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Advanced Biology Unit 3. The Monk and his peas An Austrian monk, Gregor Mendel, developed the fundamental principles that would become the modern science.

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Presentation on theme: "Advanced Biology Unit 3. The Monk and his peas An Austrian monk, Gregor Mendel, developed the fundamental principles that would become the modern science."— Presentation transcript:

1 Advanced Biology Unit 3

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3 The Monk and his peas An Austrian monk, Gregor Mendel, developed the fundamental principles that would become the modern science of genetics. Mendel demonstrated that heritable properties are parceled out in discrete units, independently inherited. These eventually were termed genes. geneticsgenes

4 Seed shape and color Flower color

5 Pod color and shape

6 Flower Position

7 Stem height

8 P1: smooth X wrinkled F1 : all smooth

9 About a 3 to 1 ratio F2 : 5474 smooth and 1850 wrinkled

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11 Q A test cross is used to determine if the genotype of a plant with the dominant phenotype is homozygous or heterozygous. If the unknown is homozygous, all of the offspring of the test cross have the __________ phenotype. If the unknown is heterozygous, half of the offspring will have the __________ phenotype.

12 answer dominant, recessive The test cross was invented by Mendel to determine the genotype of plants displaying the dominant phenotype.

13 Q Disappearance of parental phenotypes in the F1 generation A genetic cross of inbred snapdragons with red flowers with inbred snapdragons with white flowers resulted in F1-hybrid offspring that all had pink flowers. When the F1 plants were self-pollinated, the resulting F2-generation plants had a phenotypic ratio of 1 red: 2 pink: 1 white. The most likely explanation is:

14 answer Heterozygous plants have a different phenotype than either inbred parent because of incomplete dominance of the dominant allele. The features of crosses involving incomplete dominance are intermediate phenotype of heterozygous individuals, and parental phenotypes reappear in F2 when heterozygotes are crossed.

15 ABO’s of human blood Human blood type is determined by codominant alleles. There are three different alleles, known as IA, IB, and i. The IA and IB alleles are co- dominant, and the i allele is recessive. The possible human phenotypes for blood group are type A, type B, type AB, and type O. Type A and B individuals can be either homozygous (IAIA or IBIB, respectively), or heterozygous (IAi or IBi, respectively). A woman with type A blood and a man with type B blood could potentially have offspring with which of the following blood types? A, B, AB, or O

16 Q What are the possible blood types of the offspring of a cross between individuals that are type AB and type O? (Hint: blood type O is recessive)

17 answer A, B, AB, O all of the above But if the man was type O rather than type B, offspring of type B and type AB would not be possible.

18 answer In this problem there is no uncertainty about the genotype of either parent. A parent of blood type AB has the codominant IA and IB alleles. A parent of blood type O is homozygous recessive for the i allele. The Punnett square for their offspring is shown to the right. The genotypes of their offspring could be either IAi or IBi. Their children could be in blood groups A or B, but not AB or O.

19 Manx cats are heterozygous for a dominant mutation that results in no tails (or very short tails), large hind legs, and a distinctive gait. The mating of two Manx cats yields two Manx kittens for each normal, long-tailed kitten, rather than three-to-one as would be predicted from Mendelian genetics. Therefore, the mutation causing the Manx cat phenotype is likely a(n) __________ allele.

20 answer lethal The predicted segregation pattern in the F2 generation is 1/4 normal (homozygous), 1/2 Manx phenotype (heterozygous), an 1/4 embryonic lethal (homozygous for the Manx allele).

21 Monohybrid test cross To identify the genotype of yellow-seeded pea plants as either homozygous dominant (YY) or heterozygous (Yy), you could do a test cross with plants of genotype _______. YY or Yy x ___?___

22 Answer: A cross with the homozygous recessive (yy) is a test cross. If the parent of unknown genotype is heterozygous (Yy), half of the offspring will have the recessive trait. The unknown genotype could also be determined by a cross with a known heterozygote (Yy).

23 When true-breeding tall stem pea plants are crossed with true-breeding short stem pea plants, all of the ________ plants, and 3/4 of the __________ plants had tall stems. Therefore, tall stems are dominant.

24 Answers F1, F2. The F1 plants are all Tt hybrids. The recessive trait (tt) reappears in the F2 generation in about 25% of the plants.

25 Q A genetic cross between two F1-hybrid pea plants having yellow seeds will yield what percent green-seeded plants in the F2 generation? Yellow seeds are dominant to green.

26 Answers 25% Among the F2 plants of a Yy x Yy cross, 25% will be yy with the recessive, green- seeded phenotype.

27 Q In Mendel's "Experiment 1," true-breeding pea plants with spherical seeds were crossed with true-breeding plants with dented seeds. (Spherical seeds are the dominant characteristic.) Mendel collected the seeds from this cross, grew F1-generation plants, let them self-pollinate to form a second generation, and analyzed the seeds of the resulting F2 generation. The results that he obtained, and that you would predict for this experiment are:

28 Answer All the F1 and 3/4 of the F2 generation seeds were spherical. All of the F1 plants were true hybrids with a phenotype of Ss. The recessive trait reappears in the F2 generation.

29 Q A phenotypic ratio of 3:1 in the offspring of a mating of two organisms heterozygous for a single trait is expected when:

30 answer the alleles segregate during meiosis. Mendel first proposed that alleles segregate from one another during the formation of gametes.

31 Q In pea plants, spherical seeds (S) are dominant to dented seeds (s). In a genetic cross of two plants that are heterozygous for the seed shape trait, what fraction of the offspring should have spherical seeds?

32 answer 3/4 One fourth of the offspring will be homozygous dominant (SS), one half will be heterozygous (Ss), and one fourth will be homozygous recessive (ss).

33 Q The gametes of a plant of genotype SsYy should have the genotypes: A. Ss and YySs and Yy B. SY and sySY and sy C. SY, Sy, sY, and sySY, Sy, sY, and sy D. Ss, Yy, SY and sySs, Yy, SY and sy E. SS, ss, YY, and yySS, ss, YY, and yy

34 answer The gametes will receive one of each pair (Ss and Yy) of alleles. All combinations of alleles will occur with equal probability (SY, Sy, sY, and sy) because alleles of different genes are assorted independently during gamete formation (meiosis.) Each gamete has one seed shape (S=spherical or s=dented) and one color (Y=yellow or y=green) allele.

35 Dihybrid cross A cross that involves two sets of characteristics Instead of 4 possible genotypes from a monohybrid cross, dihybrid crosses have as many as 16 possible genotypes.

36 Dihybrid test cross Which of the following genotypes would you not expect to find among the offspring of a SsYy x ssyy test cross: A. ssyyssyy B. SsYySsYy C. SsyySsyy D. ssYyssYy E. SsYYSsYY

37 Answer SsYY Offspring could not be homozygous for the dominant yellow seed color (YY), because one recessive y allele must be inhereited from the ssyy parent.

38 Q The expected phenotypic ratio of the progeny of a SsYy x ssyy test cross is: A. 9:3:3:19:3:3:1 B. 3:13:1 C. 1:1:1:11:1:1:1 D. 1:2:11:2:1 E. 3:1:1:33:1:1:3

39 C. 1:1:1:1. SsYy, ssYy, Ssyy, ssyy are predicted to occur in a ratio of 1:1:1:1

40 Q In a dihybrid cross, AaBb x AaBb, what fraction of the offspring will be homozygous for both recessive traits? A. 1/161/16 B. 1/81/8 C. 3/163/16 D. 1/41/4 E. 3/43/4

41 There are four possible combinations of gametes for the AaBb parent. Half of the gametes get a dominant A and a dominant B allele; the other half of the gametes get a recessive a and a recessive b allele. Both parents produce 25% each of AB, Ab, aB, and ab.

42 Answer There is only one of 16 possible combinations with this genotype. The predicted fraction is therefore 1/16.

43 What does 3:1 indicate? A cross between parents who are both heterozygous produces a phenotypic ratio of 3:1 Aa x Aa

44 Give an example of a monohybrid test cross T is tall in pea plants T is short Crossing a T plant (could be TT or Tt) with a short (tt) plant could produce: 1:1 tall and short, if Tall parent is Tt All tall is tall parent is TT

45 Sex-linked traits In a cross between a pure bred, red-eyed female fruit fly and a white-eyed male, what percent of the male offspring will have white eyes? (white eyes are X-linked, recessive) A. 100%100% B. 75%75% C. 50%50% D. 25%25% E. 0%0%

46 A All of the males and all of the females are red-eyed.

47 Q A white-eyed female fruit fly is crossed with a red-eyed male. Red eyes are dominant, and X-linked. What are the expected phenotypes of the offspring?

48 A All of the females eggs will contain an X chromosome with the white-eye mutation. The sperm will contain either a normal X chromosome or a Y chromosome. We use a punnett square to predict the outcome of this cross. Female offspring receive an X chromosome from both the sperm and egg. All females receive the dominant, red-eyed allele from their fathers and the recessive, white-eyed allele from their mothers.

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50 Hemophilia Hemophilia results from a mutated gene on an X chromosome. Mothers always contribute an X chromosome to their offspring, while fathers contribute either an X or a Y chromosome. If hemophilia is found on any of these X chromosomes, it can be passed on to the child.

51 Hemophilia Normal Mother + Father with Hemophilia Each pregnancy has a 50% chance of resulting in a female carrier and a 50% chance of resulting in a normal male. Sons of hemophiliac fathers and normal mothers will not have hemophilia.

52 But… Carrier Mother + Normal Father Each pregnancy has a 25% chance of resulting in a normal female, a 25% chance of resulting in a female carrier, a 25% chance of resulting in a normal male, and a 25% chance of resulting in a male with hemophilia

53 Carrier Mother + Father with Hemophilia Each pregnancy has a 25% chance of resulting in a female carrier, a 25% chance of resulting in a female with hemophilia, a 25% chance of resulting in a normal male, and a 25% chance of resulting in a male with hemophilia.

54 Mother with Hemophilia + Father with Hemophilia All hemophilia.

55 Mother with, Dad normal Each pregnancy has a 50% chance of resulting in a female carrier and a 50% chance of resulting in a male with hemophilia (actual occurrence is extremely rare).

56 In summary, knowledge of genetics lets us make the following statements about hemophilia: Nearly all affected people are male. Hemophilia may represent a new mutation in the affected male. An affected male never transmits the trait to his sons. All daughters of an affected male will be carriers (if the mom is not a carrier). A carrier female transmits the trait to her sons 50 percent of the time. No daughters of a carrier female will show the trait, but a daughter in this case (if the dad is not affected) will be a carrier 50 percent of the time.

57 The ABCs of Hemophilia TypeHemophilia AHemophilia BHemophilia C Other Names Classical hemophilia Standard hemophilia Factor VIII deficiency Christmas disease (named after Stephen Christmas, a young British boy who was the first person diagnosed with the disorder) Factor IX deficiency Factor XI deficiency Frequency About 1 in 5,000 U.S. male births About 80 percent of people with hemophilia About 13,500 people in the U.S. About 1 in 30,000 U.S. male births Up to 20 percent of people with hemophilia More than 3,000 people in the U.S. About 1 in 100,000 U.S. male births About 200 cases reported worldwide since its discovery in the 1950s Gender AffectedMales almost exclusively Males and females equally Missing Factor ProteinFactor VIIIFactor IXFactor XI

58 Q Hemophilia in humans is due to an X- chromosome mutation. What will be the results of mating between a normal (non-carrier) female and a hemophilac male? A. half of daughters are normal and half of sons are hemophilic.half of daughters are normal and half of sons are hemophilic. B. all sons are normal and all daughters are carriers.all sons are normal and all daughters are carriers. C. half of sons are normal and half are hemophilic; all daughters are carriers.half of sons are normal and half are hemophilic; all daughters are carriers. D. all daughters are normal and all sons are carriers.all daughters are normal and all sons are carriers. E. half of daughters are hemophilic and half of daughters are carriers; all sons are normal.half of daughters are hemophilic and half of daughters are carriers; all sons are normal.

59 A The eggs of the mother will all contain the normal X chromosome. The sperm of the father will contain either the X chromosome with the mutation causing hemophilia or the Y chromosome. All of the daughters inherit an X chromosome with the mutation from their father, and will be carriers. All the sons inherit a normal X from the mother

60 Color blindness A human female "carrier" who is heterozygous for the recessive, sex-linked trait causing red- green color blindness, marries a normal male. What proportion of their male progeny will have red-green color blindness? A. 100%100% B. 75%75% C. 50%50% D. 25%25% E. 0%0%

61 A The eggs of the mother will contain either a normal X chromosome or an X chromosome with the mutation causing red-green color blindness. The sperm of the father will contain either the normal X chromosome or the Y chromosome. None of the female children would be red-green color blind, but half would be "carriers." Half of the sons would inherit the allele from their mother and be afflicted.

62 Q Women have sex chromosomes of XX, and men have sex chromosomes of XY. Which of a man's grandparents could not be the source of any of the genes on his Y- chromosome? A. Father's Mother.Father's Mother. B. Mother's Father.Mother's Father. C. Father's Father.Father's Father. D. Mother's Mother, Mother's Father, and Father's Mother.Mother's Mother, Mother's Father, and Father's Mother. E. Mother's Mother.Mother's Mother.

63 A D. Mother's Mother, Mother's Father, and Father's Mother. The Y chromosome is inherited solely from father to son in each generation. E. Mother's Mother. The diagram shows how the X and Y chromosomes are inherited from the maternal and paternal grandparents to the parents to the son. The Y chromosome is passed strictly from the father to male children in each generation. Neither maternal grandparent nor the paternal grandmother can be a source of any genes on the Y chromosome.

64 Two sex linked traits… What offspring would you expect from a cross between the female Drosophila described in problem 1 (red eyes and a yellow body, homozygous recessive for the yellow body color allele and homozygous dominant for the eye color allele) and the male described in problem 2 (hemizygous for both the recessive (white) eye color allele and dominant (tan) body color allele?) A reminder that the alleles for eye color and for body color are on the X chromosome of Drosophila, but not on the Y. Red eye color (w+) is dominant to white eye color (w), and tan body color (y+) is dominant to yellow body color (y).

65 Choices are… A. Daughters would be yellow-bodied, red-eyed; the sons would be tan-bodied, white-eyedDaughters would be yellow-bodied, red-eyed; the sons would be tan-bodied, white-eyed B. Daughters would be tan-bodied, red-eyed; the sons would be yellow-bodied, white-eyedDaughters would be tan-bodied, red-eyed; the sons would be yellow-bodied, white-eyed C. Daughters would be tan-bodied, red-eyed; the sons would be yellow-bodied, red-eyedDaughters would be tan-bodied, red-eyed; the sons would be yellow-bodied, red-eyed D. Daughters would be yellow-bodied, white-eyed; the sons would be tan-bodied, white-eyedDaughters would be yellow-bodied, white-eyed; the sons would be tan-bodied, white-eyed E. Daughters would be yellow-bodied, red-eyed; the sons would be tan-bodied, red-eyedDaughters would be yellow-bodied, red-eyed; the sons would be tan-bodied, red-eyed

66 A C. Daughters would be tan- bodied, red- eyed; the sons would be yellow-bodied, red-eyed

67 Q If we mated the F1 female and male flies, what male phenotype in the F2 generation would be evidence that crossing over had occurred during gamete formation? Daughters were tan-bodied, red-eyed, heterozygous for both eye and body color. The sons were yellow-bodied, red-eyed.

68 A Evidence for recombination between X chromosomes in the F1 female would be a new combination of alleles not present on any of the X chromosomes of the parents of the F1 female. Yellow body/white eyes and tan body/red eyes F2 males would be evidence of crossing over.

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70 Karyotyping Activity Patient A's Karyotype Place this chromosome in the partially completed karyotype below by clicking on its homologous chromosome. If you match the chromosome correctly, you will proceed to the next chromosome. If you match incorrectly, a page will explain why the chromosome you chose is not the unknown's pair and you can choose again.

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73 Common Terms GeneGene - a unit of inheritance that usually is directly responsible for one trait or character. AlleleAllele - an alternate form of a gene. Usually there are two alleles for every gene, sometimes as many a three or four. HomozygousHomozygous - when the two alleles are the same. HeterozygousHeterozygous - when the two alleles are different, in such cases the dominant allele is expressed. DominantDominant - a term applied to the trait (allele) that is expressed irregardless of the second allele. RecessiveRecessive - a term applied to a trait that is only expressed when the second allele is the same (e.g. short plants are homozygous for the recessive allele). PhenotypePhenotype - the physical expression of the allelic composition for the trait under study. GenotypeGenotype - the allelic composition of an organism. Punnett squaresPunnett squares - probability diagram illustrating the possible offspring of a mating.


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