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Mendel’s Four Postulates

Mendel’s Four Postulates. Unit factors in pairs genetic characteristics are controlled by unit factors which exist in pairs genes on homologous chromosomes; diploid individuals. Mendel’s Four Postulates. Dominance/recessiveness

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Mendel’s Four Postulates

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  1. Mendel’s Four Postulates • Unit factors in pairs • genetic characteristics are controlled by unit factors which exist in pairs • genes on homologous chromosomes; diploid individuals

  2. Mendel’s Four Postulates ... • Dominance/recessiveness • When two unlike unit factors are present in a single individual, one is usually dominant to the other, which is recessive • One allele may be dominant to a second; heterozygotes express phenotype of dominant allele

  3. Mendel’s Four Postulates ... • Random segregation • During the formation of gametes, the paired unit factors separate (segregate) randomly so that each gamete receives one or the other • Homologous chromosomes segregate into different gametes

  4. Mendel’s Four Postulates ... • Independent assortment • During the formation of gametes, segregating pairs of factors assort independently of each other • Genes on non-homologous chromosomes will assort into gametes independently of one another (Remember meiosis) • Aa Bb --> AB Ab aB ab

  5. Monohybrid cross(one trait) Purple X White parentals or P generation first filial or F1 generation Purple

  6. Monohybid cross ... PurpleF1 X Purple F1 Purple White 1/4 3/4

  7. Monohybrid cross ... Purple X White PP pp Purple Pp

  8. Purple White Monohybrid cross ... Purple F1 X Purple F1 Pp Pp F2: PP Pp Pp pp

  9. Monohybrid cross ... • Mendel found same ratios no matter what trait he used • Data provide the basis for his first three postulates

  10. Review of monohybrid cross and Mendel’s first threepostulates • Heredity is controlled by unit factors in pairs • trait which was hidden in the F1 reappears • no blending, discrete units • 3:1 ratio explained by paired units • Modern translation: genes are carried on each of two homologous chromosomes

  11. Review ... • One unit factor may be dominant to another • F1 generation all have the same phenotype as one of the parents • The purple allele masks the white allele • Heterozygotes express phenotype of dominant allele

  12. Review ... • Random segregation: each gamete receives one or the other unit factor • 3:1 ratio in F2 indicate each factor equally likely to be passed on • Heterozygotes produce 2 types of gametes; 1/2 w/ one allele, 1/2 w/ other

  13. Rules for labeling genes and alleles • Use same letter for each allele of a gene • Generally use the first letter of one allele • Choose letter for which upper and lower case are easily distinguishable

  14. Rules for labeling ... • Use upper case for the dominant allele; lower case for the recessive • Example: • Free vs. attached earlobes: • Free = F; attached = f

  15. Rules for labeling ... • Sometimes, one allele is said to be the wild type • this is simply the most common allele in the population • may be dominant or recessive • in Drosophila, the flies normally have red eyes; this is wild type

  16. Rules for labeling • Wild type is sometimes symbolized with a “+” • eye color in fruit flies: • red = +; white = w • + + and + w = red • w w = white

  17. Brief review: • Gene vs. allele vs. locus • gene - segment of DNA coding for a trait • allele - specific form of the gene • locus - location on the chromosome of the gene • Gene: flower color • Alleles: purple and white

  18. Brief review ... • Genotype • The genetic make-up of the individual; what alleles it has • heterozygous • homozygous recessive • homozygous dominant

  19. Brief review ... • Phenotype • The outwardly detectable trait; what the individual looks like, etc. AA and Aa give phenotype A aa gives phenotype a

  20. Dihybrid Cross • Looking at two traits at the same time • two genes • two alleles each • Here we will see evidence for Mendel’s 4th Postulate; Independent Assortment

  21. Dihybrid Cross smooth, yellow X wrinkled, green SS YY ss yy smooth, yellow F1: Ss Yy

  22. Dihybrid Cross ... Ss Yy X Ss Yy Remember: We are now dealing with two independent genes! Each individual will pass on one allele of each gene.

  23. Punnett Squares • Step 1: what kindsof gametes will each produce? • gametes are haploid -- carry one copy of each gene • Heterozygotes produce 1/2 gametes with one allele, 1/2 with the other (random segregation) • There can be ANY combination of alleles of independent genes. INDEPENDENT ASSORTMENT

  24. Punnett Squares ... • Helpful hint: • The number of different types of gametes an individual will produce is equal to: 2n where n = the number of heterozygous loci

  25. Punnett Squares ... s S Y y • Back to our example: • Ss Yy X Ss Yy • 22 gametes for each (parents identical) • four types of gametes each • SY Sy sY sy • Review meiosis to prove this!

  26. SY Sy sY sy SY Sy sY sY Punnett Squares ... • Assign gametes from one parent to rows • Assign gametes from second parent to columns SS YY

  27. Punnett Squares ... • Ratios obtained through a Punnett square tell you the expected frequencies of those phenotypes in the offspring • 3/16 of the offspring should be smooth, green • Another way of looking at it: any given offspring has a 3/16 chance of being smooth, green • You can work backwards; ratios are clues to genotypes of parents

  28. Test Crosses • cross between an individual of unknown genotype and a homozygous recessive individual • allows us to deduce the genotype of the unknown parent

  29. Test Crosses ... DR DD RR X dd rr dr

  30. dr DR Dr Test Crosses ... DD Rr X dd rr

  31. dr DR dR Test Crosses ... Dd RR X dd rr

  32. dr DR Dr dR dr Test Crosses ... Dd Rr X dd rr

  33. Parent’s phenotypes: SA x sa Offspring’s phenotypes: SA and Sa In mice, agouti fur color is dominant to white. Spotting is recessive to no spots. A solid, agouti mouse is mated to a spotted, white mouse producing 3 solid, agouti mice and 4 solid, white mice. What are the genotypes of the parents? Offspring’s genotypes: Ss Aa andSs aa Parent’s genotypes: SS Aa and ss aa

  34. Parent’s phenotypes: sA x Sa Offspring’s phenotypes: sA and sa In mice, agouti fur color is dominant to white. Spotting is recessive to no spots. A spotted, agouti mouse is mated to a solid white mouse producing 3 spotted, agouti mice and 2 spotted white mice. What are the genotypes of the parents? Offspring’s genotypes: Parent’s genotypes:

  35. Review of phenotypic ratiosin offspring • Monohybrid • homozygote X homozygote --> heterozygote • heterozygote X heterozygote --> 3:1 • heterozygote X homozygous recessive --> 1:1

  36. Review of phenotypic ratiosin offspring • Dihybrid • homozygote X homozygote --> heterozygote • heterozygote X heterozygote --> 9:3:3:1 • heterozygote X homozygous recessive --> 1:1:1:1

  37. Reciprocal crosses • Does it matter which sex has which trait? • i.e., male red-eyed fly X brown female or male brown-eyed X red female? • NO! RR X rr --> Rr rr X RR --> Rr • For autosomal traits (traits located on non-sex chromosomes)reciprocal crosses will give identical results.

  38. Trihybrid crosses and beyond! • Trihybrid crosses deal with three independent traits at once • i.e., flower color, plant height, and seed color • each gene is independent! • each gene has two alleles

  39. Trihybrid crosses ... • The basic concepts are the same as with mono- or dihybrid crosses • Just remember: • each gamete must have ONE COPY OF EACH GENE • random segregation and independent assortment still apply

  40. Trihybrid crosses ... tall, pinched pod, axial X dwarf, inflated, terminal tt PP aa TT pp AA tall, inflated, axial F1: Tt Pp Aa F1 X F1 --> ?

  41. Trihybrid crosses ... • Step one: Figure out the gametes produced by each parent (use 2n) • Parents: Tt Pp Aa; 23 = 8 • eight possible gametes from each parent: • TPA TPa TpA tPATpa tPa tpA tpa

  42. Forked-line Method • Breaks down multi-hybrid crosses into a series of monohybrid crosses • Combine the individual ratios (multiply) to get the final ratio

  43. Forked-line Method ... Tt Pp Aa X Tt Pp Aa Break into Tt X Tt; Pp X Pp; Aa X Aa Tt X Tt --> 3/4 T: 1/4 t Pp X Pp --> 3/4 P: 1/4 p Aa X Aa --> 3/4 A: 1/4 a

  44. Forked-line ... F2 offspring: < 3/4 A --> 27/64 TPA 3/4 P 1/4 a --> 9 /64 TPa < 3/4 T < 3/4 A --> 9 /64 TpA 1/4 p 1/4 a --> 3 /64 Tpa < 3/4 A --> 9 /64 tPA 3/4 P < 1/4 a --> 3 /64 tPa 1/4 t < 3/4 A --> 3 /64 tpA 1/4 p 1/4 a --> 1 /64 tpa

  45. Forked-line ... • Can find genotypic ratios • substitute 1/4 TT; 1/2 Tt; 1/4 tt, etc. • Can go straight to one phenotype or genotype • ignore all branches except those leading to the desired phenotype or genotype

  46. Forked-line ... What proportion of the offspring of the cross Ss Nn Vv X Ss nn VV will be genotype Ss Nn Vv?

  47. Forked-line ... What proportion of the offspring of the cross Ss Nn Vv X Ss nn VV will be phenotype SNV?

  48. Forked-line ... What proportion of the offspring of the cross Cc Dd Ee Rr X cc Dd ee Rr will be phenotype cder?

  49. Aa Bb cc DD Ee Ff gg x Aa Bb Cc Dd ee ff Gg 1. What proportion of offspring phenotype AbcDEFg? 2. What proportion of offspring genotype AAbbCcDdEeFfGg?

  50. Ratios -- one more time! • Monohybrid • Aa X Aa --> 3:1 • Aa X aa --> 1:1 • Dihybrid • Aa Bb X Aa Bb --> 9:3:3:1 • Aa Bb X aa bb --> 1:1:1:1 • Trihybrid • AaBbCc X AaBbCc --> 27:9:9:9:3:3:3:1 • AaBbCc X aabbcc --> 1:1:1:1:1:1:1:1

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