DNA Replication and Cell Cycle Mitosis and Meiosis

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üDNA Replication and Cell Cycle ü Mitosis and Meiosis ü Monohybrid cross

üDNA Replication and Cell Cycle ü Mitosis and Meiosis ü Monohybrid cross

1. Replication of DNA molecule. Draw the new synthetized dna with the polarity of

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

2. How is the structure of a chromosome before and after the replication process?

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

Phase M Mitosis Cell Cycle Phase S Dna Synthesis 3. Complete the scheme of

Phase M Mitosis Cell Cycle Phase S 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

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 G 1 a PROPHASE A a METAPHASE ANAPHASE Daughter cells a A a a A A a A

Draw a schematic picture of chromosomes of a diploid cell with n=1 indifferent meiosis

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 Prophase I A a Metaphase I A a Anaphase I A a

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

Anaphase I A a Telophase I A a Metaphase II Anaphase II AA Gametes A A ½A a a ½a

Which structures migrate at the opposite poles of the spindle? a) in mitosis SISTER

Which structures migrate at the opposite poles of the spindle? a) 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

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

For each cross determine genotypic and phenotypic classes expected in the progeny and relative

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

In dogs hair length is determined by a gene, P, that can be present

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 INDIVIDUAL OF THE PROGENY SHORT HAIR LONG HAIR 100 0 PARENTAL GENOTYPE 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.

In dogs hair length is determined by a gene, P, that can be present

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 0 PP x pp b) short x long 50 50 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).

In dogs hair length is determined by a gene, P, that can be present

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 0 PP x pp b) short x long 50 50 Pp x pp c) short x short 150 50 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. Mutants with purple eyes exist. This phenotype is

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+

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 a) red x red Red purple total 125 35 160 Parental genotypes pr+ pr x pr+ pr b) purple x purple c) red x red d) purple x red 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+

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 a) red x red b) purple x purple Red purple total Parental genotypes 125 35 160 pr+ pr x pr+ pr 0 45 45 pr pr x pr pr c) red x red d) purple x red 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+

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 a) red x red b) purple x purple c) red x red Red purple total Parental genotypes 125 35 160 pr+ pr x pr+ pr 0 45 45 pr pr x pr pr 177 63 240 pr+ pr x pr+ pr d) purple x red 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+

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 Red purple total Parental genotypes 125 35 160 pr+ pr x pr+ pr 0 45 45 pr pr x pr pr c) red x red 177 63 240 pr+ pr x pr+ pr d) purple x red 45 55 100 pr pr x pr+ pr a) red x red b) purple x purple d) The first parent is purple thus homozygous recessive pr pr. The second parent is Red and its genotype could be 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. One chromosome

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. Gametocite A B a b FASE S (DNA replication) Prophase I A B a b Homologous chromosome will be separated Metaphase I A B A b a B

Metaphase I A A B A b a B B A B a a

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

Which type of gametes and in which proportions are produced by individuals that have

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 ab (1) A (1/2) b (1) Ab (1/2) a (1/2) b (1) ab (1/2) B (1/2) AB (1/4) b (1/2) Ab (1/4) B (1/2) a. B (1/4) b (1/2) ab (1/4) c) Aa Bb A (1/2) a (1/2)

For each cross determine genotypic and phenotypic classes expected in the progeny and relative

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 AA bb x aa BB Aa bb x aa Bb Aa Bb x aa bb Gametes of the Genotypes and second first individual frequency of the individual (frequency) progeny (frequency) Ab (1) a. B (1) Ab (1/2) a. B (1/2) ab (1/2) ¼ AB ¼ Ab ¼ a. B ¼ ab ab (1) Aa Bb (1) phenotypes and frequency of the progeny A B (1) ¼ ¼ ¼ ¼ Aa. Bb Aabb aa. Bb aabb ¼ ¼ ¼ ¼ AB Ab a. B ab

Now use the branch diagram to calculate the phenotypic classes. b) Aa bb X

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