Scheda 2
1
FILE 1 – DNA REPLICATION – CELL CYCLE – MITOSIS AND MEIOSIS
CROSSES WITH 1 GENE
-
1. This is one DNA molecule during its replication, draw the new synthetized strands
with their polarity and the Okazaki fragments.
5'
3'
5'
3'
2. How is the structure of a chromosome before and after the replication process?
We presume that the cell is diploid. Draw a scheme of the chromosome using a bar
and a circle as centromere.
BEFORE
AFTER
3. Make a scheme of the cellular cycle highlighting the chromosome characteristics of
each singular stage.
Scheda 2
2
4. Draw a schematic picture of chromosomes of a diploid cell with n=1 indifferent
mitosis stage. The analysed individual is heterozygous for gene A.
A
G1
a
Prophase
Metaphase
Anaphase
Daughter cells
Scheda 2
3
5. Draw a schematic picture of chromosomes of a diploid cell with n=1 indifferent
meiosis stage. The analysed individual is heterozygous for gene A.
A
Gametocyte
a
Prophase I
Metaphase I
Anaphase I
Telophase I
Metaphase II
Anaphase II
Gametes
6. Which structures migrate at the opposite poles of the spindle?
a) in mitosis
b) in meiosis, division I
c) in meiosis, division II
Scheda 2
4
7. Which types of gametes and in which proportions are produced by individuals that
have the following genotypes?
a) genotype AA; b) genotype Aa; c) genotype aa
8. For each cross determine genotypic and phenotypic classes expected in the
progeny and relative frequencies.
genotypeof
the individuals
used for the
cross
Gametes of
the first
individual
(Frequency)
Gametes of
the second
individual
(Frequency)
Genotypes
Phenotypes and
and
frequency of the
frequency of progeny
the progeny
a) AA x aa
b) Aa x aa
c) Aa x Aa
9. In dogs hair length is determined by a gene, P, that can be present in two
alternative alleles, P and p. Accordingly to the crosses reported below, determine the
genotype of the individuals used for the crosses.
Parental
phenotype
a) short x long
n° of individuals in the progeny
Short hair
Long hair
100
0
b) short x long
50
50
c) short x short
150
50
Parental genotype
10. Wild-type Drosophila melanogaster has red eyes. Mutants with purple eyes exist.
This phenotype is controlled by the pr gene, which has two allelic
states pr+ and pr. The following crosses have been done:
Parental phenotype
a) red x red
b) purple x purple
n° of individuals in the
progeny
red
purple
total
125
35
160
0
45
45
Parental
genotype
Scheda 2
5
c) red x red
177
63
240
d) purple x red
45
55
100
Scheda 2
6
FILE 2–INHERITANCE OF TWO OR MORE INDEPENDENT GENES
1. Draw a scheme of meiosis process of a diploid cell with n=2. One chromosome
carries gene A, the other carries gene B. The analyzed individual is heterozygous for
both genes. Represent the two possible relative positions of the chromosomes in
metaphase I.
A
Gametocyte
b
a
B
Prophase I
Metaphase I
oppure
Metaphase II
Gametes
2. Now use the branch diagram to determine type and frequency of the gametes
produced by the same cell.
Gene A (frequency)
Gene B (frequency)
Gameti (frequency)
………….(……..)
………….(……..)
………….(……..)
………….(……..)
………….(……..)
………….(……..)
………….(……..)
………….(……..)
………...(……..)
………...(…….)
Scheda 2
7
3. Which type of gametes and in which proportions are produced by individuals that
have the following genotype (use the branch diagram)?
a) aa bb
b) Aa bb
c) Aa Bb
4. For each cross determine genotypic and phenotypic classes expected in the
progeny and relative frequencies (A and B genes are independent)
Genotype of the Gametes of the Gametes of the
Genotypes
Phenotypes
individuals used
first individual
second
and frequency and frequency
for the cross
(frequency)
individual
of the progeny of the progeny
(frequency)
a) AA bb x aa BB
b) Aa bb x aa Bb
c) Aa Bb x aa bb
Now use the branch diagram to calculate the phenotypic classes. Example of a
scheme for exercise 4 b)
Phenotypes for A gene
Cross Aa x aa
Phenotypes for B gene
Cross bb x Bb
phenotypical classes
(frequency)
………….(……..)
………….(……..)
………….(……..)
………….(……..)
………….(……..)
………….(……..)
………….(……..)
………….(……..)
………...(……..)
………...(……..)
5. In Drosophila melanogaster, body colour is determined by the e gene: the
recessive allele is responsible for the black colour
of the body, the dominant allele e+
is responsible for the grey body. Vestigial wings
are determined by the recessive
allele vg, normal wings are determined by the
dominant allele vg+. These two genes are independent. If dihybrid
flies for these two genes are crossed and resulting progeny is
composed by 368 individuals, how many individuals are present in
every phenotypical class?
Scheda 2
8
6. In guinea pigs, the N gene determinates fur colour, while the L gene controls fur
smooth or curly. Determine the genotype of individuals used for the
crosses and verify your hypothesis with χ2 test.
Phenotype of individuals used for the cross
n° of individuals in the progeny
Black
Black
White
White
curly
smooth
curly
smooth
50
0
0
0
a) black curly
x
white smooth
b) black curly
x
black curly
185
60
57
18
c) black curly
x
white smooth
105
100
98
97
d) white curly
x
black curly
63
17
58
22
Example of table for calculation of 2
Cross b)
Number of
hypoth
Number of
Balck curly x
observed
esis
expected
black curly
individuals (xo)
individuals (xa)
Black curly
Black smooth
White curly
White smooth
xo-xa
(xo-xa)2
(xo-xa)2
xa
=
2 = ………
Degrees of freedom = …………. Probability = …………….
7 – Determine the genotype of individuals used for the crosses and verify your
hypothesis with χ2 test. For each cross create a table, according to the
scheme reported below. Gene W determines Red color, gene D
determines the plant height
Phenotype of individuals used for the cross
n° of individuals in the progeny
Red tall
Red
White
White
short
tall
short
120
0
45
0
a) red tall
x
red tall
b) red tall
x
white short
100
0
105
0
c) red tall
x
white short
45
43
48
44
red tall
175
67
182
58
265
92
93
28
d) white tall x
e) red tall
x
red tall
Scheda 2
9
8 – Crossing a pure line of melons whit white and spherical fruits with an other pure
line of melons with yellow and flat fruits the f1 progeny is melons with white and
spherical fruits.
Crossing two plants of F1 we have:
P1
P2
X
F1
Plants with white and spherical fruits148
Plants with yellow and spherical fruits 52
F2
Plants with white and flat fruits
49
Plants with yellow and flat fruits
23
Total
272
Determine the genotypes of individuals of P1, P2 e F1, make the hypothesis of traits
segregation and verify the results with X2
9. Using the branch diagram, determine the gametes produced by individuals with
the following genotype (A, B and C genes are independent):
a) AaBbCC
b) AabbCcDd
c) AaBbCc
10. For the following cross, determine the phenotypical classes expected in the
progeny and their frequencies (A, B and C genes are independent)
AaBbCc x aabbcc
Scheda 3
1
FILE 3 – HUMAN GENETICS
BLOOD GROUPS- Recognition of paternity -Family Pedigree- Sex Linkage
1 – The human blood groups system ABO is regulated by 3 alleles according to the
following scheme:
Blood Group (Phenotype)
Possible Genotype
A
IAIA; IAi
B
IBIB; IBi
AB
IA IB
0
ii
Which is the dominance relation in this series of multiple alleles?
2 – Determine parental genotypes in the following crosses:
Phenotype of individual Phenotypes of the progeny Genotype of the individuals
used for the cross
used for the cross
a) A x A
A;
O
b) A x AB
A; AB
c) B x 0
B;
d) B x A
B; A; O; AB
O
3 – Rh system is determined by the gene D: the presence of antigen (Rh+individual )
is due to the dominant allele D; the absence of antigen (Rh – individual) is
homozygote for the recessive allele d.
Determine the genotypes of crossed individuals.
Phenotype of the
Phenotypes of the progeny
Genotype of
individuals used for
the crossed
the cross
individuals
+
+
+
+
a) AB Rh x 0 Rh 3/8 A Rh ; 3/8 B Rh ; 1/8 A Rh ; 1/8 B Rh
b) B Rh-
x A Rh+ 1/4 AB Rh+; 1/4 A Rh+; 1/4 B Rh+; 1/4 0 Rh+
4 – Mother and son have the indicated blood groups. Could the man be the father of
the child?
A Rh- B Rh+
B Rh-
0 Rh-
0 Rh+ A Rh+
5 – Figure out the probability of these events.
AB Rh+
A Rh- 0 Rh-
Scheda 3
AB Rh- 0 Rh+
2
0 Rh+
A Rh+
B Rh-
0 Rh-
?
?
6 – In family trees the most used symbols are:
Male
Parents
Female
children
A roman number is assigned to every generation and the individuals of the same
generation are progressively numbered from left to right with Arabic numbers.
In the following family trees, black symbols represent a mutant phenotype. Determine
if the mutant phenotype is due to a dominant or a recessive allele, taking into
consideration that the mutant allele is rare.
a)
I
II
III
b)
I
II
III
7–Considering people that become part of the family through marriage don’t have the
mutant allele, calculate the probability that an affected (sick) child is born from cross
III,3 and III,4 of 6b.
Scheda 3
3
8–in the following family trees, black symbols represent the mutant phenotype of
recessive homozygote. Considering that mutant alleles are rare, calculate the
probability of an affected child in the indicated crossed.
a)
I
II
III
IV
?
b)
I
II
III
?
c)
I
II
III
?
d)
I
II
?
9 - If individuals of cross II3-II4 have already had an affected child, which is the
probability to have a second affected child?
Scheda 4
1
FILE 4 – SEX LINKAGE
1 – Determine for each cross if the character is sex-linked or not and assign the genotype to
the crossed individuals.
female
a) pale
Parental Phenotype
male
dark
female
dark
pale
Phenotype of the progeny
male
45
dark
0
0
pale
48
b) rough
rough
rough
smooth
87
33
rough
smooth
92
27
c) red
red
red
white
102
0
red
white
49
52
d) black
white
black
white
98
0
black
white
103
0
2 – In humans the presence of a fissure in the iris (coloboma iridis) is controlled by a
recessive sex-linked gene. An affected daughter is born from
a normal couple. The husband asks for divorce, accusing his
wife of infidelity. Is he right?
3 – A recessive sex-linked gene is responsible for the most common kind of haemophilia.
Considering the family tree shown below, answer the following questions:
a) If II.2 marries a normal man, which is the probability to have children affected
by haemophilia?
b) If II.2 already has an affected child, which is the probability to have a second
affected child?
I
II
4 – Determine if the character shown in the following family tree can be due to: a) the
dominant allele of an autosomic gene; b) the recessive allele of an autosomic gene; c) the
dominant allele of a sex-linked gene; d) the recessive allele of a sex- linked gene
Scheda 5
2
FILE 5 – LINKED GENES
1 - Draw a scheme of the meiosis process of a diploid organism with n = 1, that has
two genes, A and B, on the same chromosome. The individual is heterozygous in
trans for the A and B genes ( the two dominant alleles A and B are NOT on the same
chromosome).
Meiosis without
meiosis
recombination
with recombination
prophase I
metaphase I
telophase I
metaphase II
gametes
Supposing that meiosis frequency without crossing-over is 60% and with crossingover is 40%, show the resulting gametes and respective frequencies.
How many kinds of gametes (and with which frequency) would be expected if the cell
were double heterozygous for two independent genes?
What’s the difference between these two situations?
Scheda 5
3
2 – Mais has Zb genes (Zebra crossband). The
mutants has phenotype with striated leaf (Zb normal;
zb striated). The gene P1 (pericarp color), determine
the colour of pericarp dark red (P1 dark; p1 pale),
These genes are on the same chromosome and they are 5 mapunits distant. Which
phenotypical classes are expected in the progeny and with which frequencies?
Zb
p1
zb
p1
x
zb
P1
zb
p1
Which are the parental classes? Which are the recombinant classes? Why are they
named parental and recombinant?
3 – In drosophila gene b (black body) and gene vg (vestigial wings) are 18 mu
distant. The dominant alleles are b+ (brown color) and vg+
(normal wings), the recessive alleles b (black color) and vg
(vestigial wings). A individual b+b+ vg+vg+ is crossed with an
individual bb vgvg. Which are the phenotypes of the individuals?
How will be the F1 (genotype and phenotype)?
If an individual of F1 is crosse with an recessive homozygous, which phenotypical
classes are expected in the progeny and with which frequences?
4 - If the genes A and B are located at a distance of 20 map units, which percentage
of the cells has done crossing-over during meiosis? Explain your answer.
5 - for each of the following crosses:
- determine the genotype of the individuals used for the cross; - determine if there is
concatenation of the analyzed genes;
- if genes are concatenated, determine their map distance;
- draw a schematic picture of the map of crossed individuals.
Parental phenotype
Phenotype of the progeny
Total
a) AB x ab
AB 23 Ab 21 aB 17 ab 18
79
b) GH x gh
GH 71
Gh 26 gH 32 gh 68
197
c) VZ x vz
VZ
Vz 64 vZ 55 vz 38
198
d) DE x de
DE 0
De 78 dE 81 de 0
159
e) SF x Sf
SF
Sf
31
240
f) QW x QW
QW 178 Qw 58 qW 61 qw 17
314
41
92
87 sF 30 sf
6 Genes M and N are concatenated at 10 mu. The E gene is located on another
chromosome. Which gametes are produced by an individual: MmNnEe if genes MN
are in cis?
Which gametes if the genes M and N are in trans?
1
Scheda 6
FILE 6 – GENE INTERACTIONS
1 - Consider two independent genes, A and B; two heterozygous individuals for both
genes are crossed. Complete the scheme below
AaBb x AaBb
Phenotypes and
frequencies
Gene A
Phenotypes and
frequencies
Gene B
Phenotypes and
frequencies
Genes A and B
………………
………………….
………………
………………….
………………
………………….
………………
………………….
………………
………………
2 - Because of the gene interaction phenomenon, some phenotypic classes are
reduced in number, as two or more classes may display the same phenotype.
Which are the phenotypic classes of an F2 in which:
a) the dominant allele of a gene masks the effect of the other gene?
……………..
……………..
……………..
b) the recessive homozygote masks the effect of the other gene?
……………..
……………..
……………..
c)
the phenotype due to the dominant allele of a gene is indistinguishable from
the phenotype due to the dominant allele of the other gene? Distinguish the two
situations: when there is addictive effect and when there isn’t.
……………..
……………..
……………..
……………..
……………..
2
Scheda 6
3 – (Trifolium repens) The A gene encodes the enzyme α, the independent gene B
encodes the enzyme β; these enzymes are both involved in the
following metabolic pathway for cyanide production.
.

S1

S2
cyanide
The recessive alleles, a and b, encodes for inactive enzymes.
a) Identify the genotypes of P and F1 individuals and determine which phenotypic
classes are expected in F2 and with which frequencies.
P
not-producing plant x not-producing plant
F1
producing plants
F2
?
b) what’s the name of genes that interact in the way showed here?
c) which phenotypic classes are expected from the backcross of an F1 plant with the
recessive homozygote?
4. The two independent genes P and C control two different consecutive steps in the
same metabolic pathway.
S1
a) PPcc
P

S2
C

yellow fruit
The dominant alleles P and C encode for active enzymes, while
the recessive alleles, p and c, encode for inactive enzymes.
Which phenotype is expected for individuals with the following
genotype (intermediate substrates are colourless):
b) PpCc
c) ppCC
d) Ppcc
5. - The two independent genes, A e E, that originate through duplication, regulate
the same step in the metabolic pathway
that leads to the synthesis of a pigment.
The colourless phenotype can be seen
only in the recessive homozygote state for
both genes. Which phenotypic ratio is
expected from the cross between two dihybrids?
6. Two pure lines of pepper are corssed: first line with Red fruits and the second one
with orage fruites. The plants of F1 produce only red peppers,
but crossing two individuals of F1 we obtain:
Red
Orange
White
192
47
14
3
Scheda 6
We know that the color fruit is controlled by 2 genes (A and B), which kind of
interaction is present? Determine the genotypes and hypotize a possible action on
metabolic chain.
Scheda 7
1
FILE 7 - MUTANTS IN MICRORGANISMS - SELECTION - MUTATIONS IN GENES
THAT CONTROL METABOLIC PATHWAYS
1. The reproduction of the bacterium Escherichia coli is achieved by binary fission,
after his genome replication. Complete the following scheme representing the
chromosome in various stages.
2. A prototroph bacterium can grow on a minimum media, composed by inorganic
minerals and containing an organic source of carbon. Glucose is the most simple
source of carbon. Alternative carbon sources can be used by wild bacteria. Some
mutants loose this ability.
The ability of bacteria to grow on different media is reported in the following table;
identify their phenotype.
STRAIN
n. 1
n. 2
n. 3
n. 4
Minimum media containing
Galactose
Lactose
+
+
+
+
+
Glucose
+
+
+
+
Phenotype
Arabinose
+
+
+
-
3. A prototroph bacterium can grow on a minimum media, because it can synthesize
what it needs. Mutants that aren’t able to grow on a minimum medium (auxotrophs)
can be analyzed on media enriched with different kinds of molecules, to determine
their phenotype.
The ability of bacteria to grow on different media is reported in the following table;
identify their phenotype.
STRAIN
n. 1
n. 2
n. 3
n. 4
Complete
media
+
+
+
+
--
Minimum media + glucose and
arginine pyrimidines adenine
+
+
+
-
biotin
+
Phenotyp
e
4. A mutant bacterium isn’t able to synthesize the prolin amino acid and to
metabolize lactose. Indicate which molecules must be added to a minimum media to
grow this mutant.
Scheda 7
2
5. An Escherichia coli mutant carries a temperature-sensitive mutation in the
polymerase III gene that is necessary for DNA replication. At which temperature can
this bacterium grow?
6. The wild-type Escherichia coli can’t grow on streptomycin-enriched media. If in a
culture of 10 millions cells, there are about 100 streptomycin-resistant mutant cells,
how can I select them (i.e. grow only the resistant cells)?
7. Two Escherichia coli mutants display both the Met- phenotype, i.e. they aren’t able
to....
The two mutations are located on different genes. How can you explain this
situation?
8. Two Escherichia coli mutants can’t grow on galactose medium if galactose is the
unique source of Carbon.
- Assign the phenotypes at the mutants......................
- Are you sure that the mutations regard the same gene?
Scheda 7
3
SCHEDA 8 – GENETIC MUTATIONS – Reversion and suppression1 – In a mRNA sequence (wt) there is a triplet UUU. After a mutation the triplet changes in
UUA. What kind of mutation happened and which effects have the mutation on the protein
encoded by the gene?
2 – This is a sequence of wt mRNA:
5’- AUG AGA CCC ACC….
What kind of effects has a mutation the substitute the fifth base from G to U?
5’– AUG AUA CCC ACC…
What kind of effects has a deletion on the sixth base?
5’– AUG AUC CCA CC….
Compare the two previous mutation and the effects on the protein.
3 - A nucleotide sequence can be seen below:
5
5'
3'
10
15
20
25
30
35
ATTCGATGGGATGGCAGTGCCAAAGTGGTGATGGC
TAAGCTACCCTACCGTCACGGTTTCACCACTACCG
3'
5'
Knowing that the transcription of this sequence is from left to right, write the resulting
mRNA sequence:
…………………………………………………………………………
Knowing that this mRNA sequence contains the translation starting codon, identify the
starting codon and indicate the amino acid sequence of the resulting peptide.
………………………………………………………………………….
Individuate which consequences will have on the amino acid sequence:
- a transition of the TA base pair in position 18;
…………………………………………………………………………
…………………………………………………………………………
…………………………………………………………………………
- a transversion of the CG base pair in position 20;
…………………………………………………………………………
…………………………………………………………………………
…………………………………………………………………………
- an insertion of a base pair after pair 11 (AT)
…………………………………………………………………………
…………………………………………………………………………
…………………………………………………………………………
d) How can we abolish the insertion mutation in position 11?
Scheda 7
4
4 – This is a polypeptidic sequence of a protein.Here wild type and mutants sequences are
compared. What type of mutations did happen?
WT :
Met – Arg – Phe – Thr…..
Mutant 1:
Met – Ile – Phe – Thr …..
Mutant 2:
Met – Ser – Ile - Tyr…..
5 An Escherichia coli mutant auxotroph for tryptophan (Trp-) has an amino acid
substitution in tryptophan-synthetase: glycine in position 210 is replaced by an arginine.
On the basis of the genetic code, individuate which kind of mutation (on the DNA) you
think has caused the amino acid replacement.
6 - In Escherichia coli metA gene a base substitution occurred. Because of this mutation,
in the mRNA a UAA codon is present inside the gene.
- Which consequences will have this mutation on protein synthesis?
7 - The primary transcript of chicken ovalbumin RNA is composed by 7 introns (white) and
8 exons (black):
- If the Ovoalbumin DNA is isolated, denaturated and hybridized with its
cytoplasmatic mRNA, which kind of structure we expect?
- if a deletion of a base pair occurs in the middle of the second intron, which will be the
likely effects on the resulting polypeptide?
- if a deletion of a base pair occurs in the middle of the first exon, which will be the likely
effects on the resulting polypeptide?