DNA Replication and Cell Cycle Mitosis and Meiosis Monohybrid cross

lagging strand RNA primer leading strand 1. Replication of DNA molecule. Draw the new synthetized dna with the polarity of the strands and the Okazaki fragments. lagging strand 5’ Okazaki Fragments 5' 3' 5 3’ ' 5’ RNA primer 3’ leading strand

2. How is the structure of a chromosome before and after the replication process? Performe a scheme with the chromosome like a line and the centromere like a circle. BEFORE AFTER

Phase M Mitosis Cell Cycle Phase S DURANTE LA FASE S AVVIENE LA DUPLICAZIONE DEI CROMOSOMI Dna Synthesis 3. Complete the scheme of cell cycle of a diploid eukaryotic cell. Draw the chromosomes at different stages of the cycle.

Draw a schematic picture of chromosomes of a diploid cell with n=1 Indifferent mitosis stage. The analysed individual is heterozygous for gene A. G1 A a PROPHASE A a A a METAPHASE a A ANAPHASE A a A a Daughter cells

Draw a schematic picture of chromosomes of a diploid cell with n=1 indifferent meiosis stage. The analysed individual is heterozygous for gene A. GAMETOCITE A a A a Prophase I A a Metaphase I Anaphase I A a

Anaphase I Telophase I Metaphase II Anaphase II Gametes A a A a A a A

Which structures migrate at the opposite poles of the spindle? in mitosis SISTER CHROMATIDS b) in meiosis, division I HOMOLOGOUS CHROMOSOMES c) in meiosis, division II SISTER CHROMATIDS

Which types of gametes and in which proportions are produced by individuals that have the following genotypes? genotype AA; ONLY GAMETES A b) genotype Aa; ½ GAMETES A ½ GAMETES a c) genotype aa ONLY GAMETES A

For each cross determine genotypic and phenotypic classes expected in the progeny and relative frequencies. Genotype of the individuals used for the cross Gametes of the first individual (Frequency) Gametes of the second individual (Frequency) Genotypes and frequency of the progeny Phenotypes and frequency of the progeny AA x aa Aa x aa Aa x Aa A (1) a (1) Aa (1x1=1) A (1) ½ A ½ a Aa (½ x 1= ½) aa (½ x 1= ½) A (½) a (½) a (1) A (¼+2/4=3/4) a (¼) AA (½ x ½= ¼ ) Aa ( ¼ + ¼ = 2/4) aa (½ x ½= ¼ ) ½ A ½ a

n° OF INDIVIDUAL OF THE PROGENY 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 n° OF INDIVIDUAL OF THE PROGENY PARENTAL GENOTYPE SHORT HAIR LONG HAIR a) SHORT x LONG 100 PP x pp X a) The parents have different phenotypes then different genotypes. The progeny is homogeneous (short hair) then short hair (P) is dominant over long hair (p). The parent with long hair will be homozygous recessive (pp) while the parent with short hair coul be PP o Pp. In order to determine the genotype of the first parent I observed the phenotypes of the progeny: all individuals with short hair. Then the first parent will be homozygous dominant PP.

n° OF INDIVIDUAL OF THE PROGENY 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 n° OF INDIVIDUAL OF THE PROGENY PARENTAL GENOTYPE SHORT HAIR LONG HAIR a) short x long 100 b) short x long 50 PP x pp Pp x pp X b) We have established that short is dominant over long: the parent with long hair is homozygous recessive pp while the parent with short hair could be PP o Pp. In the progeny we have long and short hair individuals in the same proportion. The parent with the short hair will be heterozygous (Pp).

n° OF INDIVIDUAL OF THE PROGENY 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 n° OF INDIVIDUAL OF THE PROGENY PARENTAL GENOTYPE SHORT HAIR LONG HAIR a) short x long 100 b) short x long 50 c) short x short 150 PP x pp Pp x pp Pp x Pp X c) The parents have the same genotypes (short hair) but in the progeny we have an alternative phenotype (long hair): both individuals will be heterozygous to produce homozygous recessive(with frequency of ¼).

Wild-type Drosophila melanogaster has red eyes 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.

This phenotype is controlled by the pr gene, which has two allelic states pr+ and pr. The following crosses have been done: Parental phenotypes n° of individuals in the progeny Parental genotypes Red purple total a) red x red 125 35 160 b) purple x purple c) red x red d) purple x red pr+ pr x pr+ pr a) Crossing two individuals with Red phenotypes we obtain individual with purple phenotype. The parent is heterozygous and Red is the dominant character (3/4 Red, ¼ Purple).

This phenotype is controlled by the pr gene, which has two allelic states pr+ and pr. The following crosses have been done: Parental phenotypes n° of individuals in the progeny Parental genotypes Red purple total a) red x red 125 35 160 b) purple x purple 45 c) red x red d) purple x red pr+ pr x pr+ pr pr pr x pr pr b) In the progeny we have only purple individuals. The parents are homozygous recessive.

This phenotype is controlled by the pr gene, which has two allelic states pr+ and pr. The following crosses have been done: Parental phenotypes n° of individuals in the progeny Parental genotypes Red purple total a) red x red 125 35 160 b) purple x purple 45 c) red x red 177 63 240 d) purple x red pr+ pr x pr+ pr pr pr x pr pr pr+ pr x pr+ pr c) In the progeny we observe purple individuals.. The parents are heterozygous and the progeny is distributed: 3/4 red ¼ purple.

This phenotype is controlled by the pr gene, which has two allelic states pr+ and pr. The following crosses have been done: Parental phenotypes n° of individuals in the progeny Parental genotypes Red purple total a) red x red 125 35 160 b) purple x purple 45 c) red x red 177 63 240 d) purple x red 55 100 pr+ pr x pr+ pr pr pr x pr pr pr+ pr x pr+ pr pr pr x pr+ pr d) The first parent is purple thus homozygous recessive pr pr. The second parent is Red and its genotype could be pr+ pr+ o pr+ pr. In the progeny we observe individuals homozygous recessive, thus the second parent is heterozygous pr+ pr.

Draw a scheme of meiosis process of a diploid cell with n=2 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 A B b Gametocite FASE S (DNA replication) Prophase I A B a b Metaphase I Homologous chromosome will be separated a A B b a A B b

Metaphase I A B A b a b a B A B a b A b a B Gametes A B a b A b a B A B a b A b a B ¼ AB ¼ ab ¼ Ab ¼ aB

Gene A (frequency) Gene B (frequency) Gametes Now use the branch diagram to determine type and frequency of the gametes produced by the same cell. Gene A (frequency) Gene B (frequency) Gametes B ½ AB ¼ ………….(……..) ………….(……..) ………...(…….) A ½ b ½ Ab ¼ B ½ aB ¼ a ½ b ½ ab ¼

Which type of gametes and in which proportions are produced by individuals that have the following genotype (use the branch diagram)? a) aa bb ab (1) a (1/2) b (1) A (1/2) Ab (1/2) ab (1/2) b) Aa bb c) Aa Bb a (1/2) B (1/2) b (1/2) A (1/2) ab (1/4) AB (1/4) Ab (1/4) aB (1/4)

For each cross determine genotypic and phenotypic classes expected in the progeny and relative frequencies (A and B genes are independent) genotype of the individuals used for the cross Gametes of the first individual (frequency) Gametes of the second individual (frequency) Genotypes and frequency of the progeny phenotypes and frequency of the progeny AA bb x aa BB Aa bb x aa Bb Aa Bb x aa bb Ab (1) aB (1) Aa Bb (1) A B (1) ¼ AaBb ¼ Aabb ¼ aaBb ¼ aabb ¼ AB ¼ Ab ¼ aB ¼ ab Ab (1/2) ab (1/2) aB (1/2) ab (1/2) ¼ AB ¼ Ab ¼ aB ¼ ab ¼ AaBb ¼ Aabb ¼ aaBb ¼ aabb ¼ AB ¼ Ab ¼ aB ¼ ab ab (1)

Now use the branch diagram to calculate the phenotypic classes. b) Aa bb X aa Bb Phenotype for A gene Phenotypes for B gene Phenotypical classes (cross Aa X aa) (cross bb X BB) B ½ AB ¼ ………….(……..) ………….(……..) ………...(…….) A ½ b ½ Ab ¼ B ½ aB ¼ a ½ b ½ ab ¼