MENDEL AND THE GENE IDEA

1. MENDEL AND THE GENE IDEA

2. Introduction 1. Mendel brought an experimental and quantitative approach to genetics 2. By the law of segregation, the two alleles for a character are packaged into separate gametes 3. By the law of independent assortment, each pair of alleles segregates into gametes independently 4. Mendelian inheritance reflects rules of probability 5. Mendel discovered the particulate behavior of genes

3. Mendel

4. The current theory for the mechanism for the transmission of genetic material was the “blending” hypothesis. This hypothesis proposes that the genetic material contributed by each parent mixes in a manner analogous to the way blue and yellow paints blend to make green. Introduction • Mendel proposed an alternative model, “particulate” inheritance • This proposes that parents pass on discrete heritable units - genes - that retain their separate identities in offspring. • Genes can be sorted and passed on, generation after generation, in undiluted form.

5. Blending Hypothesis: 1800s -suggested that traits of parents mix to form intermediate traits in offspring. Parents Offspring Red flower x White flower Pink flower Tall height x Short height Medium height Blue bird x Yellow bird Green birds Fair skin x dark skin Medium skin color If blending always occurred, eventually all extremecharacteristics would disappear from the population.

6. Father of Modern Genetics: Gregor Mendel -abbey garden, where Mendel documented mechanisms of inheritance. 1857

7. Mendel grew up on a small farm in what is today the Czech Republic. In 1843- entered monastery. He studied at the University of Vienna from 1851 to 1853. The monks at this monastery had a long tradition of interest in the breeding of plants, including peas. Around 1857, Mendel began breeding garden peas to study inheritance. mendel

8. JOKE BREAK

9. Pea plants have several advantages for genetics. Pea plants are available in many varieties with distinct heritable features (characters) with different variants (traits). Short generation time, little space needed, cross or self fertilize

10. Overview of Mendel’s experiments. -Mendel looked at 7 traits-

11. Cross fertilization and self fertilization

12. Pollen hybridize X • He took a true breeding purple flower plant and crossed it with a true breeding white flower plant. • He called these the parent generation (P generation) • What do you think the offspring looked like? ALL PURPLE

13. Allowed to self-pollinate F1 SAY WHAT??????? F2

14. Something is being passed from parent to offspring. He called these “Factors” Sometimes you can see “it” and sometimes you can’t see “it”. If you can see it- it is dominant.(T) If it’s there and you can’t see it- it’s recessive.(t) Mendel concluded: Purple flower is a dominant trait and white flower is a recessive trait.

15. Each Version of the factor, now known as gene, is called an Allele.

16. 1. Alternative versions of genes (different alleles) account for variations in inherited characters. Different alleles vary somewhat in the sequence of nucleotides at the specific locus of a gene Mendel developed a hypothesis to explain these results that consisted of four related ideas. 2. For each character, an organism inherits twoalleles, one from each parent. • 3. If two alleles differ, then one, the dominant allele, is fully expressed in the organism’s appearance. • The other, the recessive allele, has no noticeable effect on the organism’s appearance.

17. 4. The two alleles for each character segregate (separate) during gamete production (meiosis).

18. Mendel’s quantitative analysis of F2 plants revealed the two fundamental principles of heredity: the law of segregation&the law of independent assortment.

19. If the blending model were correct, the F1 hybrids from a cross between purple-flowered and white-flowered pea plants would have pale purple flowers. Instead, the F1 hybrids all have purple flowers, just as purple as the purple-flowered parents. Law of segregation- the two alleles for a character are packaged into separate gametes

20. The white trait, absent in the F1, reappeared in the F2. Based on a large sample size, Mendel recorded in the F2plants: 705 purple-flowered 224 white-flowered 3:1 ratio(phenotype) Law of segregation- the two alleles for a character are packaged into separate gametes The reappearance of white-flowered plants in the F2 generation indicated that the heritable factor for the white trait was not diluted or “blended” by coexisting with the purple-flower factor in F1 hybrids.

21. Law of segregation-the two alleles for a character are packaged into separate gametes This segregation of alleles corresponds to the distribution of homologous chromosomes to different gametes in meiosis. packaged into separate gametes

22. A Punnett squarepredicts the results of a genetic cross between individuals of known genotype. Two heterozygotes - 3:1 ratio of dominant: recessive phenotypes 1:3:1 genotypes USING THE Law of segregation……….. • PRACTICE: • DRAGON Genetics • #1 & 2 monohybrid

23. In cats, long hair is recessive to short hair. A true-breeding (homozygous) short-haired male is mated to a long-haired female. What will their kittens look like?

24. In fruit flies, the gene for wing shape has an unusual allele called ‘curly’ (designated ‘Cy’). The normal (wild type) allele is designated ‘cy.’ A fly homozygous for cy (cy cy) has normal, straight wings. The heterozygote (Cy cy) has wings which curl up on the ends (and, incidentally, can’t really fly). The homozygote for the Cy allele (Cy Cy) never hatches out of the egg. In other words, this allele is lethal in the homozygous condition. If two curly winged flies are mated, and the female lays 100 eggs, predict the following, : • How many eggs will produce living offspring? • How many straight winged flies do you expect among the living offspring? • What percentage of the living offspring do you expect to be curly winged like the parents?

25. When crossing two pure breeding plants, Mendel found similar 3 to 1 ratios among F2 offspring when he conducted crosses for six other characters, each represented by two different varieties.

26. Predicting the genotype of an organism with the dominant phenotype: The organism must have at least one dominant allele, but it could be homozygous dominant or heterozygous. what’s the genotype????? A testcross, breeding a homozygous recessive with dominant phenotype, but unknown geneotype, can determine the identity of the unknown allele.

27. Testcross • A test cross is a way to explore the genotype of an organism. It is used to determine if the genotype of a plant with the dominant phenotype is homozygous or heterozygous. • EX: "What's the genotype of this guinea pig with respect to its fur color?"

28. The law of independent assortment- each pair of alleles segregates into gametes independently • Alleles that are not on the same chromosome • Ex. The alleles for height segregate independently from the alleles of a gene for color. • As long as the genes are on separate chromosomes, the separation during meiosis is RANDOM.

29. Random Orientation of Chromosomes During Meiosis The law of independent assortment

30. The law of independent assortment- each pair of alleles segregates into gametes independently -During meiosis, the chromosomes line up randomly during metaphase II. -So, depending on how they line, this determines the alleles in each gamete.

31. Mendel’s experiments that followed the inheritance of flower color or other characters focused on only a single character via monohybrid crosses. He conducted other experiments in which he followed the inheritance of two different characters, a dihybrid cross. The law of independent assortment- each pair of alleles segregates into gametes independently animation

32. If no independent assortment: The F2 offspring (in dihybrid) would only produce two phenotypes in a 3:1 ratio, just like a monohybrid cross. This was not consistentwith Mendel’s results. animation

33. Mendel crossed true-breeding plants that had yellow, round seeds (YYRR) with true-breeding plants that has green, wrinkled seeds (yyrr). When sperm with four classes of alleles and ova with four classes of alleles combined, there would be 16 equally probable waysin which the alleles can combine in the F2 generation.

34. When crossing a homozygous dominante with a homozygous recessive… These combinations produce four distinct phenotypes in a 9:3:3:1 ratio. This was consistent with Mendel’s results.

35. Cross two purebred guinea pigs. Brown-long hair x Black-short hair Give phenotype and genotype ratios. DO IT NOW……… Black-BBrown-bLong hair-SShort-s

36. Complete 1st pg of • Dragon Genetics……

37. The probability scale ranged from zero (an event with no chance of occurring) to one (an event that is certain to occur). The probability of tossing heads with a normal coin is ½. The probability of rolling a 3 with a six-sided die is 1/6, and the probability of rolling any other number is 1 - 1/6 = 5/6. SO……. What’s the probablility of getting a dominant allele from a parent that is Rr ? ½…50%, right? Mendelian inheritance reflects rules of probability

38. ALSO…When tossing a coin, the outcome of one toss has no impact on the outcome of the next toss. Each toss is an independent event, just like the distribution of alleles into gametes. Like a coin toss, each ovumfrom a heterozygous parent has a ½ chance of carrying the dominant allele and a ½chance of carrying the recessive allele. The same odds apply to the sperm.

39. Rule of multiplication to determine the chance that two or moreindependent events will occur together in some specific combination. Compute the probability of each independent event. Multiply the individual probabilities. The probability that two coins tossed at the same time will land heads up is: ½ x ½ = ¼. What is the probability that two heterozygous pea plants (Pp) will produce a white-flowered offspring (pp) ? ½ x ½ = ¼

40. Rule of multiplicationalso applies to dihybrid crosses. For a heterozygous parent (YyRr) the probability of producing a YR gamete is ½ x ½ = ¼. We can use this to predict the probability of a particular F2 genotype without constructing a 16-part Punnett square. • The probability that an F2 plant will have a YYRRgenotype from heterozygous parents is • 1/16 (¼ chance for a YR ovum and ¼ chance for a YR sperm).

41. Examples: An organism had three independently assorting traits: AaBbCc. What fraction of its gamete will contain ABC? 1/8 What about an organism with AABcCc? What fraction of its gamete will contain ABC? 1/4

42. Do these problems now…. • If a YyPp plant is crossed to a Yypp plant, what is the probability that the resulting plant will have the genotype Yypp? • A true-breeding plant with green seeds and white flowers is crossed with a plant that is heterozygous for both traits. What is the probability that the cross will yield a plant with green seeds and white flowers? (flower color- purple dominant, seed color- yellow dominant)

43. EXAMPLES • If a YyPp plant is crossed to a Yypp plant, what is the probability that the resulting plant will have the genotype Yypp?

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45. 2. A true-breeding plant with green seeds and white flowers is crossed with a plant that is heterozygous for both traits. What is the probability that the cross will yield a plant with green seeds and white flowers? (flower color- purple dominant, seed color- yellow dominant)

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47. EXAMPLE: Determine the probability of an offspring having two recessive phenotypes for at least two of three traits resulting from a trihybrid cross between pea plants that are PpYyRr and Ppyyrr. 1st –Determine the number of possibilities that will fulfill the requirement: There are five possible genotypes that fulfill this condition: ppyyRr, ppYyrr, Ppyyrr, PPyyrr, and ppyyrr. Determine probability of each and then add together. Rule of addition: When more than one arrangement of the events producing the specified outcome is possible, the individual probabilities are added.

48. For ppYyrr: 1/4 × 1/2 × 1/2 = 1/16. For Ppyyrr: 1/2 × 1/2 × 1/2 = 1/8 or 2/16. For PPyyrr: 1/4 × 1/2 × 1/2 = 1/16. For ppyyrr: 1/4 × 1/2 × 1/2 = 1/16. For ppyyRr 1/4 × 1/2 × 1/2 = 1/16. Therefore, the chance that a given offspring will have at least two recessive traits is… 1/16 + 2/16 + 1/16 + 1/16 + 1/16 = 6/16.

49. If a homozygous dominant long neck, heterozygous fire breathing, purple bodied dragon was crossed with a heterozygous long neck, heterozygous fire breathing, heterozygous green bodied dragon, what would be the chances of an offspring being at least homozygous recessive for 2 of the traits? Dragon Genetics ….back page 

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