Biology Chapter 11 Introduction to Genetics Mendel and

Biology Chapter 11 Introduction to Genetics: Mendel and Meiosis

INTRODUCTION TO GENETICS I. The work of Gregor Mendel A. : the scientific study of heredity Genetics B. Heredity: Passing genes from generation to generation II. Gregor Mendel's Peas A. In the 1800's, _______________ (an Gregor Mendel Austrian Monk) conducted the first scientific study of heredity using pea plants. B. Pea plants contain both male pollen (sperm) and female (eggs) reproductive parts.

Flowering Plant Structures: Pea Plant C. ________ = Joining of male and female reproductive cells Fertilization

D. _________= a pea plant whose pollen Self-pollination fertilizes the egg cells in the very same flower. 1. Mendel discovered that some plants ______ for certain traits “Bred True” 2. Trait= Specific Characteristics Example: seed color, plant height 3. True breeding (a. k. a. pure)= Peas that are allowed to self-pollinate produce offspring identical to themselves Example: Short plants that self pollinate for generations always produce offspring that were pure for shortness.

Cross Pollination Self pollination

E. ________= male sex cells from one Cross-pollination flower pollinate a female sex cell on a different flower. F. Mendel manually cross pollinated pea plants, removing the male parts to ensure no selfpollination would occur. G. Through a series of experiments, Mendel was able to make discoveries of basic principles of heredity. 1. principle of Dominance 2. principle of Segregation 3. principle of Independent Assortment

III. Experiments Mendel performed 7 A. Mendel studied __ different traits in pea plants each with 2 contrasting features. B. Each trait Mendel studied was controlled by one gene. C. Different forms of a gene (trait) = Alleles Example: Gene for plant height has 2 alleles Dominant: T = tall Recessive: t = short

Figure 11 -3 Mendel’s Seven F 1 Crosses on Pea Plants Mendel’s Seven Crosses on Pea Plants Section 11 -1 Go to Section: Seed Shape Seed Color Round Yellow Seed Coat Color Gray Pod Shape Pod Color Flower Position Smooth Green Axial Tall Short Wrinkled Green White Constricted Yellow Terminal Round Yellow Gray Smooth Green Axial Plant Height Tall

Mendel Experiment #1: Parent Offspring pure bred tall x pure bred tall All plants are TALL pure bred short x pure bred short All plants are Pure bred tall x pure bred short All plants are SHORT TALL

Conclusion: genes · individual factors (now known as _____) · the factors did not blend ________________= some alleles Principle of Dominance are dominant (expressed trait - written as a capital letter. Some are recessive (hidden/masked trait; written as a lower case letter) From these conclusions, Mendel wanted to continue his experiments to see what happened to the recessive trait

Principles of Dominance Section 11 -1 P Generation Tall Go to Section: Short F 1 Generation Tall F 2 Generation Tall Short

Principles of Dominance Section 11 -1 P Generation Tall Go to Section: Short F 1 Generation Tall F 2 Generation Tall Short

Principles of Dominance Section 11 -1 P Generation Tall Short F 1 Generation Tall F 2 Generation Tall 3 tall : 1 short Go to Section: Short

Conclusion: Principle of Segregation · ______________: The reappearance of the recessive allele indicated that at some point the allele for shortness separated from the allele for tallness. Mendel suggested that the alleles separated during the formation of the sex cells during meiosis.

IV. PROBABILITY AND PUNNETT SQUARES A. Each coin flip is independent of the next. Past outcomes do not affect future ones Similar to alleles that segregate randomly, like a coin flip. B. The larger the number of trials the closer you get to the expected outcomes C. The principles of probability can be used to predict the outcomes of genetic crosses.

IV. PUNNETT SQUARES Use of Punnett squares help determine the probable outcomes of genetic crosses. · Vocabulary to help with Punnett squares -Homozygous = Having 2 identical alleles (TT, tt) -Heterozygous= Having 2 different alleles (Tt) -Genotype= Genetic makeup of an organism (TT, tt, Tt) -Phenotype= Physical appearance (tall or short) -Hybrids= The offspring resulting from a cross between parents of contrasting traits

Example of a Punnett square: Parent (P) cross homozygous tall( ) x homozygous short( ) tt TT t t · T Tt Tt Tt F 1 offspring Probability of producing homozygous tall offspring? 0/4 Probability of producing hybrid? 4/4

IV. PROBABILITY AND SEGREGATION A. For fun, lets cross F 1’s to see if Mendel’s assumption about segregation are correct: Tt x Tt T T TT Tt t t Tt tt If the alleles segregate during meiosis, then the probable outcomes will be: TT= 1/4 Tall= 3 Tt= 2/4 Short= 1 tt= 1/4 Ratio tall: short= 3: 1

Conclusion: Mendel was correct in his assumptions about Segregration IV. PROBABILITY AND INDEPENDENT ASSORTMENT A. Mendel wondered if one pair of alleles affected the segregation of another pair of alleles. Do round seeds have to be yellow? B. The two factor cross: Mendel crossed RRYY x rryy (P)(aka: two trait cross) All offspring are Hybrid (Rr. Yy) (F 1)

A. Then he crossed the hybrids (F 1): Rr. Yy x Rr. Yy Punnett square formatting rules for 2 trait crosses 1. Determine the possible gametes produced by the parents. a. F- irst two Rr. Yy (RY) (Ry) O- utside two (r. Y) I- nside two (ry) L- ast two

2. Place one parent’s gametes at the top of a 16 Punnett square and the other parent’s gametes on the side of the 16 -Punnett square. RY Ry r. Y Rr. YY ry Rr. YY Rr. Yy Rryy rr. YY Rryy Rr. Yy Ry RRYy RRyy RY RRYy ry r. Y rr. Yy rryy

Section 11 -3 Probability: RY (round and yellow)= 9/16 Ry (round and green = 3/16 r. Y (wrinkled and yellow)= 3/16 ry (wrinkled and green)= 1/16 Phenotype Ratio= 9: 3: 3: 1 Conclusion= Alleles for seed shape independently assort. Go to Section:

In humans hitchhiker’s thumb is a dominant trait to having a straight thumb, and being able to roll your tongue is a dominant trait compared to not being able to. Complete the following dihybrid cross: Male: Homozygous dominant for hitchhiker’s thumb and heterozygous for tongue rolling Female: Has a straight thumb and cannot roll her tongue

In humans hitchhiker’s thumb is a dominant trait to having a straight thumb, and being able to roll your tongue is a dominant trait compared to not being able to. Complete the following dihybrid cross: Male: Heterozygous for hitchhiker’s thumb and heterozygous for tongue rolling Female: Heterozygous for hitchhiker’s thumb and cannot roll her tongue

Independent assortment Genes for different traits can segregate independently during the formation of gametes ****This is true if the traits you are studying are located on different chromosomes Just by chance all 7 of Mendel’s traits were on different chromosomes.

Beyond Dominant and Recessive Alleles Key idea: Some alleles are neither dominant nor recessive, and many traits are controlled by multiple alleles or multiple genes.

Incomplete Dominance in Four O’clock Flowers Incomplete Dominance: One allele is not completely ________ dominant over another. Therefore the phenotype in the heterozygous is somewhere _____ in between the two homozygous phenotypes.

Incomplete Dominance in Four O’clock Flowers

equally Codominance: both alleles contribute _____ to the phenotype. more than two Multiple Alleles: Genes that have _______ alleles. This does not mean an individual can have more than two alleles, but that there are more than two population alleles in the ________ for a given trait. Ex. Rabbit coat color, blood type

Multiple Alleles and Codominance 3 Alleles: i. A, i. B, I Blood Type/Phenotype AO i. A and i. B are codominant AA i. A, i. B both dominate over i BB BO

Polygenic Inheritance: The interaction of many genes controls one trait. It is usually recognized in traits that show a __________ such as skin color, height, range of phenotypes and body weight.

Section 11 -4: Meiosis I. MEIOSIS A. Meiosis= process of _____________ in which the number of Reduction Division chromosomes per cell is cut in 1/2 and the homologous chromosomes that exist in a diploid cell are separated. (and produce haploid cells) B. Purpose= Form gametes (egg and sperm)

II. DIPLOID AND HAPLOID CHROMOSOME NUMBER A. During ________ the genetic material from fertilization one parent combines with genetic material from another Example: A fruit fly has 8 chromosomes A set of 4 came from the female fly A set of 4 came from the male fly B. The two sets of chromosomes are said to be homologous = a female chromosome has a corresponding male chromosome.

C. = contain both sets of Diploid (2 n) homologous chromosomes D. = contain 1 set only Haploid (n) Male gamete Female gamete Sperm (n) = 23 chromosomes Egg (n) = 23 chromosomes

Question: If we start with a diploid cell, how do we get an organism that produces haploid gametes? Answer: Meiosis (aka: reduction division) 1 replication; 2 divisions Example: what if: 8 46 Human Fruit fly 16 92 46 Duplicated 8 46 chromosomes 23 23 4 Duplicated chromosomes 4 4 8 4

III. PROCESS OF MEIOSIS (DIVIDED INTO 2 STAGES: MEIOSIS I & II INTERPHASE: growth, DNA synthesis, protein production, organelle production Meiosis I prophase I 1. homologous chromosomes pair up (Form tetrads) 2. nucleoli disappear 3. nucleus disappears 4. crossing-over occurs: portions of chromatids exchange genetic material 2 n (diagram 277) A.

Crossing-Over Crossing Over: exchange of genetic material between homologous chromosomes Go to Section:

Crossing Over Go to Section:

Crossing-Over Crossing Over Go to Section:

metaphase I 1. homologous pairs (tetrads) line up at the equator 2. spindles attach to chromosomes and independent assortment occurs anaphase I 1. spindles pull the homologous chromosomes toward opposite ends of the cell Key point: homologous pairs separate, cell now haploid

Telophase I 1. Nuclear membranes reform n 2. cell begins to separate into two new haploid cells 3. 2 haploid daughter cells

Figure 11 -15 Meiosis Section 11 -4 Meiosis I Interphase I Prophase I Cells undergo Each chromosome pairs with its corresponding homologous chromosome to form a tetrad. a round of DNA replication, forming duplicate Chromosomes. Go to Section: Metaphase I Anaphase I Spindle fibers attach to the The fibers pull the chromosomes. homologous chromosomes toward the opposite ends of the cell.

Figure 11 -15 Meiosis Section 11 -4 Meiosis I Interphase I Prophase I Metaphase I Cells undergo a round of DNA replication, forming duplicate Chromosomes. Each chromosome pairs with its corresponding homologous chromosome to form a tetrad. Spindle fibers attach to the The fibers pull the chromosomes. homologous chromosomes toward the opposite ends of the cell. Go to Section: Anaphase I

Figure 11 -15 Meiosis Section 11 -4 Meiosis I Interphase I Prophase I Metaphase I Anaphase I Cells undergo a round of DNA replication, forming duplicate Chromosomes. Each chromosome pairs with its corresponding homologous chromosome to form a tetrad. Spindle fibers attach to the chromosomes. The fibers pull the homologous chromosomes toward the opposite ends of the cell. Go to Section:

Figure 11 -15 Meiosis Section 11 -4 Meiosis I Interphase I Prophase I Metaphase I Anaphase I Cells undergo a round of DNA replication, forming duplicate Chromosomes. Each chromosome pairs with its corresponding homologous chromosome to form a tetrad. Spindle fibers attach to the chromosomes. The fibers pull the homologous chromosomes toward the opposite ends of the cell. Go to Section:

B. Meiosis II (similar process as mitosis; no replication) Prophase II Metaphase II Anaphase II Telophase II/ Cytokinesis n n n ***RESULT: 4 haploid daughters that are genetically different!! n

Figure 11 -17 Meiosis II Prophase II Metaphase II Anaphase II Meiosis I results in two The chromosomes line up in a The sister chromatids haploid (N) daughter cells, similar way to the metaphase separate and move toward each with half the number of stage of mitosis. opposite ends of the cell. chromosomes as the original. Go to Section: Telophase II Meiosis II results in four haploid (N) daughter cells.

Figure 11 -17 Meiosis II Prophase II Meiosis I results in two haploid (N) daughter cells, each with half the number of Go to chromosomes as Section: the original. Metaphase II Anaphase II The chromosomes line up in a The sister chromatids similar way to the metaphase separate and move toward stage of mitosis. opposite ends of the cell. Telophase II Meiosis II results in four haploid (N) daughter cells.

Meiosis II Section 11 -4 Prophase II Go to Section: Figure 11 -17 Meiosis II Metaphase II Anaphase II Meiosis I results in The two haploid (N) chromosomes line The sister chromatids separate and move toward up in a similar way opposite ends of the cell. daughter cells, each with half the to the metaphase stage of mitosis. number of chromosomes as the original. Telophase II Meiosis II results in four haploid (N) daughter cells.

Meiosis II Figure 11 -17 Meiosis II Section 11 -4 Prophase II Metaphase II Anaphase II Meiosis I results in The sister chromosomes line two haploid (N) chromatids up in a similar way daughter cells, separate and each with half the to the metaphase move toward stage of mitosis. number of opposite ends of Go to chromosomes as the cell. Section: the original. Telophase II Meiosis II results in four haploid (N) daughter cells.

Meiosis II Section 11 -4 Prophase II Figure 11 -17 Meiosis II http: //www. sumanasinc. com/webcontent/anisampl es/majorsbiology/meiosis. html Metaphase II Meiosis I results in The chromosomes line two haploid (N) up in a similar way daughter cells, each with half the to the metaphase stage of mitosis. number of Go to chromosomes as Section: the original. Anaphase II The sister chromatids separate and move toward opposite ends of the cell. Telophase II Meiosis II results in four haploid (N) daughter cells.

IV. GAMETE FORMATION (refer to page 278) A. Males The 4 haploid cells (gametes) = sperm 1. 2. male gametes produced by a process called _________ spermatogenesis B. Females 1. 4 haploid cells are produced but only 1 -haploid cell is a viable egg 3 -produce polar bodies caused by uneven cytoplasmic division 2. female gametes produced by a process called ________ oogenesis

V. COMPARING MITOSIS AND MEIOSIS A. Mitosis results in the production of two genetically identical diploid cells, whereas meiosis produces four genetically different haploid cells. Mitosis Meiosis Number of daughter cells Type of cells produced Number of divisions Number of replications 2 diploid cells 4 haploid cells Body cells gametes 2 1 1 Purpose of division Sexual reproduction Growth, replacement, repair, asexual reproduction

11 -5: Gene Linkage and Gene Maps Standards addressed: CA B 1 3. b students know the genetic basis for. Mendel’s laws of segregation and independent assortment. *B 1 3. d. Students know how to use data on frequency of recombination at meiosis to estimate genetic distances between loci and to interpret genetic maps of chromosomes. Key concept: What structures actually assort independently?

Actually ____________ do assort the chromosomes independently just as Mendel had suggested but the _______ on the chromosomes can be ______. linked together genes A. Linked genes 1. Genes located on the _____ same chromosome together 2. Inherited _______ 3. Do not undergo __________; independent assortment they don't follow Mendel's law (Just by chance all the traits Mendel studied were located on separate chromosomes. . . none were linked. )

B. Linkage group= all the genes on a _______ chromosome * If there are ___ pairs of chromosomes then there are 4 4 23 ____ linkage groups. Humans have ____ pairs of chromosomes therefore ____ linkage groups 23

III. Crossing Over A. If two genes are found on the same chromosome, does it mean that they are linked forever? NO! recombinants. Crossing over produces __________ B. Recombinants= individuals with new combinations _________ of genes

IV. Gene Mapping A. Sturtevant stated that: · crossing over occurs ________ along the randomly linkage groups. · the ________ the genes are from each further other the _______ they will cross over more likely frequency of recombination · using the ____________ (how often crossing over occurs), a gene _______ can be made map for each chromosome

B. Gene map= the _________ on a positions of genes chromosome Example: gene a and gene b cross over 20% gene a and gene c cross over 5% gene b and gene c cross over 75% chromosome: C A B

Figure 11 -19 Gene Map of the Fruit Fly Exact location on chromosomes Chromosome 2