGCATTGGGGAACCCTGTGCGGA -3’
-
mRNA
5’- AGGAAAAUGCAUUGGGGAACCCUGUGCGGA -3’
-
peptide sequence
Met-His-Trp-Gly-Thr-Leu-Cys-Gly
2.
The following sequence is located at the beginning of a gene.
The underlined part corresponds to an intron, which flanks two exons.
ATTAGCCATGCTCTCCGTCCCAACTGTAAGTATGCGCGAGATCGTTACCAGGATGATTGT…
Determine the sequence of the corresponding mRNA and peptide.
Solution
-
mRNA
5’- AUUAGCCAUGCUCUCCGUCCCAACUGAUGAUUGU -3’
-
peptide
Met-Leu-Ser-Val-Pro-Thr-Asp-Asp-Cys
3.
The following sequence (positive strand) is located in the middle of a bacterial gene:
CTCTCCGTCCCAACTGATGATTGT
What are the possible peptides corresponding to this sequence?
Solution
1. Leu-Ser-Val-Pro-Thr-Asp-Asp-Cys
2. fMet-Ile
4.
Write the 3 possible correct codes for the following polypeptides (Multiple answers several
codons code for different amino acids),
a.
Leu – Val – Cys – Lys - Stop
5’- CUUGUAUGCAAAUGA -3’
5’- CUCGUAUGCAAAUGA -3’
5’- CUCGUAUGUAAAUGA -3’
b. His – Met – Ser – Thr – Stop
5’- CAUAUGAGUACUUAA -3’
5’- CAUAUGAGCACUUAA -3’
5’- CAUAUGAGUACAUAA -3’
2
5. Here is the sequence of a gene:
PROMOTER: TTCCCTAGATAGAGATACTTTGCGCGCACACACATGCAAACGCGCGCAAAAAGG
AAAGCCCACCTATAAACTCCAGCCGCAAAGAGAAAACCGGAGCAGCCGCAGCTC
EXON 1:
ACCTGGCCGCGGGGCGGCGCGCTCGATCTACGCGTCCGGGGCCCCGCGGGGCCGG
GCCCGGAGTCGCCATG AAT CGC TGC TGG GCG CTC TTC CTG TCT CTC
TGC TGC TAC CTG CGT CTG GTC AGC GCC GAG
INTRON 1:
GTGAGTTGCCACGGCGGCATGCAGTTGGTTCGCCCCTTTTGGTGTCTGCCCGGCAG
EXON 2:
GGG ACC CCA TTC CCG AGG AGC TTT ATG AGA TGC TGA GTG ACC
ACT CGA TCC GCT CCT TTG ATG ATC TCC AAC GCC TGC TGC ACG GAG ACC
CCG GA
INTRON 2:
GTAAATGGAATCCTCGCCCCGCGCTCCGGCCCTCCGAGGAGGTGGGG
CCGCCTGGTGTCTGACTGTGACTTCTCCTGCAG
EXON3:
A GGA AGA TGG GGC CGA GTT GGA CCT GAA CAT GAC CCG CTC CCA
CTC TGG AGG CGA GCT GGA GAG CTT GGC TCG TGG AAG AAGGAGCCTGG
a) Where does transcription start?
At the first nucleotide of exon 1
b) What is the sequence of the messenger RNA precursor and of the mature messenger RNA?
Precursor : exon1+intron1+exon2+intron2+exon3
Mature: exon1+exon2+exon3
(Replacing T by U)
c) Where does translation start and what is the peptide sequence (first ten residues)?
Asn-Arg-Cys-Trp-Ala-Leu-Phe-Leu-Ser-Leu
d) Where does translation stop? TGA in exon 2
6. A somatic cell that has 2n=24 chromosomes undergo mitosis 5 times continously. Please
calculate:
- The number of cells after the whole process? 25
- The number of chromosomes that the environment did provide to the whole process? 24*(25-1)
7. Complete the following table:
Nucleotide Components and Function
Nucleic Acid Type
DNA
deoxyribose
Name of the sugar present
in nucleotides
mRNA
ribose
3
tRNA
ribose
Name of bases present in
nucleotides
Function
Where can you find each
of these in a eukaryotic
cell?
Guanine
Cytosine
Thymine
Adenine
Store genetic
material of
organism
Nucleus
Mitochondria
Chloroplast
Guanine
Cytosine
Uracil
Adenine
Directs aa sequence of
protein
Guanine
Cytosine
Uracil
Adenine
Transport aa to site of
protein synthesis
Nucleus
Mitochondria
Chloroplast
Cytosol (free ribosome)
ER rough
Nucleus
Mitochondria
Chloroplast
Cytosol(free ribosome)
ER rough
Annex
Figure 1.The genetic code.
A
Alanine
Ala
I
Isoleucine
Ile
C
Cysteine
Cys
K
Lysine
Lys
D
Aspartate
Asp
L
Leucine
Leu
M
Methionine
Met
E
(aspartic acid)
Glutamate
N
Asparagine
Asn
P
Proline
Pro
F
(glutamic acid)
Phenylalanine
Phe
Q
Glutamine
Gln
G
Glycine
Gly
R
Arginine
Arg
H
Histidine
His
S
Serine
Ser
Glu
4
T
Threonine
Thr
V
Valine
Val
W
Tryptophan
Trp
Y
Tyrosine
Tyr
Figure 2.Amino acid names and abbreviations.
5
CHAPTER 2
GENE VARIANTS &POLYMORPHISMS
Problems
1. This is the sequence of the first exon of a mouse gene.
CGGGCACC ATG AGC GAC GTG GCT ATT GTG…
a) What would be the consequence of deleting the boxed nucleotides? Please write down the
protein sequences before and after the mutation.
Solution:
Deleting the boxed nucleotides results in a frame-shift that changes the reading of subsequent
codons, so it will alters the entire amino acid sequence.
Before: Met-Ser-Asp-Val-Ala-Ile-Val
After: Met-Ser-Arg-Gly-Tyr-Cys
b) What would be the expected impact of this deletion on the protein function?
The protein might loss its function because its function associates with its high-ordered
structure which depends on the sequence of amino acid. This often disrupts the biochemistry
process in the cell.
2.
The following sequence is located at the beginning of a gene.
The underlined part corresponds to an intron, which flanks two exons.
ATTAGCCATGCTCTCCGTCCCAACTGTAAGTATGCGCGAGATCGTTACCAGGATGATTGT…
What would be the expected impact of the following mutations:
a) ATTAGCCATGCTGTCCGTCCCAACTGTAAGTATGCGCGAGATCGTTACCAGGATGAT…
C to G
There is no change in the aa sequence because both codons CUG and CUC coded for Leu
b) ATTAGCCATGCTCTCCGTCCCAACTGTAAGTATGAACGAGATCGTTACCAGGATGAT…
CG to AA
There is no change in aa sequence because the mature mRNA doesn’t consist of intron.
c) ATTGGCCATGCTCTCCGTCCCAACTGTAAGTATGCGCGAGATCGTTACCAGGATGAT…
A to G
6
There is no change in aa sequence because the mRNA is translated from the first ATG
codon which stands after the mutation.
3. A research team sequenced a human gene and the corresponding mRNA. Here is the sequence of
the genomic DNA. The parts that are identical between the genomic DNA and the mRNA are
written using uppercase letters.
agcgaaatttaatgagcgtgtaacaggggactgaaaatcctgatttctcaAGCTATCAAA
del 1
GGTTTATAAAGCCAATATCTGGGAAAGAGAAAACCGTGAGACTTCCAGATCTTCTCTGGT
GAAGTGTGTTTCCTGCAACGATCACGAACATAAACATCAAAGGATCGCCATGGAAAGgta
del 2
agtgtgacaactcactgcgttggtggctcgcgttcttatgagctaagGGTCCCTCCTGCT
GCTGCTGGTGTCAAACCTGCTCCTGTGCCAGAGCGTGACCCCCTTGCCCATCTGTCCCGG
Del 3
CGGGGCTGCCCGATGCCAGGTGACCCTTCGAGACCTGTTTGACCGCGCCGTCGTCCTGTC
CCACTACATCCATAACCTCTCCTCAGAAATGTTCAGCGAATTCgtaagtaccatgcttct
ggcttcctattgaatttgtctcatcatttccagGATAAACGGTATACCCATGGCCGGGGG
del 4
TTCATTACCAAGGCCATCAACAGCTGCCACACTTCTTCCCTTGCCACCCCCGAAGACAAGGA
GCAAGCCCAACAGATGAATgtgagtccttcatccaggctttgca
a) Which parts of that sequpromoterence correspond to exons, introns and promoter?
b) What is the sequence of the peptide encoded by this gene (first ten residues)?
c) What will be the impact of each of the deletions indicated by boxed nucleotides on the
protein sequence and length?
Solution:
a) Promoter: first part with lowercase letter
Exons: all parts with the uppercase letter
Introns: all parts with the lowercase letter except the first one
b) Met-Glu-Arg-Val-Pro-Pro-Ala-Ala-Ala-Gly
c) - del 1: May disrupt the normal processes of gene activation by disturbing the
ordered recruitment of transcription factors at the promoter.
- del 2 : The mutation happens before the start codon, so it does not affect the coded
message. As a result, the related products do not change.
- del 3: Deleting the boxed nucleotides results in a frame-shift that changes the
reading of subsequent codons, so it will alters the entire amino acid sequence. The
polypeptide length (shorter) and chemical composition will be changed, resulting in
a non-functional protein that often disrupts the biochemical processes of a cell. 8
- del 4 : Mutation in introns: It does not cause change in the coded message, but may
result in a defect in RNA splicing and so affect the corresponding protein. Intronic
mutation however can introduce novel splice sites, activate novel promoters (which
may direct sense or antisense transcription causing alterations in mRNA, miRNA or
lncRNA expression), or introduce/eliminate enhancer activity, so these could affect
gene expression levels (affect regulatory region)
4.
a) Gene B has 390 Guanine and the total number of hydrogen bonds is 1670, is substitution mutated
in 1 pair of nucleotides to gene b. Gene b has 1 hydrogen bond more than gene B. Calculate each
type of nucleotide in gene b.
Solution:
- In allele B
7
The number of Cytosine = Guanine= 390
The number of Thymine = Adenine =(1670-390*3):2=250
- In allele b
There is 1 hydrogen bond more than allele B. Therefore, the mutation in 1 pair of nucleotide is the
substitution of 1 pair A-T into 1 pair G-C
The number of C=G=390+1=391
The number of T=A=250-1=249
b) Gene X has 3600 hydrogen bonds and the number of nucleotide Adenine is equal to 30% the
number nucleotide of the whole gene. Gene X is mutated type deletion 1 pair of nucletode A-T to
gene x.
A cell with heterozygote genotype Xx undergoes mitosis to 2 daughter cells. Please calculate each
type of nucleotide that the environment need to provide to the process.
Solution:
- In allele X:
We have
+ The number of hydrogen bonds=2*A+3*G=3600
+ %A=%T=30% => %G=%C=20%
The number of A=T=900
The number of G=X=600
-In allele x:
A=T=900-1=899
G=C=600
The number of Nu provided by the environment is 600+900+899+600=2999
5.
a) A gene D has 4800 hydrogen bonds and the ratio of nucleotide number Adenine/Guanine =
1/2, , is substitution mutated in 1 pair of nucleotides to gene d with 4801 hydrogen bonds.
Calculate each type of nucleotide in gene D and d.
Solution:
-In allele D
A=T=600
G=C=1200
-In allele d
A=T=600-1=599
G=C=1200+1=1201
b) Gene A with 3000 hydrogen bonds and the number of nucleotide Adenine is equal to
Guanine, is mutated to gene a. When gene a underwent DNA replication, the environment
provided 2398 nucleotides. Which kind of mutation that gene A had?
Solution:
- In allele A:
A=T=600
G=C=600
- In allele a:
The number of nucleotides provided by environment after 1 time of DNA replication is
equal to the number of nucleotide of allele a
Therefore, the allele a has loss 2 nucleotide which means the mutation is deletion.
6. This is the result of a paternity test of an alleged father (Jim Doe) and child (John). Please check
if Jim is biological father of John, in case yes, point out which allele the child inherited from the
father.
8
Solution:
Jim is the biological father of John
7. Using the genetic code table (attached), fill in the amino acid sequence starting
with the first ATG of the coding strand. It helps to mark off the codons by threes.
Y
14
16.3 or 17.3
17
9
10 or 16
12
13
23
10
9
6
15
29 or 30
11
11
9 or 11
13 or 14
18
14
18
DNA (5')G G A T A G C A T G A A A C C C G C A T A A (3')
amino acid : Met-Lys-Pro-Ala
How would the amino acid sequence above change with the following changes (mutations) in the
DNA code (changes are marked in bold-face):
a.
(5')G G A T A G C A T G A A A C C A G C A T A A (3') change to T
amino acid:
Met-Lys-Pro-Ala
b.
(5')G G A T A G C A T G A A A C C C C C A T A A (3') change to A
amino acid:
Met-Lys-Thr-Ala
c. (5')G G A T A G C A T G A A A - C C G C A T A A (3') delete
amino acid :
Met-Lys- Pro-His
d. (5')G G A T A G C A T G T A A C C C G C A T A A (3') change to C
amino acid :
Met-Gln-Pro-Ala
9
8. A diploid organism with 4000 spermatocytes undergoes meiosis to produce gametes. During
meiosis there are 40 cells with 1 pair of unseparated chromosomes in meiosis I, meiosis II took
place normally, the rest of the cells meiosis normally. Please indicate:
a) What is the percentage of non-mutated gametes? 90%
b) What percentage of mutant gametes that have more 1 chromosome than normal? 5%
c) What percentage of mutant gametes that have less 1 chromosome than normal? 5%
Solution:
a)
normal gametes : (4000-40) x 4= 15 840
percentage of non-mutated gametes: 15 840/ 16 000 = 99%
b)
after meiosis II, each mutated cell produces 2 gametes that have more 1 chromosome than normal
=> 40 x 2 = 80 gametes that have more 1 chromosome than normal
=> 80/ 16 000 = 0.5%
c)
after meiosis II, each mutated cell produces 2 gametes that have LESS 1 chromosome than normal
=> 40 x 2 = 80 gametes that have less 1 chromosome than normal
=> 80/ 16 000 = 0.5%
9. In Drosophila (2n=8), an individual has two chromosomal structure mutations: a chromosome in
the 1st pair carries deletion mutation, and a chromosome in the 3rd pair carries inversion mutation. If
the meiosis process takes place normally, please calculate:
a. Percentage of the normal gametes? 1/4
b. Percentage of the gametes carry the deletion chromosome? 1/2
c. Percentage of the gametes carry both the deletion chromosome and the inversion chromosome? ¼
Solution:
We have 2n = 8 => n=4
=> There are four pairs of chromosomes with drosophila.
Assume each different chromosome by an alphabet letter
So we have
1st pair is A//a
2nd pair B//b
3rd pair D//d
4th pair E//e
Assume that A is the deletion one D is the inversion one.
a. Percentage of the normal gametes?
[1(a) x 2(B/b) x 1(d) x 2(E/e) ]/[2(A/a) x 2(B/b) x 2(D/d) x 2(E/e)] = 1/4
b. Percentage of the gametes carry the deletion chromosome?
[1(A) x 2(B/b) x 2(D/d) x 2(E/e) ]/[2(A/a) x 2(B/b) x 2(D/d) x 2(E/e)] = 1/2
c. Percentage of the gametes carry both the deletion chromosome and the inversion
chromosome?
[1(A) x 2(B/b) x 1(D) x 2(E/e) ]/[2(A/a) x 2(B/b) x 2(D/d) x 2(E/e)] = 1/4
10
CHAPTER 3
MENDEL’S LAWS
Problems
1. Here are four pedig rees. The black symbols rep resent patients suffering from a rare disease that
is transmitted according to Mendel’s law.
a) For each family, determine whether
1, recessive because the normal parents have
the trait is dominant or recessive.
b) Determine the genotypes of the
affected individuals and their parents.
affected child
Child : aa
Each parent: Aa
2, dominant because the affected parents have
normal child
Affected children in second generation: Aa
because one of their parents is normal (aa)
and the affected parent have genotype Aa
because they have normal child
The 3 middle affected children in third
generation have genotype either AA or Aa
because their parents are all affected and have
genotype Aa(since one of their children is
normal)
The last affected child in the third generation
:Aa because he has normal children
3. recessive because the normal parents have affected children
parents: Aa
affected child: aa
4. same as 3
11
2. Crossing two drosophila flies (both) with normal wings produces 27 individuals with short
wingsand 79 with normal wings.
a) How is the short wings trait transmitted?
b) What are the parents genotypes?
c) If flies with short wings are crossed with one of the parent flies, how many normal flies
are expected in an offspring of 120 flies?
Solution:
a. Short wings : normal wings=1:3
And the parents are all normal wings
 the short wings trait is recessive and transmitted according to Melden’s law
b. A: dominant, a: recessive
Parents: AaxAa
c. Short wings: aa
Aaxaa => 1Aa:1aa
 Normal flies in 120 offsprings: ½ x 120=60
3. Albinism in plants is the inability to make chlorophyll. It is a recessive trait. In an
experiment, heterozygous tobacco plants undergo auto-fertilization and produce 600 seeds,
of which 80% germinate.
a) How many albino plants will be obtained?
b) How many plants will show the parental genotype?
Solution:
a. Assume that dominant allele is A; recessive allele is a.
Heterozygous parents
Aa x Aa => ¼(AA) : 2/4(Aa) : ¼(aa) => ¾ normal : ¼ albinism.
Albino seeds: 600 x ¼ = 150 (seeds)
Albino plants: 150 x 80% = 120 ( plants)
b. Parental genotype seeds: 600 x ½ = 300 (seeds)
Parental genotype plants: 300 x 80% = 240 (plants)
4. The ability to taste phenylthiourea (or phenylthiocarbamide) is a genetic trait that is
transmitted according to Mendel’s laws. Tasters are able to recognize a solution containing
only 0.005% of this product, by contrast to non-tasters, who are not able to detect this
molecule even at much higher concentrations.
Two parents who are both able to taste phenylthiourea have four children, of whom two are
unable to taste the molecule.
a) How is this trait transmitted?
b) What are the parents’ genotypes?
c) If this couple has a fifth child, what is the probability that the child is able to taste
phenylthiourea?
12
Solution:
Two parents who are both able to taste phenylthiourea have four children, of whom two are
unable to taste the molecule. Able to taste x Able to taste => Unable to taste => Able to taste
is Dominant; Unable to taste is recessive; Assume that dominant allele is A; recessive allele is
=> Parents : Aa x Aa.
a) Able to taste is Dominant; Unable to taste is recessive
b) Aa x Aa.
c) Aa x Aa => ¼(AA) : 2/4(Aa) : ¼(aa) => ¾ able to taste : ¼ unable to taste.
=> 3/4 is the probability that the child is able to taste phenylthiourea
5. Normal human skin pigmentation is dominant over the albino trait. If an albino man expects
a child with a woman who has normal skin pigmentation and whose father was albino, what
are the possible ge notypes and phenotypes of the child?
Solution:
Assume that dominant allele is A; recessive allele is a.
The albino man : aa
The normal woman has normal skin pigmentation whose father was albino is received
one ‘a’ allele from her father => Her genotype: Aa
=> Aa x aa => ½(Aa) : ½(aa) => ½ normal : ½ albino.
6.
In this pedigree, patient represented by a black symbol are born deaf.
a) What can be concluded in term of transmission of this trait?
b) What are the possible genotypes of individual III2?
Solution:
a) I3 normal x I4 normal => II3 deaf => Normal is dominant ; Deaf is recessive.
Assume that dominant allele is A; recessive allele is a.
b) II3: aa. III2 is normal but received one ‘a’ allele from the II3 father => Aa.
7.
The dominant allele A in Ayrhirecows produces a cut in the ears. In the following pedigree,
in which black symbols represent individuals with cut ears, determine thefrequency of this
trait in the offspring from the following matings: (a) III1 X III3 (b) III2 X III3 (c) III3 X III4
(d) III1 X III5 (e) III2 X III5
13
Solution
All the normal individuals in the pedigree have the genotype : aa.
III1 : aa ; III2: Aa ( Because receive one allele ‘a’ from II1); III3: aa; III4: aa; III5: Aa ( Because
receive one allele ‘a’ from II6)
(a) III1 X III3 : aa x aa
(b) III2 X III3 : Aa x aa
(c) III3 X III4 : aa x aa
(d) III1 X III5 : Aa x aa
(e) III2 X III5 : Aa x Aa
8. In Linumusitatissimum (flax, used to make linen textile), the blue color of the flowers (b) and
the long fiber length (l) are recessive traits. A farmer mate true breeding plants with short fibers
and white flowers with plants with blue flowers and long fibers.
a) What are the genotypes of these plants?
b) What is the geno type of the F1generation?
c) If plants from F1 are auto-fertilized and 800 F2 plants are obtained,
1. calculate the expected number of plants with white flowers and long fibers
2. determine the percentage of the different genotypes among these (white flowers and long
fibers).
Solution:
a) Short fibers and white flowers => Dominant ,
dominant and this is true breeding plants => BBLL.
Blue and long fibers => Recessive , Recessive => bbll.
b)P: BBLL x bbll => F1: BbLl
c) If plants from F1 are auto-fertilized and 800 F2 plants are obtained,
F1:BbLl x BbLl => (3 white : 1 blue) x ( 3 short : 1 long) => (9 white short : 3 white long : 3 blue
short : 1 blue long)
1. calculate the expected number of plants with white flowers and long fibers :
3/16 x 800 = 150 ( plants)
2. determine the percentage of the different genotypes among these (white flowers and long
fibers):
White long genotypes : BBll and Bbll => We need to compare the proportion of BB
and Bb
=> We have Bb x Bb
=> F2 genotypes proportion : (1BB:2Bb:1bb) => BB:Bb=1/2 => 1BBll : 2Bbll
9.
In a bovine population, two co-dominant alleles control the coat color: R (red) and r (white).
Heterozygous individuals present an intermediate color named “rouan”. The “strait hair”
14
trait is dominant over the “curly” trait (s). What is the result of mating a red curly bull with a
homozygous white cow with strait hair?
a) What is the offspring phenotype?
b) What is the offspring and parents genotype?
c) What is the result of mating a F1individual with a rouan curly bull?
Solution:
RRss x rrSS => RrSs ( Rouan, strait)
a) Rouan, strait hair.
b) Offspring: RrSs ; parents : RRss x rrSS.
c) F1: RrSs x Rrss ( Nhân độc lập từng kiểu hình của từng tính trạng với kiểu hình của tính trạng
còn lại sẽ được kiểu hình của offsprings)
=>F2: (1 Red : 2 Rouan : 1 White) x ( 1 Strait : 1 Curly)
=> ( 1 Red Strait : 1 Red Curly : 2 Rouan Strait : 2 Rouan Curly : 1 White Strait : 1 White Curly)
10. Short hair is a dominant trait in guinea pigs and is controlled by a single gene (allele L),
while the allele l corresponds to long hair. The fur color is controlled by a gene which has
two co-dominant alleles, such that: the YY genotype = yellow, YW = cream and WW =
white. If individuals that have the genotype Ll YW are mated, what will be the possible
phenotypes in the F1 generation and the percentages?
Solution:
Ll YW x Ll YW
=> (1LL:2Ll:1ll) x (1YY : 2YW : 1WW)
=> (1 LLYY : 2LLYW : 1LLWW : 2LlYY : 4LlYW : 2LlWW : 1llYY : 2llYW : 1llWW)
= > (3 Short Yellow : 6 Short Cream : 3 Short White : 1 Long Yellow : 2 Long Cream : 1 Long
White)
11. A man with Huntington disease (autosomal dominant – rare diseases) and the AB blood
group married a healthy woman who is O. The genes are independent. What is the
probability that their child is healthy and A?
Solution:
We have Huntington disease is a rare disease, so we can assume that the man carried that disease
have heterozygous genotype .
Denote :
H – Huntington
h – normal
Ia – Blood type A
Ib – Blood type B
Io – Blood type O
The man genotype : HhIaIb
The healthy woman genotype : hhIoIo
The child is healthy and A must have hhIaIo genotype
=> The probability is: ½(hh)x1/2(IaIo)=1/4.
15
12. Two male drosophila flies (1 & 2) are mated with two females (3 & 4). These parents have
long wings and a grey body.The results of the mating are:
1 x 3 : 153 grey individualswith long wings
58 grey individuals with wing remnants
1 x 4 : 26 black individuals with long wings
73 grey individuals with long wings
2 x 3 : 112 grey individuals with long wings
2 x 4 : 89 grey individuals with long wings
What are the parent genotypes?
Solution:
All parents have long wings and grey body producing wing remnant and black individuals
 The dominant alleles:
A : Grey
B: Long
 The recessive alleles:
a: Black
b: remnants
- The color of body trait:
1x4: grey : black=3:1=> 1 and 4 : Aa
1x3: 100% grey=> 3 is AA
2x4:100% grey=> 2 is AA
- The wings trait:
1x3: long : remnants = 3:1 => 1 and 3 : Bb
1x4: 100% long => 4 is BB
2x3: 100% long=> 2 is BB
 1 is AaBb /2 is AABB/ 3 is AABb/ 4 is AaBB
16
CHAPTER 4
PARTICULAR CASES INCLUDING CROSSING OVER
Problems
1. A farmer crosses two true breeding types of peonies: one has red flowers with large leaves (R
and G dominant alleles, respectively), the other is white with small leaves.
a) If these two traits are controlled by two genes, what will be the phenotype of the F1
offspring?
b) If F1 individuals undergo auto-fertilization, what will be the percentages of each
phenotype in F2 if the two genes are independent?
c) Same if the two genes are tightly linked?
d) Same if the two genes are separated by a distance of 10 Morgan units (centimorgan)?
Solution:
a. Consider the flower color trait , true breeding type cross with each other :
RR (red flower ) x rr ( white flower ) => Rr ( red flower )
Consider the leaves trait , true breeding type cross with each other : GG (large leaves) x rr
(small leaves) => Gg ( large leaves )
=> Combine those two traits, we have the F1 phenotype : Red flower, large leaves.
b. The genes are independent => F1 genotype : RrGg
F1 goes auto-fertilization => F1 x F1 : RrGg x RrGg
Gametes F1: ( 1RG : 1Rg : 1rG : 1rg ) x ( 1RG : 1Rg : 1rG : 1 rg )
F2 Genotypes:
1 RRGG: 2RRGg: 1 RRgg : 2RrGG: 4 RrGg: 2Rrgg: 1:rrGG: 2rrGg: 1rrgg
F2 Phenotypes:
9 red large : 3 red small : 3 white large : 1 white small
c. P true breeding : RG // RG x rg // rg
Gametes P : (1RG) x (1rg)
F1 genotype : RG // rg
F1 goes auto-fertilization => F1 x F1 : RG // rg x RG // rg
Gametes F1: (1RG : 1 rg) x (1RG : 1 rg)
F2 genotype : 1RG//RG: 2RG//rg: 1rg//rg
F2 phenotype : 3 red large : 1 white small
d. We have two genes are separated by a distance of 10 Morgan units
=> The recombination frequency is 10%.
F1 genotype : RG // rg
F1 goes auto-fertilization => F1 x F1 : RG // rg x RG // rg
Gametes F1: (45%RG : 45%rg : 5%Rg: 5%rG) x (45%RG : 45%rg : 5%Rg: 5%rG)
F2 phenotype :
70.25% Red large : 4.75% Red small : 4.75% White large : 20.25% White small
17
2.
The success of renal transplantation depends on three human histocompatibility genes, HLAA, HLA-B and HLA-C, which must match between the donor and the receiver. A single
mismatch may cause the kidney rejection. Each gene has multiple co-dominant alleles. These
three genes are located very close to each other on chromosome 6, so that the recombination
rate is very low (below 1%).
The father has the following genotype: A1, A2, B24, B10, Cw4 and Cw7 and the mother is
A1, A1, B11, B7, Cw5 and Cw8. Their first boy is A1, A1, B24, B11, Cw7 and Cw8. What
is the probability that the second child is compatible with his/her brother?
Solution:
The first boy received A1 from both father and mother , received B24 and Cw7 from
father, received B11 and Cw5 from mother.
=> The first child genotype : A1B24Cw7 // A1B11Cw8.
=> The father genotype : A1B24Cw7 // A2B10Cw4.
=> The mother genotype : A1B11Cw8 // A1B7Cw5.
We need the second child compatible with his/her brother => The second child
genotype A1B24Cw7 // A1B11Cw8.
The probability is : 50% x 50% = 25%
3.
Wild type drosophila are mated with mutants with yellow eyes (s phenotype) and no antenna
(bobbed or B phenotype). F1 offspring have wild type eyes and “bobbed” antennas. When
F1 individuals are mated, the F2 result is the following:
101 individuals are
32 individuals are
31 individuals are
13 individuals are
[wt B]
[wt wt]
[s B]
[s wt]
a) Are these genes linked?
b) What are the genotypes of the parents and the F1 individuals?
Solution:
a) eye’s color: wt/s = (101+32)/(31+13) = (3:1)
antenna present: B/wt = (101+31)/(32+13) = (3:1)
( 3wt : 1s ) x (3B : 1wt) = 9wtB : 3wt wt : 3sB : 1swt
→ independent → no linked
b)
Because:The ratio of wt/s=3:1 => wt is dominant => wt : A , yellow: a
The ratio of B/wt=3:1 => Bobbed is dominant => Bobbed: B, wt: b
=> P:
AAbb
x
aaBB
(wild, wild )
( yellow, bobbed)
F1:
AaBb ( wild,bobbed)
4. A man inherited cataract from his mother (normal fingers and toes) and polydactyly
from his father (normal eyes). The two diseases are caused by dominant alleles of
18
genes that are tightly linked. This man marries a healthy woman. What will be the
phenotype of the children?
Solution:
The two diseases are caused by dominant alleles
=> A: cataract, a: NO cataract
B: polydactyly, b: NO polydactyly
Man inherited from mother ( Ab) and from father ( aB)
=> Man : Ab/aB
Man + woman ( healthy) -> Ab/aB + ab
-> children : ½ Ab ½ aB
=> ½ cataract
½ polydactyly
5. Here is a map of the locus for the a, b and c genes:
a/A
b/B
c/C
10 cM
15 cM
What will be the percentages of gametes produced by an individual who is A-b-C ?
a-B-c
Solution:
The double recombination frequency is : 0.1 x 0.15 = 0.015
=> The double recombination gametes are : A-B-C = a-b-c = 0.015 /2 = 0.0075
The single recombination gametes at A are : a-b-C = A-B-c = (0.1 - 0.015 ) /2 =0.0425
The single recombination gametes at C are : A-b-c = a-B-C = ( 0.15 - 0.015 ) /2 =0.0675
The normal gametes are : A-b-C=a-B-c= 0.5 - 0.0075 - 0.0425 - 0.0675 = 0.3825
6. Two types of drosophila are studied:
Wild type flies:
mutants:
red eye (red)
Grey body (grey)
Normal wings (norm)
chestnut eye (ch)
black body (bl)
indented wings (ind)
Wild type males are mated with mutant females.The F1 population is wild type.Then,
a) a F1 female is mated with a mutant male (offspring = F2a)
b) a F1 male is mated with a mutant female (offspring = F2b)
19
Here are the phenotypes of the two populations F2a and F2b:
1. F2a
Eye
Body
Wings
red
grey
norm
red
grey
ind
red
bl
norm
Red
Bl
Ind
Ch
Grey
Norm
ch
grey
ind
Ch
Bl
norm
ch
bl
ind
616
65
170
147
144
172
66
620
red
grey
norm
ch
bl
ind
542
551
2. F2b
eye
body
wings
a) How would you explain the F2b results?
b) Determine the genetic map of the locus.
Solution:
P : Wild types x mutants => 100% wild type => Wild type of 3 genes are all dominant and F1 is
heterozygous in all genes.
Denote :
A - red eye >> a - chestnut eye
B - Grey body >> b - black body
D - Normal wings >> d - indented wings .
If the genes are independent, the ratio of phenotype in F2a and F2b should be similar
=> The genes are not independent
=> The genes are linked.
=> F1 genotype : ABD // abd
The mutant phenotype ( both female and male ) : abd // abd
a. We know that, the male drosophila do not have the gene interchange =>
Mating B : male ABD // abd x female abd // abd
Gamete F1B : (1 ABD : 1 abd ) x (1 abd )
F2B genotypes : 1 ABD//abd : 1 abd//abd
F2b phenotypes : 1 red,grey,normal : 1 chestnut,black,indented.
=> AS THE RESULT IN REALITY.
b. We know that, the female drosophila occurs gene interchange .
We have to find the double recombination phenotype in the table
The double recombination frequency is the product of two recombination
frequencies , so it’s phenotype will most likely be the smallest, these are :
%Aa,Bb,dd = %aa,bb,Dd=> The double recombination gamete that F1 female create is :
ABd and abD
=> The middle gene is D/d.
The distance between A/a and D/d is : (65 + 66 + 144 + 147) / 2000 = 21.1 cM
The distance between B/b and D/d is : ( 65 + 66 + 170 + 172 ) / 2000= 23.65 cM
7.
In experiments using drosophila confuse, researchers observed that the vermillon trait (v
allele) is responsible for dark red eyes. If homozygous a wild type female (light red eyes) is
mated with a vermillon male (mating A), the offspring includes 61 wild type females and 57
20
wild type males. If a vermillon female is mated with a wild type male (mating B), the
offspring is 88 wild type females and 79 vermillon males.
a) How would you interpret these results?
b) What are the parents and F1genotypes?
c) If F1individuals are mated, what phenotypes and percentages will be observed in F2?
Solution:
a)
We have that in mating A, homozygous wild type female mated with vermillion male =>
100% wild type => Wild type is dominant against vermillion.
Also in mating B , the recessive trait ( vermillion) in mating B only occurs in males => This
trait is a X-linked trait.
b)
Denote that :
wild type : V
V>>v
mating A
P : XV XV x Xv Y
Gametes P : ( 1 XV ) x (1Xv :1 Y)
F1 genotypes: 1 XV X v :1 XVYmating B
P: Xv Xv x XV Y
Gametes P : ( 1 Xv ) x (1XV :1 Y)
F1 genotypes: 1 XV X v :1 XvY
c) If F1 Individuals are mated, what phenotypes and percentages will be observed in F2?
mating A’
F1: XVXv x XVY
Gametes F1: ( 1XV : 1Xv ) x ( 1XV :1Y )
F2 genotype: 1XVXV : 1 XVXv : 1XVY: 1XvY
=> F2 phenotype : 2 normal female : 1 normal male : 1 vermillon male
mating B’
F1: XVXv x XvY
Gametes F1: ( 1XV : 1Xv ) x ( 1Xv :1Y )
F2 genotype: 1XVXv : 1 XvXv : 1XVY: 1XvY
=> F2 phenotype : 1 normal female : 1 vermillon female : 1 normal male : 1 vermillon male
8. A healthy woman, whose father died from hemophilia (X-linked recessive), married a healthy
man. Could their children suffer from hemophilia?
Solution:
Heathy woman has died father from hemophilia (X-linked recessive)
Hemophilia is recessive: a=> normal A
=>Her father genotype: XaY
A healthy woman: XAXa
Her healthy husband: XAY
=> XAX a x XAY
Gametes: (1XA : 1Xa) x (1XA : 1Y)
Children phenotypes: 2 Healthy female : 1 healthy male: 1 hemophilia male
=> their children could suffer from hemophilia with probability P=1/4
21
CHAPTER 5
POPULATION GENETICS
Problems
1. In the course of a study dealing with blood group frequencies in a population of 2000
individuals, part of the results were lost. We know that 800 individuals were O and 780 were
A.
a) What are the allelic frequencies?
b) What are the phenotype frequencies?
Solution:
a)
Assume that allele frequency of IA = p
Assume that allele frequency of IB = q
Assume that allele frequency of IO = r
There are 800 blood O individuals => IOIO = 800 / 2000 = 0.4
√10
=> r^2 = 0.4 => r = 5
There are 780 blood A individuals => IAIO + IAIA = 780 / 2000 = 0.39
=> p^2 + 2.p.r = 0.39 => p =
q=1–p–r=
b)
Blood A : 0.39
Blood O : 0.4
√79−2√10
10
10−√79
10
√79−2√10 10−√79
x 10
10
10−√79
√10 10−√79
Blood B : ( 10 )2+2x 5 x 10
Blood AB : 2x
2.
A. If the frequency of albino (recessive trait) in a given population is 0.0009, what will be
the frequency of heterozygous individuals?
B. Huntington disease (autosomal dominant) has a frequency of 1 case in 25 000. Calculate
the frequency of the mutated allele in the population.
Solution:
A. The allelic frequency of albino = √0.009= 0.03
=> The frequency of heterozygous individuals = 2x0.03x(1-0.03)=5.82%
1
B. The % of normal allele=√1 − 25000=99.99%
The % of mutated allele=100%-99.99%=0.01%
22
3.
A. If, in 5% of men are color blind (recessive X-linked trait) in a given population, what
will be the frequency of color blind women?
B. If allele frequencies corresponding to the blood group gene was: p(A) = 0.3 ; q(B) = 0.1
; r(o) = 0.6, what will be the phenotype percentages?
Solution:
Assume that : XA : normal ; Xa : Color blind
XX: p2 (XAXA) + 2pq(XAXa ) + q2 (XaXa ) = 1
XY: p(XAY) + q(XaY) = 1
=> Xa = 0.05 ; XA = 1-0.05 = 0.95
=> The frequency of color blind women : q^2 = 0.05^2 = 0.25%
4.
A. The ability to taste phenylthiocarbamide is a dominant trait in humans controlled by
allele G of a gene. In a given population, 2800 individuals are able to taste this molecule
while 1200 do not.
a) What are the allele frequencies?
b) What are the genotype frequencies?
Solution:
a) What are the allele frequencies?
Assume that :
G : able to taste (p)
g : unable to taste (q)
1200 unable to taste => q^2 = 1200/(2800+1200) = 0.3
=> q = sqrt(0.3) => p = 1 – sqrt(0.3) .
b) What are the genotype frequencies?
GG = p^2 = (1 – sqrt(0.3))^2
Gg = 2pq = 2sqrt(0.3)(1-sqrt(0.3))
gg = q^2 = 0.3
B. In humans, the Rh factor genetic information is inherited from our parents, but it is
inherited independently of the ABO blood type alleles. In humans, Rh+ individuals have
the Rh antigen on their red blood cells, while Rh− individuals do not. There are two
different alleles for the Rh factor known as Rh+ and rh. Assume that a dominant gene Rh
produces the Rh+ phenotype, and that the recessive rh allele produces the Rh− phenotype.
In a population that is in Hardy-Weinberg equilibrium, if 190 out of 200 individuals are
Rh+, calculate the frequencies of both alleles.
Solution:
Assume that : Rh+ : p ; Rh- : q ( allelic frequency)
190 out of 200 are Rh+ = > 10 out of 200 are Rh=> q^2 = 10/200 = 0.05 => q = sqrt(0.05)
=> p = 1- sqrt(0.05)
23
5.
Premature baldness is a trait influenced by sex: it is dominant in men and recessive in
women. In a study of 10000 men, 7225 were not bald. How many individuals are expected
to have normal hair in a population of 10000 women?
Solution:
Assume that :
H : Bald (p)
h : Normal (q)
Trait is influenced by sex => Hh in men : Bald ; Hh in women : Normal
10000 men , 7225 normal => hh = 7225/10000 = 0.7225 = q^2 => q = 0.85 => p = 1-0.85 = 0.15
In women ; Normal are : Hh + hh = 2x0.15x0.85 + 0.85^2 = 0.9775
=> 9775 women are expected to have normal hair.
6.
A. An allele W, for white wool, is dominant over allele w, for black wool in a sheep. In a
sample of 900 sheep, 891 are white and 9 are black. Calculate the allelic frequencies within
this population, assuming that the population is in H-W equilibrium.
Solution:
White allelic frequency = p ; black allelic frequency = q
9 are black => ww = 9/900 = 0.01 => q^2 = 0.01 => q= 0.1 => p = 0.9
B. In the United States, approximately one child in 10,000 is born with PKU
(phenylketonuria), a syndrome that affects individuals homozygous for the recessive allele
(aa).
(a) Calculate the frequency of this allele in the population.
(b) Calculate the frequency of the normal allele.
(c) Calculate the percentage of carriers of the trait within the population.
Solution:
(a) Calculate the frequency of this allele in the population.
Allelic frequency : Normal : p ; PKU : q.
1/10000 PKU => q^2 = 1/10000 => q = 0.01
(b) Calculate the frequency of the normal allele.
p = 1-0.01 = 0.99
(c) Calculate the percentage of carriers of the trait within the population.
A hereditary carrier (genetic carrier or just carrier), is a person or other organism that has
inherited a recessive allele for a genetic trait or mutation but usually does not display that
trait or show symptoms of the disease.
Carriers of the trait : Aa = 2pq = 2x0.01x0.99 = 1.98 %
7. In cats, the gene for hair color is located on a non-homologous region of the X
chromosome: the dominant allele for black hair is completely dominant over the
recessive allele for tabby hair, individuals with the heterozygous genotype has calico
hairs. . A cat population is in genetic equilibrium with 10% male tabby cats.
Theoretically, if the population has 1000 cats, how many calico cats are expected ?
Solution:
24
Asumme that XA : black, Xa: tabby
In XY: % tabby cat =2x10%(% of male tabby cat in population)=20%
 % Xa=20%=> %XA=80%
In XX: %XAXa=2x20%x80%=32%
 In population: % calio cat=32%:2=16%
 160 calio cat
8. In sheep, allele D codes for horns, allele d for no horns; genotype Dd for horns in
males and no horns in females. In a population in genetic equilibrium, sheep have
50% horns. All hornless sheep in this population are randomly mated to obtain F1.
Theoretically, in F1, what is the ratio of hornless sheep that can be expected ?
Solution:
% sheep having horn=%DD+%Dd/2=50%
 %𝐷2 + %𝐷xx(1-%D)=1/2 => %D=50%=> %d=50%
 All hornless sheep in this population are randomly mated to obtain F1:
𝑃:(♂)100%dd x
(♀)(50%Dd+25%dd)
Gamates: (♂) 100%d x (♀) (1/3 D+ 2/3 d)
1
2
F1: 1/3 Dd +2/3 dd=> hornless sheep: 3 : 2 + 3=5/6
25