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Mendelian Genetics

Mendelian Genetics. DiHybrid Crosses Independent Assortment Vs. Linkage. The Dihybrid Cross. As the name suggests, a dihybrid cross is a cross between parents that differ from each other with regard to two traits.

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Mendelian Genetics

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  1. Mendelian Genetics DiHybrid Crosses Independent Assortment Vs. Linkage

  2. The Dihybrid Cross As the name suggests, a dihybrid cross is a cross between parents that differ from each other with regard to two traits. The question that arises is the degree to which the inheritance of one trait influences the inheritance of the other.

  3. Independent Assortment The name implies that the traits will mix (assort) independently of each other. The independent assortment hypothesis assumes that the inheritance of one trait is separate from the other – the inheritance of one is random with regard to the other

  4. Independent Assortment Results If we were to run a monohybrid cross, we would expect a 3:1 ratio in the F2 generation If we ran 2 monohybrid crosses simultaneously, and they assorted independently of each other, we could assume that they would mix and match in every combination

  5. Independent Assortment Results Given as a probability, a 3:1 ratio is really a ¾ chance of dominant expression and a ¼ chance of recessive expression Using the product law of probability the likelihood of dominant expression of both traits is 9/16, one dominant and one recessive 3/16, and both recessive 1/16

  6. Results expressed as a ratio The probabilities described in the previous frame are more commonly referred to as a 9:3:3:1 ratio This was Gregor Mendel’s hypothesis. Mendel hypothesized that traits were inherited independently, and his hypothesis was supported by the results of his dihybrid cross experiments with peas

  7. Mendel’s Dihybrid Cross

  8. Analyzing Results In the Punnett square given, the inheritance of each allele was consistent with the law of segregation, alleles separated with equal probability in gamete formation Also, the R allele was as likely to be inherited with the Y allele as with the y allele. Seed color and seed shape were inherited independently

  9. Conclusions Mendel performed dihybrid crosses with several combinations of traits. Each time he observed the predicted 9:3:3:1 ratio. His conclusions form the Law of Independent Assortment A modern understanding of Independent Assortment recognizes the fact that genes are on chromosomes

  10. Chromosome Theory If genes are on chromosomes, then allele pairs are a result of pairing of homologous chromosomes Homologous chromosomes separate during meiosis, just as alleles separate during gamete formation

  11. Chromosome Theory We remember that homologous chromosomes synapse, forming a tetrad, attach to a spindle fiber, align and separate during meiosis 1 Each tetrad is a different homologous pair, attached to a different spindle fiber The division of pair #1 has no effect on the division of any other pair

  12. Modern view of Mendel So Independent Assortment is the result of genes for one trait being located on a different chromosome than the genes for the other trait. But what if they aren’t? Peas demonstrated Independent Assortment, but will all pairs of genes?

  13. No. They won’t. Let’s take humans for example. We have 23 homologous pairs. We also have way more than 23 genes. It stands to reason that some of those genes will be on the same chromosome Genes on the same chromosome are “linked” genes. They will be inherited together

  14. What would linkage look like? To use Mendel’s peas as an example, the dihybrid cross we referenced earlier studied seed color and seed shape P1: Round, Yellow x wrinkled, green F1: all Round, Yellow P2: Round, Yellow x Round, Yellow F2: Make a prediction

  15. F2 results, Linked Genes First we need to consider what we started with. If the genes are linked, they should remain in the same combination they started in (back in the P1 generation) P1: round, yellow x wrinkled, green So if round is linked, it’s linked to yellow and similarly wrinkled would be linked to green (“parental types”)

  16. Or to look at it another way So basically, 2 linked genes travelling together behave pretty much like a single gene That means linkage for a dihybrid cross would give the same kind of results as a monohybrid cross We would expect an F2 result of 3 Round, Yellow : 1 wrinkled, green Note, all offspring are parental types

  17. Comparing Predictions Independent assortment Linkage 3:1 ratio in the F2 Only the parental combinations are represented, no recombinant offspring are produced Genes are on the same chromosome so they always stay together • 9:3:3:1 ratio in the F2 • All phenotype combinations are represented, both parental and recombinant • Genes are on different chromosomes, so there is no impediment to recombination

  18. Dihybrid Cross – Determine if the genes for wing shape and eye color are Linked or Independent P1 = Straight wings, Red eyes X Curly wings, Orange eyes F1 = all Straight wings, Red eyes P2 = Straight wings, Red eyes x Straight wings, Red eyes F2 = 36 straight wing, red eye 12 straight wing, orange eye 13 curly wing, red eye 4 curly wing, orange eye

  19. Which F2 phenotypes are parental? Which are recombinant? P1 = Straight wings, Red eyes X Curly wings, Orange eyes F1 = all Straight wings, Red eyes P2 = Straight wings, Red eyes x Straight wings, Red eyes F2 = 36 straight wing, red eye 12 straight wing, orange eye 13 curly wing, red eye 4 curly wing, orange eye

  20. Which F2 phenotypes are parental? Which are recombinant? P1 = Straight wings, Red eyes X Curly wings, Orange eyes F1 = all Straight wings, Red eyes P2 = Straight wings, Red eyes x Straight wings, Red eyes F2 =36 straight wing, red eye 9 12 straight wing, orange eye 3 13 curly wing, red eye 3 4 curly wing, orange eye 1

  21. Conclusion • Wing shape and eye color are independent. • Straight wings were not exclusively inherited with red eyes, they appeared in the F2 generation paired with orange eyes in statistically significant proportions • The gene for wing shape must be located on a different chromosome than the gene for eye color

  22. Dihybrid Cross – Determine if the genes for eye color and bristles are Linked or Independent P1 = red eyes, normal bristles X brown eyes, stubble bristles F1 = all red eye, normal bristle P2 = red eyes, normal bristles X red eyes, normal bristles F2 = 28 red eyes, normal bristles 0 red eyes, stubble bristles 0 brown eye, normal bristles 9 brown eye, stubble bristle

  23. Are the genes for eye color and bristles on the same chromosome or on different chromosomes? P1 = red eyes, normal bristles X brown eyes, stubble bristles F1 = all red eye, normal bristle P2 = red eyes, normal bristles X red eyes, normal bristles F2 = 28 red eyes, normal bristles 0 red eyes, stubble bristles 0 brown eye, normal bristles 9 brown eye, stubble bristle

  24. Parental vs Recombinant P1 = red eyes, normal bristles X brown eyes, stubble bristles F1 = all red eye, normal bristle P2 = red eyes, normal bristles X red eyes, normal bristles F2 =28 red eyes, normal bristles 3 0 red eyes, stubble bristles 0 0 brown eye, normal bristles 0 9 brown eye, stubble bristle 1

  25. Parental vs Recombinant P1 = red eyes, normal bristles X brown eyes, stubble bristles F2 =28 red eyes, normal bristles 3 0 red eyes, stubble bristles 0 0 brown eye, normal bristles 0 9 brown eye, stubble bristle 1 All the F2 offspring were parental types, no recombination occurred

  26. Are the genes for eye color and bristles on the same chromosome or on different chromosomes? F2 = 28 red eye, normal bristles 0 red eyes, stubble bristles 0 brown eye, normal bristles 9 brown eye, stubble bristle These results are what you would expect from Linked genes (genes located on the same chromosome), but they do not take into account the real occurrence of Crossing Over. Real life scenarios will produce some level of genetic recombination even when genes are linked

  27. Characteristic Data from Linked Genes F2 = 22 red eyes, normal bristles 2 red eyes, stubble bristles 3 brown eye, normal bristles 8 brown eye, stubble bristle In this data set, we have 30 “parental type” F2 offspring and 5 “recombinant” type offspring. This represents 14.3% recombination. This is far too small to be a result of independent assortment, so the genes must be linked. Recombination is the result of Crossing Over

  28. Barbara McClintock • Barbara McClintock won a Nobel Prize for her research in Cytogenetics. • She used maize (corn) for her research

  29. Why Corn? • Easy to control crosses • Every kernel of corn is a genetically unique offspring

  30. Lab - DiHybrid Cross in Corn • Dihybrid = crossbreed for 2 traits • Seed color and seed shape • P1: Purple Smooth x Yellow Wrinkled • Fi: all Purple Smooth • P2: Purple Smooth x Purple Smooth • F2: We will need to make alternative predictions for the F2 offspring based on Linkage vs. Independent Assortment

  31. F2 Predictions - Linked

  32. F2 Predictions - Linked

  33. F2 Predictions – Independent Assortment

  34. F2 Predictions – Independent Assortment

  35. F2 Results • Working in pairs, count and record the number of each phenotype • Compile the data to get class totals • Evaluate the results and determine if the genes are linked or independent • Purple, Smooth • Purple, Wrinkled • Yellow, Smooth • Yellow, Wrinked

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