Genetic Processes:

1. Eye colour may change with age or can exhibit odd phenotypes such as the heterochromia seen in this cat. Genetic Processes: These organisms are important because they contributed to our earliest understanding of inheritance. Dogs Cats Cats Peas Horses Fruit flies Wheat

2. Gregor Mendel (1822-1884) • Genetics is the study of heredity and variations that attempts to predict inheritance patterns and explain why an offspring resembles their parents and why differences appear. • Gregor Mendel lived as a monk in Austria his entire adult life and observed inherited (hereditary) traits in the 1850s. • For seven years, he bred and tested more than 29 000 pea plants in order to formulate his transmission ideas and develop a theory of inheritance patterns. • He hypothesized that each pea plant parent passed down separate and distinct factors (now known as genes) to its offspring for generations that are responsible for hereditary traits. • Mendelian Laws and the Chromosome Theory help explain inheritance patterns and now form the basis for genetics.

3. Mendel’s Experiments • Mendel’s experiments on garden peas started with selectively pure-breeding or true-breeding plants through generations to produce offspring with one distinct trait that descended from and is identical to their parental ancestors. • These offspring were created by self-pollination through cross-fertilization which uses eggs and sperm from different parents who have been identified as pure-breeding or true-breeding.

4. Pure-Breeding or True-Breeding: • Parents that are pure-breed or true-breed are homozygous and exhibit only one trait and produce the same trait when self-pollinated through cross-fertilization. • Offspring that are hybrids are heterozygous and exhibit one or more traits when different pure/true-bred plants are mated. • Homozygous is a zygote which results from the union of two gametes with identical alleles of a gene and thus are represented by the same symbols (TT or tt). • Heterozygous is a zygote which results from the union of two gametes with different alleles of a gene and thus are represented by contrasting symbols (Tt).

5. A gene is the basic unit of heredity and is made up of a piece of DNA at a specific location on a chromosome called a locus which codes for protein(s) that produce a certain inherited trait. An allele is an alternate form of a gene which controls a given characteristic such as the colour of a flower and are found at the same positions (loci) on corresponding (homologous) chromosomes.

6. Dominant is the allele that is expressed when in the heterozygous form and is represented by a capital letter. • Recessive is the allele that is masked when in the heterozygous form and is represented by the corresponding lower case letter. • Genotype is the description of the actual genetic make-up represented as the combination of alleles for a gene. • Phenotype is the description of the visible appearance or expression of the characteristic trait which depends upon its genotype and its interaction with its environment. T T T t t t tt TT Tt

7. Pea Plant Inherited (Hereditary) Traits:

8. The Mendelian Model of Inheritance: • Mendel summarized his work without knowing about genes. • Genes are the mechanism of how these traits were passed down. • Genes can exist in different forms, known as alleles. • Alleles for pea seed shape include round and wrinkled. • Offspring inherit one allele of a gene from each parent. • If both alleles are the same, then homozygous for that gene. • If alleles are different, then heterozygous for that gene. • Alleles can be dominant or recessive. • Dominant alleles are expressed in heterozygous offspring. • Recessive alleles are only expressed in homozygous offspring. • Each gamete (sex cell) carries only one allele for each trait.

9. Laws of Heredity:Mendel's Law of Segregation • The principle of Mendel's Law of Segregation is explained by or describes what happens to homologous chromosomes (allele pairs) that make up a gene during the formation of gametes in meiosis. • During meiosis, alleles segregate or separate into gametes, where only one allele for a gene trait appears in each gamete.

10. Laws of Heredity:Mendel's Law of Independent Assortment • The principle of Mendel's Law of Independent Assortment suggests that alleles of different genes assort independently of other genes during gamete formation in meiosis – metaphase I. • Any plant contains many different kinds of genes, for example, one gene determines flower color, a second gene determines length of stem, a third gene determines shape of pea pods, and so on. • Mendel discovered that the way in which alleles from different genes separate and then recombine during fertilization is unconnected to other genes.

11. Laws of Heredity: The Chromosomal Theory of Inheritance proposes that genes are contained within chromosomes in the cell's nucleus and have five basic points: • Nuclei contain two sets of homologous chromosomes (one maternal and one paternal). • Chromosome retain identity and are genetically continuous through the life cycle. • The two sets of homologous chromosomes are functionally equivalent. • Maternal and paternal homologous chromosomes synapse during meiosis then move to opposite poles. • Maternal and paternal homologous chromosomes segregate independently.

12. Monohybrid Inheritance A monohybrid cross is a cross between two individuals differing in ONE characteristic trait. (c) McGraw Hill Ryerson 2007

13. Mendel’s Monohybrid Crosses: P = parental generation (from different flowers) F1 = first filial generation of offspring (called hybrids) F2 = second filial generation of offspring (from F1 parents) • Mendel studied 7 traits in pea plants using monohybrid crosses. • In each monohybrid cross, one trait would disappear in F1, then re-appear in about 25% of the F2 generation.

14. A Punnett square is a simple lattice diagram that shows the segregation and combination of male and female gametes produced as a result of meiosisand is used to illustrate the possible results of a cross between the gametes of two individuals.

16. In the cross-fertilization shown, the parents are both known to be true-breeding and exhibit only one trait difference, therefore this is a monohybrid cross. • The offspring exhibit definite traits (tall or short), and not a combination (medium). • Traits that “skipped” the F1 generation were not lost, but appeared in the F2 generation. • The F1 generation is heterozygous, displaying the dominant trait. • The F2 phenotypes are 75% dominant to 25% recessive (3:1) trait ratio.

17. Probability and the Punnett Square: • Mendel’s range of accurate data allowed him to realize the rules of probability could be used. • A Punnett square is used to determine the probabilities of allele combinations when the genotypes of the parents are known. • The given Punnett square shows a monohybrid cross of two plants heterozygous for the height allele. • What is the probability that an offspring of this cross will have the genotype Tt? • What is the probability that an offspring of this cross will have the phenotype tall (T)? 3 Tall : 1 Short

18. Constructing a Punnett Square forOne Gene with Two Alleles at One Locus: Genotype1 BB:2Bb:1bb Phenotype3 Blue: 1 green

19. Example: A geneticist crosses two pea plants. One of the plants is heterozygous for the dominant inflated pea pod trait, and the other plant has constricted pea pods. What would be the expected genotypic and phenotypic proportions of the offspring? Given Inflated pod (I) is dominant to constricted pod (i) II or Ii = Inflated pods ii = constricted pods The cross is heterozygous inflated x constricted: Iixii • Solution • • The Ii parent produces gametes I and i. • The ii parent produces gametes i and i. Required • expected proportions of F1 plants with genotypes II, Ii, and ii • expected proportions of plants with inflated pods and plants with constricted pods Paraphrase From the Punnett square, ½ of the offspring would be heterozygous (Ii) and ½ would be homozygous recessive (ii). Therefore, ½ of the offspring would have inflated pods and ½ would have constricted pods. Analysis • Determine the gametes from the parental genotypes. • Construct a Punnett square for this cross to determine the allele combinations and phenotypes of the offspring.

20. Practice Problem 1:Use the Punnett square method to analyze the following crosses. What would be the expected genotypic and phenotypic proportions of the offspring in each cross? A brown snake heterozygous for the skin-colour gene (Aa) is crossed with an albino (colourless) snake (aa). • Solution • • The Aa parent produces gametes A and a. • The aa parent produces gametes a and a. Given Brown (A) is dominant to albino pod (a) AA or Aa = brown snake aa = albino snake The cross is heterozygous brown x albino: Aaxaa Required • expected proportions of F1 plants with genotypes AA, Aa, and aa • expected proportions of brown snakes with albino snakes Paraphrase From the Punnett square, ½ of the offspring would be heterozygous (Aa) and ½ would be homozygous recessive (aa). Therefore, ½ of the offspring would be brown and ½ would be albino. Analysis • Determine the gametes from the parental genotypes. • Construct a Punnett square for this cross to determine the allele combinations and phenotypes of the offspring.

21. Practice Problem 2:Use the Punnett square method to analyze the following crosses. What would be the expected genotypic and phenotypic proportions of the offspring in each cross? A man with dimples (DD) has children with a woman who is heterozygous for the dimples / no dimples gene (Dd). • Solution • • The DD parent produces gametes D and D. • The Dd parent produces gametes D and d. Given Dimples (D) is dominant to no dimples (d) DD or Dd = dimples dd = no dimples The cross is homozygous dimples x heterozygous dimples: DDxDd Required • expected proportions of F1 plants with genotypes DD, Dd, and dd • expected proportions of dimples with no dimples Paraphrase From the Punnett square, ½ of the offspring would be heterozygous (Aa) and would be homozygous dominant (AA). Therefore, all of the offspring would have dimples. Analysis • Determine the gametes from the parental genotypes. • Construct a Punnett square for this cross to determine the allele combinations and phenotypes of the offspring.

22. Practice Problem 3:Use the Punnett square method to analyze the following crosses. What would be the expected genotypic and phenotypic proportions of the offspring in each cross? Some people have the ability to taste the bitter chemical PTC. A PTC tasting woman (Tt) has children with a PTC-tasting man (Tt). Given PTC tasting (T) is dominant to no PTC tasting (t) TT or Tt = PTC tasting tt = no PTC tasting The cross is heterozygous PTC tasting x heterozygous PTC tasting: TtxTt Solution • The Tt parents produce gametes T and t. Required • expected proportions of F1 plants with genotypes TT, Tt, and tt • expected proportions of PTC tasting with no PTC tasting Paraphrase From the Punnett square, ½ of the offspring would be heterozygous (Tt), ¼ would be homozygous dominant (TT) and ¼ would be homozygous recessive (tt). Therefore, ¾ of the offspring would taste PTC and ¼ would not. Analysis • Determine the gametes from the parental genotypes. • Construct a Punnett square for this cross to determine the allele combinations and phenotypes of the offspring.

23. The Monohybrid Test Cross: • The phenotype can be seen, but how can the genotype be determined? • A test cross is a back cross involving breeding an individual with the dominant phenotype with a homozygous recessive individual to determine whether the unknown genotype is homozygous dominant or heterozygous. • If offspring are all dominant phenotype, then the parent was homozygous dominant. • If offspring are half dominant phenotype and half recessive phenotype (1:1 ratio), then the parent was heterozygous dominant.

24. Dihybrid Inheritance A dihybrid cross is a cross between two individuals differing in TWO characteristic traits. (c) McGraw Hill Ryerson 2007

25. Mendel’s Dihybrid Crosses: • In each dihybrid cross, a 9:3:3:1 phenotypic ratio was observed where there are two loci in which there are two alleles per locus. • There is independent assortment between loci where the gene at one locus does not affect the phenotype of another locus. • There is a dominance-recessive relationship between the alleles of a gene found at each locus. P Generation F1 Generation F2 Generation

28. Constructing a Punnett Square forTwo Genes with Two Alleles at Two Loci: Genotype9:3:3:1 Phenotype9:3:3:1

29. Example: In mice, the normal long-tail phenotype is dominant to the short-tail trait, and black coat colour is dominant to brown coat colour. If two long-tailed black mice, heterozygous for both traits, are mated, what proportion of their offspring will be brown with short tails? Given Long tail (L) is dominant. Short tail (l) is recessive. LL or Ll produce long tails. ll produces short tails. Black coat colour (B) is dominant. Brown coat colour (b) is recessive. BB or Bb produce black coat colour. bb produces brown coat colour. The cross is BbLl XBbLl. Solution • The BbLl parents produce gametes BL, Bl, bL, bl. Required • expected proportion of offspring that will be brown with short tails Paraphrase Mice with brown coats and short tails have the genotype bbll. From the Punnett square, 1/16 of the offspring will be brown with short tails. Analysis • Determine the gametes from the parental genotypes. • Construct a Punnett square for this cross to determine the allele combinations and phenotypes of the offspring.

30. Practice Problem 1: A black mouse with a short tail (BBtt) is mated with a brown mouse with a short tail (bbtt). What proportion of their offspring will be black with short tails? Given Long tail (T) is dominant. Short tail (t) is recessive. TT or Tt produce long tails. tt produces short tails. Black coat colour (B) is dominant. Brown coat colour (b) is recessive. BB or Bb produce black coat colour. bb produces brown coat colour. The cross is BBtt Xbbtt. • Solution • • The BBtt parent produces gametes Bt. • The bbtt parent produces gametes bt. Required • expected proportion of offspring that will be black with short tails Analysis • Determine the gametes from the parental genotypes. • Construct a Punnett square for this cross to determine the allele combinations and phenotypes of the offspring. Paraphrase All of the offspring, 100% or 1, have black coats and short tails.

31. Practice Problem 2: A tall, purple-flowered pea plant (TTPP) is crossed with a short, purple-flowered pea plant (ttPp). What proportion of the offspring will be heterozygous for flower colour? What proportion of the offspring will be tall with white flowers? Given Tall (T) is dominant. Short (t) is recessive. TT or Tt produce tall plants tt produces short plants. Purple flower colour (P) is dominant. White flower colour (p) is recessive. PP or Pp producs purple flowers. pp produces white flowers. The cross is TTPP XttPp. • Solution • • The TTPP parent produce gametes TP. • The ttPp parent produce gametes tP and tp. Required • expected proportion of offspring that will be tall with white flowers Paraphrase Half or 8/16 of the offspring are heterozygous for flower colour, Pp. There are no plants with white flowers produced; every plant has purple flowers. Analysis • Determine the gametes from the parental genotypes. • Construct a Punnett square for this cross to determine the allele combinations and phenotypes of the offspring.

32. Practice Problem 3: A black-eyed, long-haired rabbit (BBss) is mated with a red-eyed, short-haired rabbit (bbSS). What proportion of the offspring will have black eyes and short hair? What proportion will have red eyes and long hair? Given Black eyes (B) is dominant. Red eyes (b) is recessive. BB or Bb produce black eyes. bb produces red eyes. Short hair (S) is dominant. Long hair (s) is recessive. SS or Ss produce short hair. ss produces long hair. The cross is BBss XbbSS. • Solution • • The BBss parent produces gametes Bs. • The bbSS parent produces gametes bS. Required • expected proportion of offspring that will have black eyes and short hair Analysis • Determine the gametes from the parental genotypes. • Construct a Punnett square for this cross to determine the allele combinations and phenotypes of the offspring. Paraphrase 100% of the offspring or 1, have black eyes and short hair. None have red eyes or long hair.

33. The Dihybrid Test Cross: • The phenotype can be seen, but how can the genotype be determined? • A test cross is a back cross involving breeding an individual with the dominant phenotype with a homozygous recessive individual to determine whether the unknown genotype is homozygous dominant or heterozygous. • If offspring are all dominant phenotype, then the parent was homozygous dominant for both traits. • If offspring are a 1:1:1:1 phenotype ratio for all genotype combinations, then the parent was heterozygous dominant for both traits.

34. Applying Mendel’s Laws: • Mendel’s laws of inheritance apply to animals and plants. • Thomas Morgan used fruit flies, Drosophila melanogaster, to test Mendel’s theories on an animal species. • The fruit fly has easy to identify traits and reproduces quickly. • Many (but not all) traits in plants and animals follow Mendel’s laws. • Fruit fly reproduction and inheritance patterns help us better understand genetic processes.