1 / 73

Advanced Biology

Advanced Biology. Unit 3. The Monk and his peas .

chantrea
Télécharger la présentation

Advanced Biology

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. AdvancedBiology Unit 3

  2. 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.

  3. Seed shape and colorFlower color

  4. Pod color and shape

  5. Flower Position

  6. Stem height

  7. P1: smooth X wrinkled • F1 : all smooth

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

  9. 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.

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

  11. 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:

  12. 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.

  13. 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

  14. 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)

  15. 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.

  16. 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.

  17. 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.

  18. 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).

  19. 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 ___?___

  20. 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).

  21. 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.

  22. 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.

  23. 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.

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

  25. 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:

  26. 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.

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

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

  29. 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?

  30. 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).

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

  32. 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.

  33. 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.

  34. 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. ssyy B. SsYy C. Ssyy D. ssYy E. SsYY

  35. 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.

  36. Q The expected phenotypic ratio of the progeny of a SsYy x ssyy test cross is: • A. 9:3:3:1 • B. 3:1 • C. 1:1:1:1 • D. 1:2:1 • E. 3:1:1:3

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

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

  39. 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.

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

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

  42. 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

  43. 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% • B. 75% • C. 50% • D. 25% • E. 0%

  44. A 0% All of the males and all of the females are red-eyed.

  45. 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?

  46. 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.

  47. 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.

More Related