10 2 Inheritance Dihybrid Crosses Polygenic Inheritance Gene

10. 2 Inheritance Dihybrid Crosses, Polygenic Inheritance & Gene Linkage 1

Mendel’s Law of Independent Assortment “Can you remember it? ” 10. 2 Dihybrid Crosses & Gene Linkage 2

Mendel’s Law of Independent Assortment “The presence of an allele of one of the genes in a gamete has no influence over which allele of another gene is present. ” This only holds true for unlinked genes (genes on different chromosomes). 10. 2 Dihybrid Crosses & Gene Linkage 3

Mendel’s Law of Independent Assortment “The presence of an allele of one of the genes in a gamete has no influence over which allele of another gene is present. ” Key to alleles: Y = yellow y = green S = smooth s = rough This only holds true for unlinked genes (genes on different chromosomes). SY meiosis s. Y Sy sy 10. 2 Dihybrid Crosses & Gene Linkage 4

Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: What is the predicted phenotype ratio for a cross between two pea plants which are heterozygous at both loci? F 0 Y = yellow y = green S = smooth s = rough Phenotype: Genotype: Punnet Grid: Heterozygous at both loci gametes F 1 10. 2 Dihybrid Crosses & Gene Linkage 5

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough What is the predicted phenotype ratio for a cross between two pea plants which are heterozygous at both loci? F 0 Phenotype: Genotype: Punnett Grid: Smooth, yellow Heterozygous at both loci Ss. Yy gametes F 1 10. 2 Dihybrid Crosses & Gene Linkage 6

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough What is the predicted phenotype ratio for a cross between two pea plants which are heterozygous at both loci? F 0 Phenotype: Genotype: Punnett Grid: Smooth, yellow Heterozygous at both loci Ss. Yy gametes SY Sy s. Y sy SY SSYy Ss. YY Ss. Yy Sy SSYy SSyy Ss. Yy Ssyy s. Y Ss. Yy ss. YY ss. Yy sy Ss. Yy Ssyy ss. Yy ssyy F 1 10. 2 Dihybrid Crosses & Gene Linkage 7

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough What is the predicted phenotype ratio for a cross between two pea plants which are heterozygous at both loci? F 0 Phenotype: Genotype: Punnett Grid: F 1 Smooth, yellow Heterozygous at both loci Ss. Yy gametes SY Sy s. Y sy SY SSYy Ss. YY Ss. Yy Sy SSYy SSyy Ss. Yy Ssyy s. Y Ss. Yy ss. YY ss. Yy sy Ss. Yy Ssyy ss. Yy ssyy Phenotypes: 9 Smooth, yellow : 3 Smooth, green : 3 Rough, yellow : 1 Rough, green 10. 2 Dihybrid Crosses & Gene Linkage 8

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Calculate the predicted phenotype ratio for: F 0 Key to alleles: Y = yellow y = green S = smooth s = rough Phenotype: Genotype: Heterozygous at both loci Heterozygous for S, homozygous dominant for Y Punnett Grid: F 1 Phenotypes: 10. 2 Dihybrid Crosses & Gene Linkage 9

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough Calculate the predicted phenotype ratio for: F 0 Phenotype: Genotype: Smooth, yellow Heterozygous at both loci Ss. Yy Smooth, yellow Heterozygous for S, homozygous dominant for Y Ss. YY Punnett Grid: F 1 Phenotypes: 10. 2 Dihybrid Crosses & Gene Linkage 10

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough Calculate the predicted phenotype ratio for: F 0 Phenotype: Genotype: Punnett Grid: Smooth, yellow Heterozygous for S, homozygous dominant for Y Heterozygous at both loci Ss. YY Ss. Yy gametes SY s. Y SY Sy s. Y sy F 1 Phenotypes: 10. 2 Dihybrid Crosses & Gene Linkage 11

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough Calculate the predicted phenotype ratio for: F 0 Phenotype: Genotype: Punnett Grid: F 1 Smooth, yellow Heterozygous for S, homozygous dominant for Y Heterozygous at both loci Ss. YY Ss. Yy gametes SY s. Y SY SSYY Ss. YY Sy SSYy Ss. Yy s. Y Ss. YY sy Ss. Yy ss. Yy Phenotypes: 10. 2 Dihybrid Crosses & Gene Linkage 12

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough Calculate the predicted phenotype ratio for: F 0 Phenotype: Genotype: Punnett Grid: F 1 Smooth, yellow Heterozygous for S, homozygous dominant for Y Heterozygous at both loci Ss. YY Ss. Yy gametes SY s. Y SY SSYY Ss. YY Sy SSYy Ss. Yy s. Y Ss. YY sy Ss. Yy ss. Yy 6 Smooth, yellow : 2 Rough, yellow Phenotypes: 3 Smooth, yellow : 1 Rough, yellow Present the ratio in the simplest mathematical form. 10. 2 Dihybrid Crosses & Gene Linkage 13

Dihybrid Crosses Common expected ratios of dihybrid crosses. Ss. Yy Heterozygous at both loci Ss. YY Heterozygous at one locus, homozygous dominant at the other SY s. Y SY SSYY Ssyy Sy SSYy Ss. Yy ss. YY ss. Yy s. Y Ss. YY ss. Yy ssyy sy Ss. Yy ss. Yy SY Sy s. Y sy SY SSYy Ss. YY Ss. Yy Sy SSYy SSyy Ss. Yy s. Y Ss. Yy sy Ss. Yy Ssyy 3: 1 9: 3: 3: 1 Ssyy Ss. Yy Heterozygous at both loci Heterozygous/ Homozygous recessive Sy sy SY SSYy Ss. Yy Sy SSyy Ssyy s. Y Ss. Yy sy Ssyy ssyy SSyy ss. YY = All Ss. Yy SSYY ssyy = all Ss. Yy Ssyy ss. Yy =1: 1: 1: 1 4: 3: 1 10. 2 Dihybrid Crosses & Gene Linkage 14

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A rough yellow pea is test crossed to determine its genotype. F 0 Phenotype: Rough, yellow Genotype: Punnett Grid: F 1 Phenotypes: 10. 2 Dihybrid Crosses & Gene Linkage 15

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A rough yellow pea is test crossed to determine its genotype. F 0 Phenotype: ss. Yy Genotype: Punnett Grid: F 1 Rough, yellow gametes s. Y sy Phenotypes: 10. 2 Dihybrid Crosses & Gene Linkage 16

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A rough yellow pea is test crossed to determine its genotype. F 0 Phenotype: ss. Yy or ss. YY Genotype: Punnett Grid: F 1 Rough, yellow gametes s. Y sy s. Y Phenotypes: 10. 2 Dihybrid Crosses & Gene Linkage 17

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A rough yellow pea is test crossed to determine its genotype. F 0 Phenotype: Genotype: Punnett Grid: Rough, yellow ss. Yy or ss. YY ssyy s. Y gametes sy s. Y All sy F 1 Phenotypes: ith a w n w o e unkn e. essiv c e r s u ygo oz m o h n know is th s s o r c t r: A tes be Remem 10. 2 Dihybrid Crosses & Gene Linkage 18

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A rough yellow pea is test crossed to determine its genotype. F 0 Phenotype: Genotype: Punnett Grid: F 1 Rough, yellow ss. Yy or ss. YY ssyy gametes s. Y sy s. Y All sy ss. Yy ssyy ss. Yy Phenotypes: 10. 2 Dihybrid Crosses & Gene Linkage 19

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A rough yellow pea is test crossed to determine its genotype. F 0 Phenotype: Genotype: Punnett Grid: F 1 Rough, yellow ss. Yy or ss. YY ssyy gametes s. Y sy s. Y All sy ss. Yy ssyy ss. Yy Phenotypes: Some green peas will be present in the offspring if the unknown parent genotype is ss. Yy. No green peas will be present in the offspring if the unknown parent genotype is ss. YY. 10. 2 Dihybrid Crosses & Gene Linkage 20

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring. F 0 Phenotype: Smooth, green Genotype: Punnett Grid: F 1 Phenotypes: 10. 2 Dihybrid Crosses & Gene Linkage 21

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring. F 0 Phenotype: Genotype: Punnet Grid: Smooth, green ssyy gametes All sy F 1 Phenotypes: 10. 2 Dihybrid Crosses & Gene Linkage 22

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring. F 0 Phenotype: Genotype: Punnet Grid: F 1 Smooth, green SSyy ssyy gametes Sy Sy All sy Ssyy Phenotypes: 10. 2 Dihybrid Crosses & Gene Linkage 23

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring. F 0 Phenotype: Genotype: Punnet Grid: F 1 Smooth, green SSyy or Ssyy ssyy gametes Sy Sy Sy sy All sy Ssyy ssyy Phenotypes: 10. 2 Dihybrid Crosses & Gene Linkage 24

Dihybrid Crosses Consider two traits, each carried on separate chromosomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring. F 0 Phenotype: Genotype: Punnet Grid: F 1 Smooth, green SSyy or Ssyy ssyy gametes Sy Sy Sy sy All sy Ssyy ssyy Phenotypes: No rough peas will be present in the offspring if the unknown parent genotype is SSyy. The presence of rough green peas in the offspring means that the unknown genotype must be Ssyy. The expected ratio in this cross is 3 smooth green : 1 rough green. This is not the same as the outcome. Remember that each reproduction event is chance and the sample size is very small. With a much larger sample size, the outcome would be closer to the expected ratio, simply due to probability. 25

Thomas Hunt Morgan was an embryologist in the early 20 th century. When he realized that his field had run dry of new things to learn, he switched to genetics, as he felt there was more to learn there. He crossed fruit flies to determine patterns of inheritance, until one cross did not fit the Mendelian Ratio: Heterozygous Fly X Homozygous Recessive Fly +vg Bb vgvg bb Expected vg b +B +vg Bb +b +vg bb vg B vgvg Bb vg b Vgvg bb vg b 25% +B +vg Bb 25% +b +vg bb 25% vg B vgvg Bb 25% vg b Vgvg bb 41. 5% 8. 5% 41. 5% Observed Why the difference? His graduate assistant figured out this interaction. 26

Skill: Use of chi-squared analysis on data from dihybrid crosses. In order to determine the actual difference between the expected and observed results, a chi-squared analysis must be performed to determine if the difference is significant Chi Squared – χ2, is a variable that represents the difference between expected and observed results in an experiment. Expected vg b Observed vg b +B +vg Bb +b +vg bb vg B vgvg Bb vg b Vgvg bb 250 - 25% +B +vg Bb +b +vg bb vg B vgvg Bb vg b Vgvg bb 415 - 41. 5% 85 - 8. 5% 415 - 41. 5%

Skill: Use of chi-squared analysis on data from dihybrid crosses. +B +b All we need to do +vg Bb +vg bb Expected vg b is to use the numbers of the 250 - 25% phenotypes and +B +b plug them in to +vg Bb +vg bb the table to Observed vg b compare the O 415 - 41. 5% 85 - 8. 5% and E data sets. Finally, we compare the Chi -Squared Value with the Critical Value. A higher value implies significance, while a lower value is not significant. Observed Phenotypes (o) Expected (e) vg B vgvg Bb vg b Vgvg bb 250 - 25% vg B vgvg Bb vg b Vgvg bb 85 - 8. 5% 415 - 41. 5% (o-e)2 X 2 = Degrees of freedom = # of potential phenotypes - 1 (o-e)2 e

Skill: Use of chi-squared analysis on data from dihybrid crosses. Finally, we compare the Chi -Squared Value with the Critical Value. A higher value implies significance, while a lower value is not significant. Chi Squared Value = Degrees of freedom = This test will tell you if the genes are linked or not: # of potential phenotypes - 1 A significant result = Gene Linkage A non-significant result = No Linkage And will enable you to make a statement about your hypothesis: Significant result = Accept Alternative Hypothesis Non-significant result = Accept Null Hypothesis 10. 2 Dihybrid Crosses & Gene Linkage 29

Autosomal Gene Linkage Linked genes are pairs or groups of genes which are inherited together, carried on the same chromosome. SCN 5 A (voltage-gated sodium channel) The SCN 5 A, PDCD 10 and SOX 2 genes are all linked by being on chromosome 3. They are a linkage group, and alleles of each will therefore be inherited together. Independent assortment does not occur between linked genes. PDCD 10 (programmed cell death) SOX 2 (transcription factor - promoter region) Chromosome 3 from: http: //en. wikipedia. org/wiki/Chromosome_3_%28 human%29 10. 2 Dihybrid Crosses & Gene Linkage 30

Autosomal Gene Linkage Linked genes are pairs or groups of genes which are inherited together, carried on the same chromosome. Standard notation for linked genes: A B “heterozygous at both loci” a b SCN 5 A (voltage-gated sodium channel) Locus 1 The SCN 5 A, PDCD 10 and SOX 2 genes are all linked by being on chromosome 3. Locus 2 The line denotes the chromosome, or the fact that the two genes are linked. They are a linkage group, and alleles of each will therefore be inherited together. Independent assortment does not occur between linked genes. Syllabus examples of Linkage Groups: PDCD 10 (programmed cell death) SOX 2 (transcription factor - promoter region) Chromosome 3 from: http: //en. wikipedia. org/wiki/Chromosome_3_%28 human%29 Sweet peas (Lathyrus odoratus): flower color (P/p) linked with pollen grain shape (L/l) Corn (Zea mays): Kernel color (C/c) linked with Waxiness of kernels (W/w) 10. 2 Dihybrid Crosses & Gene Linkage 31

Autosomal Gene Linkage vs Sex-Linked Disorders Sex-linked disorders are carried on the non-homologous regions of the X chromosome. SCN 5 A (voltage-gated sodium channel) Linked genes are pairs or groups of genes which are inherited together, carried on the same chromosome. A B a b Locus 1 Locus 2 PDCD 10 Alleles are expressed whether they are dominant or recessive, as there is no alternate allele carried on the Y chromosome. Gene-related disorders which are sex-linked include red-green color blindness and hemophilia. Males are more frequently affected by sex-linked disorders. (programmed cell death) Y X SOX 2 (transcription factor - promoter region) Chromosome 3 from: http: //en. wikipedia. org/wiki/Chromosome_3_%28 human%29 There about 2000 genes on X and 86 on Y. Gene linkage is therefore also common on X and Y. 10. 2 Dihybrid Crosses & Gene Linkage 32

Linkage Groups Are carried on the same chromosome and are inherited together. They do not assort independently. In sweet peas (Lathyrus odoratus), the genes for flower colour and pollen grain shape are carried on the same chromosome. Plants which are heterozygous at both loci are test-crossed. What ratio of phenotypes is expected? Key to alleles: P = purple p = white L = long l = short Genotype: Phenotype: Image: 'Sweet Pea' http: //www. flickr. com/photos/69166981@N 00/3600419425 10. 2 Dihybrid Crosses & Gene Linkage 33

Linkage Groups Are carried on the same chromosome and are inherited together. They do not assort independently. In sweet peas (Lathyrus odoratus), the genes for flower colour and pollen grain shape are carried on the same chromosome. Plants which are heterozygous at both loci are test-crossed. What ratio of phenotypes is expected? Genotype: Phenotype: p l Locus 1 Locus 2 Key to alleles: P = purple p = white L = long l = short White; Short Image: 'Sweet Pea' http: //www. flickr. com/photos/69166981@N 00/3600419425 10. 2 Dihybrid Crosses & Gene Linkage 34

Linkage Groups Are carried on the same chromosome and are inherited together. They do not assort independently. In sweet peas (Lathyrus odoratus), the genes for flower colour and pollen grain shape are carried on the same chromosome. Key to alleles: P = purple p = white L = long l = short Plants which are heterozygous at both loci are test-crossed. What ratio of phenotypes is expected? Genotype: Phenotype: p l P L p l Locus 1 Locus 2 White; Short Purple; Long Image: 'Sweet Pea' http: //www. flickr. com/photos/69166981@N 00/3600419425 10. 2 Dihybrid Crosses & Gene Linkage 35

Linkage Groups Are carried on the same chromosome and are inherited together. They do not assort independently. In sweet peas (Lathyrus odoratus), the genes for flower colour and pollen grain shape are carried on the same chromosome. Key to alleles: P = purple p = white L = long l = short Plants which are heterozygous at both loci are test-crossed. What ratio of phenotypes is expected? Genotype: p l P L p l Locus 1 Locus 2 Phenotype: White; Short Punnet Grid: Possible Gametes Purple; Long PL pl All pl Phenotypes: Ratio: Image: 'Sweet Pea' http: //www. flickr. com/photos/69166981@N 00/3600419425 10. 2 Dihybrid Crosses & Gene Linkage 36

Linkage Groups Are carried on the same chromosome and are inherited together. They do not assort independently. In sweet peas (Lathyrus odoratus), the genes for flower colour and pollen grain shape are carried on the same chromosome. Key to alleles: P = purple p = white L = long l = short Plants which are heterozygous at both loci are test-crossed. What ratio of phenotypes is expected? Genotype: p l P L p l Locus 1 Locus 2 Phenotype: White; Short Punnet Grid: Possible Gametes PL pl All pl Pp. Ll ppll Purple; Long White; Short Phenotypes: Ratio: Purple; Long 1 : 1 Image: 'Sweet Pea' http: //www. flickr. com/photos/69166981@N 00/3600419425 10. 2 Dihybrid Crosses & Gene Linkage 37

Linkage Groups Are carried on the same chromosome and are inherited together. They do not assort independently. In sweet peas (Lathyrus odoratus), the genes for flower color and pollen grain shape are carried on the same chromosome. Plants which are heterozygous at both loci are test-crossed. A small number of purple; short and white; long individuals have appeared in the offspring. Explain what has happened. Key to alleles: P = purple p = white L = long l = short Image: 'Sweet Pea' http: //www. flickr. com/photos/69166981@N 00/3600419425 10. 2 Dihybrid Crosses & Gene Linkage 38

Recombination of alleles occurs as a result of crossing-over between non-sister chromatids. Exchange of alleles gives new genotypes of gametes. Plants which are heterozygous at both loci are test-crossed. A small number of purple; short and white; long individuals have appeared in the offspring. Explain what has happened. Key to alleles: P = purple p = white L = long l = short Diploid cell Heterozygous at both loci 10. 2 Dihybrid Crosses & Gene Linkage 39

Recombination of alleles occurs as a result of crossing-over between non-sister chromatids. Exchange of alleles gives new genotypes of gametes. Key to alleles: P = purple p = white L = long l = short Plants which are heterozygous at both loci are test-crossed. A small number of purple; short and white; long individuals have appeared in the offspring. Explain what has happened. The test cross individual is homozygous recessive at both loci, so only one type of gamete is produced. Possible gametes: Test individual: p l Heterozygous individual: Diploid cell Heterozygous at both loci Chromosomes replicate in Synthesis phase 10. 2 Dihybrid Crosses & Gene Linkage 40

Recombination of alleles occurs as a result of crossing-over between non-sister chromatids. Exchange of alleles gives new genotypes of gametes. Key to alleles: P = purple p = white L = long l = short Plants which are heterozygous at both loci are test-crossed. A small number of purple; short and white; long individuals have appeared in the offspring. Explain what has happened. The test cross individual is homozygous recessive at both loci, so only one type of gamete is produced. Alleles segregate in meiosis, giving two possible gametes: P p L Possible gametes: Test individual: p l Heterozygous individual: P L p l l Diploid cell Heterozygous at both loci Chromosomes replicate in Synthesis phase 10. 2 Dihybrid Crosses & Gene Linkage 41

Recombination of alleles occurs as a result of crossing-over between non-sister chromatids. Exchange of alleles gives new genotypes of gametes. Key to alleles: P = purple p = white L = long l = short Plants which are heterozygous at both loci are test-crossed. A small number of purple; short and white; long individuals have appeared in the offspring. Explain what has happened. Possible gametes: Test individual: p l Heterozygous individual: Diploid cell Heterozygous at both loci Chromosomes replicate in Synthesis phase P L p l Crossing Over Prophase I Alleles are exchanged Crossing-over occurs occasionally. It is more likely to happen between linked genes which are further apart. 10. 2 Dihybrid Crosses & Gene Linkage 42

Recombination of alleles occurs as a result of crossing-over between non-sister chromatids. Exchange of alleles gives new genotypes of gametes. Key to alleles: P = purple p = white L = long l = short Plants which are heterozygous at both loci are test-crossed. A small number of purple; short and white; long individuals have appeared in the offspring. Explain what has happened. Possible gametes: Test individual: p l Heterozygous individual: Diploid cell Heterozygous at both loci Chromosomes replicate in Synthesis phase P L p l Recombinants: Crossing Over Sister chromatids are Prophase I separated in anaphase II. Alleles are exchanged Recombined gametes are produced. Crossing-over occurs occasionally. It is more likely to happen between linked genes which are further apart. P l p L 10. 2 Dihybrid Crosses & Gene Linkage 43

Recombination of alleles occurs as a result of crossing-over between non-sister chromatids. Exchange of alleles gives new genotypes of gametes. Key to alleles: P = purple p = white L = long l = short Plants which are heterozygous at both loci are test-crossed. A small number of purple; short and white; long individuals have appeared in the offspring. Explain what has happened. Normal gametes (majority) Possible Gametes P L Possible gametes: Test individual: p p l l Heterozygous individual: All p l P L p l Recombinants: P l p L Crossing-over occurs occasionally. It is more likely to happen between linked genes which are further apart. 10. 2 Dihybrid Crosses & Gene Linkage 44

Recombination of alleles occurs as a result of crossing-over between non-sister chromatids. Exchange of alleles gives new genotypes of gametes. Key to alleles: P = purple p = white L = long l = short Plants which are heterozygous at both loci are test-crossed. A small number of purple; short and white; long individuals have appeared in the offspring. Explain what has happened. Normal gametes (majority) Possible gametes: Test individual: p l Possible Gametes P L p l All p l Pp. Ll ppll P L Purple; long White, short p l Heterozygous individual: Recombinants: P l p L Crossing-over occurs occasionally. It is more likely to happen between linked genes which are further apart. 10. 2 Dihybrid Crosses & Gene Linkage 45

Recombination of alleles occurs as a result of crossing-over between non-sister chromatids. Exchange of alleles gives new genotypes of gametes. Key to alleles: P = purple p = white L = long l = short Plants which are heterozygous at both loci are test-crossed. A small number of purple; short and white; long individuals have appeared in the offspring. Explain what has happened. Normal gametes (majority) Recombinant gametes (small number) Possible gametes: Test individual: p l Possible Gametes P L p l P l p L All p l Pp. Ll ppll Ppll pp. Ll P L Purple; long White, short Purple; short White, long p l Heterozygous individual: Recombinants: P l p L Crossing-over occurs occasionally. It is more likely to happen between linked genes which are further apart. 10. 2 Dihybrid Crosses & Gene Linkage 46

Crossing-Over Synapsis Homologous chromosomes associate Increases genetic variation through recombination of linked alleles. Chiasma Formation Neighboring non-sister chromatids are cut at the same point. A Holliday junction forms as the DNA of the cut sections attach to the open end of the opposite non-sister chromatid. Recombination As a result, alleles are swapped between nonsister chromatids. 10. 2 Dihybrid Crosses & Gene Linkage 47

Crossing-Over Increases genetic variation through recombination of linked alleles. 10. 2 Dihybrid Crosses & Gene Linkage 48

Gene Linkage & Recombination SCN 5 A The further apart a pair of alleles are on a chromosome, the more likely it is that crossing over may occur between them - leading to recombination. (voltage-gated sodium channel) Crossing-over is more likely to occur between SCN 5 A and PDCD 10 than between PDCD 10 and SOX 2. Knowing this, researchers can map the position of genes on a chromosome based on the frequency of recombination between gene pairs: the further apart they are, the more often they cross over. PDCD 10 (programmed cell death) SOX 2 (transcription factor - promoter region) Chromosome 3 from: http: //en. wikipedia. org/wiki/Chromosome_3_%28 human%29 Animation and quiz from: http: //www. csuchico. edu/~jbell/Biol 207/animations/recombination. html 10. 2 Dihybrid Crosses & Gene Linkage 49

Gene Linkage & Recombination Which description best fits this image? a. b. c. d. Four chromosomes, four chiasmata Four chromatids, two chiasmata, two centromeres Two chromosomes, four chiasmata A pair of sister chromatids 10. 2 Dihybrid Crosses & Gene Linkage 50

Gene Linkage & Recombination Which description best fits this image? a. b. c. d. Four chromosomes, four chiasmata Four chromatids, two chiasmata, two centromeres Two chromosomes, four chiasmata A pair of sister chromatids 10. 2 Dihybrid Crosses & Gene Linkage 51

Gene Linkage & Recombination Which description best fits this image? chiasmata Chromosome 1 a Sister chromatids Chromosome 1 b Sister chromatids centromeres a. b. c. d. Four chromosomes, four chiasmata Four chromatids, two chiasmata, two centromeres Two chromosomes, four chiasmata A pair of sister chromatids 10. 2 Dihybrid Crosses & Gene Linkage 52

Gene Linkage & Recombination The genes for kernel color and waxiness are linked in the corn plant (Zea mays). In a cross between a plant that is homozygous dominant at both loci with a plant that is heterozygous at both loci, identify the following genotypes as: a: regular b: recombinants c: impossible Cc. Ww CCWw Cc. WW CCww cc. WW 10. 2 Dihybrid Crosses & Gene Linkage 53

Gene Linkage & Recombination The genes for kernel color and waxiness are linked in the corn plant (Zea mays). In a cross between a plant that is homozygous dominant at both loci with a plant that is heterozygous at both loci, identify the following genotypes as: a: regular b: recombinants c: impossible Cc. Ww C C CCWw Cc. WW CCWW Key to alleles: C = colored c = no color W = waxy w = not waxy CCww cc. WW W W Regular gametes (majority) Recombinant gametes (small number) Possible Gametes All C W 10. 2 Dihybrid Crosses & Gene Linkage 54

Gene Linkage & Recombination The genes for kernel color and waxiness are linked in the corn plant (Zea mays). In a cross between a plant that is homozygous dominant at both loci with a plant that is heterozygous at both loci, identify the following genotypes as: a: regular b: recombinants c: impossible Cc. Ww C C CCWw W W Cc. WW C W c w Regular gametes (majority) Possible Gametes C W CCWW Key to alleles: C = colored c = no color W = waxy w = not waxy CCww cc. WW Recombinant gametes (small number) c w All C W 10. 2 Dihybrid Crosses & Gene Linkage 55

Gene Linkage & Recombination The genes for kernel color and waxiness are linked in the corn plant (Zea mays). In a cross between a plant that is homozygous dominant at both loci with a plant that is heterozygous at both loci, identify the following genotypes as: a: regular b: recombinants c: impossible Cc. Ww C C CCWw W W Cc. WW C W c w Regular gametes (majority) Possible Gametes C W c w All C W CCWW Cc. Ww CCWW Key to alleles: C = colored c = no color W = waxy w = not waxy CCww cc. WW Recombinant gametes (small number) 10. 2 Dihybrid Crosses & Gene Linkage 56

Gene Linkage & Recombination The genes for kernel color and waxiness are linked in the corn plant (Zea mays). In a cross between a plant that is homozygous dominant at both loci with a plant that is heterozygous at both loci, identify the following genotypes as: a: regular b: recombinants c: impossible Cc. Ww C C CCWw W W Cc. WW CCWW Key to alleles: C = colored c = no color W = waxy w = not waxy CCww C W C w c W Regular gametes (majority) cc. WW Recombinant gametes (small number) Possible Gametes C W c w C w c W All C W CCWW Cc. Ww CCWw Cc. WW 10. 2 Dihybrid Crosses & Gene Linkage 57

Gene Linkage & Recombination Two genes are linked as shown here E m e M The genes are far apart such that crossing-over between the alleles occurs occasionally. Which statement is true of the gametes? A. All of the gametes will be Em and e. M B. There will be equal numbers of EM, e. M and em C. There will be approximately equal numbers of EM and e. M gametes D. There will be more Em gametes than em gametes 10. 2 Dihybrid Crosses & Gene Linkage 58

Gene Linkage & Recombination Two genes are linked as shown here E m e M The genes are far apart such that crossing-over between the alleles occurs occasionally. Which statement is true of the gametes? A. All of the gametes will be Em and e. M B. There will be equal numbers of EM, e. M and em C. There will be approximately equal numbers of EM and e. M gametes D. There will be more Em gametes than em gametes 10. 2 Dihybrid Crosses & Gene Linkage 59

Gene Linkage & Recombination Two genes are linked as shown here E m e M The genes are far apart such that crossing-over between the alleles occurs occasionally. Which statement is true of the gametes? A. All of the gametes will be Em and e. M B. There will be equal numbers of EM, e. M and em C. There will be approximately equal numbers of EM and e. M gametes D. There will be more Em gametes than em gametes E m E M e m Regular gametes (majority) E m e M Recombinant gametes (small number) E M e m 10. 2 Dihybrid Crosses & Gene Linkage 60

Polygenic Inheritance (AHL) 61

Polygenic Inheritance A single characteristic controlled by multiple genes. 62

Polygenic Inheritance A single characteristic controlled by multiple genes. Polygenic inheritance gives rise to continuous variation in the phenotype. Use these two examples in the exam. Human Skin Color Wheat kernel color Watch this quick video on the mechanisms behind skin color/kernel color. 63

Polygenic Inheritance A single characteristic controlled by multiple genes. Polygenic inheritance gives rise to continuous variation in the phenotype. Use these two examples in the exam. Human Skin Color Wheat kernel color Other examples: • Susceptibility* to heart disease, certain types of cancer, mental illnesses. • The Autism Spectrum. Autism is a pervasive developmental disorder that presents on a scale (known as the Childhood Autism Rating Scale). It is not as clearly polygenic as the above examples - it is suspected that gene interactions and environmental factors play a large role. *susceptibility is not deterministic, but it is beneficial to know if you are at elevated genetic risk of these illnesses. 64

Polygenic Inheritance of Skin Color Polygenic inheritance gives rise to continuous variation in the phenotype. Globally we observe continuous variation in skin colors. Skin color is the result of pigments, such as melanin, being produced - the darker the skin, the greater the protection against the harmful effects of the Sun. Skin color is though to be controlled by up to four separate genes, each with their own alleles. This is too large for us to deal with simply, so we'll look at two genes with two alleles each. Image from: http: //www. danacentre. org. uk/events/2007/05/03/259 65

Polygenic Inheritance of Skin Color Polygenic inheritance gives rise to continuous variation in the phenotype. Globally we observe continuous variation in skin colors. Skin color is the result of pigments, such as melanin, being produced - the darker the skin, the greater the protection against the harmful effects of the Sun. Skin colour is though to be controlled by up to four separate genes, each with their own alleles. This is too large for us to deal with simply, so we'll look at two genes with two alleles each. Image from: http: //www. danacentre. org. uk/events/2007/05/03/259 Watch this TED Talk and think about the following questions: • • • What is melanin and what purpose does it serve? What skin tone were early humans most likely to have? Why does this change with latitude as humans migrated towards the poles? What are the relative advantages and disadvantages of light and dark skin, depending on climate? TOK/ Aim 8: Why have people historically discriminated based on skin color? How could the Natural Sciences educate people to think twice about their prejudices? Nina Jablosnki breaks the illusion of skin colour, via TED. http: //www. ted. com/talks/lang/eng/nina_jablonski_breaks_the_illusion_of_skin_color. html 66

Polygenic Inheritance of Skin Color Example: 2 genes (A and B), 2 alleles each Assume: genes are not linked (separate chromosomes) In polygenics, alleles can be: • Contributing (they add to the phenotype) • Non-contributing (they do not add to the phenotype) How many genotypes are possible? Key to alleles: A = add melanin a = don’t add melanin B = add melanin b = don’t add melanin 67

Polygenic Inheritance of Skin Color Example: 2 genes (A and B), 2 alleles each Assume: genes are not linked (separate chromosomes) In polygenics, alleles can be: • Contributing (they add to the phenotype) • Non-contributing (they do not add to the phenotype) How many genotypes are possible? Key to alleles: A = add melanin a = don’t add melanin B = add melanin b = don’t add melanin Remember that alleles segregate during meiosis. or or or Alleles of unlinked chromosomes orient randomly. There is also random fertilization of gametes. So many combinations! gametes 68

Polygenic Inheritance of Skin Color Example: 2 genes (A and B), 2 alleles each Assume: genes are not linked (separate chromosomes) In polygenics, alleles can be: • Contributing (they add to the phenotype) • Non-contributing (they do not add to the phenotype) How many genotypes are possible? Key to alleles: A = add melanin a = don’t add melanin B = add melanin b = don’t add melanin Nine: Notice that the possible combinations of genotypes gives rise to continuous variation in the phenotype. This population follows a normal distribution. 69

Polygenic Inheritance of Skin Color Is it possible for twins to be: a. Different colors? Key to alleles: A = add melanin a = don’t add melanin B = add melanin b = don’t add melanin 70

Polygenic Inheritance of Skin Colour Is it possible for twins to be: a. Different colors? YES. As long as they are non-identical twins. Two eggs will have been fertilized by individual sperm cells. Key to alleles: A = add melanin a = don’t add melanin B = add melanin b = don’t add melanin Each gamete carries a different combination of alleles, so it is possible that the twins have noticeably differently-colored skin. https: //www. youtube. com/watch? v=-e 5 ODsf_TEk 71

Polygenic Inheritance of Skin Color Is it possible for twins to be: b. Lighter or darker than both parents? Key to alleles: A = add melanin a = don’t add melanin B = add melanin b = don’t add melanin 72

Polygenic Inheritance of Skin Color Is it possible for twins to be: b. Lighter or darker than both parents? F 0 F 1 Phenotype: Genotype: Aa. Bb Punnett Grid: gametes AABb Key to alleles: A = add melanin a = don’t add melanin B = add melanin b = don’t add melanin Genotypes: Phenotypes: 73

Polygenic Inheritance of Skin Color Is it possible for twins to be: b. Lighter or darker than both parents? F 0 Phenotype: Genotype: Aa. Bb Punnett Grid: gametes AABb AB AB Ab Key to alleles: A = add melanin a = don’t add melanin B = add melanin b = don’t add melanin Ab AB Ab a. B ab F 1 Genotypes: Phenotypes: 74

Polygenic Inheritance of Skin Color Is it possible for twins to be: b. Lighter or darker than both parents? F 0 F 1 Phenotype: Genotype: Aa. Bb Punnett Grid: gametes AABb AB AB Ab Ab AB AABB AABb Ab AABb AAbb a. B Aa. BB Aa. Bb ab Aa. Bb Aabb Key to alleles: A = add melanin a = don’t add melanin B = add melanin b = don’t add melanin Genotypes: Phenotypes: 75

Polygenic Inheritance of Skin Color Is it possible for twins to be: b. Lighter or darker than both parents? YES. F 0 Phenotype: Genotype: Aa. Bb Punnett Grid: gametes darker F 1 AABb AB AB Ab Ab AB AABB AABb Ab AABb AAbb a. B Aa. BB Aa. Bb ab Aa. Bb Aabb Key to alleles: A = add melanin a = don’t add melanin B = add melanin b = don’t add melanin lighter Genotypes: Phenotypes: 76

Global Evolution of Skin Color What are the advantages of: a. Dark skin in hot climates? b. Pale skin in cold climates? Map of global skin color distribution from: http: //en. wikipedia. org/wiki/Human_skin_color Human skin color is influenced by genes, but also by how much light a person receives. Human height is similar, in that it is influenced by genes, but also by the amount of nutrition an individual receives during growth periods. 77

Global Evolution of Skin Color Map of global skin colour distribution from: What are the advantages of: http: //en. wikipedia. org/wiki/Human_skin_color a. Dark skin in hot climates? Protection against UV and therefore skin damage and cancer. b. Pale skin in cold climates? Increased production of vitamin D in low-sunlight conditions. 78

Global Evolution of Skin Color Map of global skin colour distribution from: What are the advantages of http: //en. wikipedia. org/wiki/Human_skin_color a. Dark skin in hot climates? Protection against UV and therefore skin damage and cancer. Dark-skinned people in cold climates should use vitamin D supplements. b. Pale skin in cold climates? Increased production of vitamin D in low-sunlight conditions. Pale-skinned people in hot climates should slip-slop-slap-seek-slide. In this case, the correlation between skin color and latitude does suggest causality. 79

Polygenic Inheritance of Wheat Kernel Color Inheritance of color of wheat kernels works in a similar way to human skin color. A wheat plant which is homozygous dominant for both genes is crossed with one which is heterozygous for both genes. What is the predicted ratio of phenotypes in the cross? F 0 Key to alleles: A = add red a = don’t add red B = add red b = don’t add red Genotype: Punnett Grid: gametes F 1 80

Polygenic Inheritance of Wheat Kernel Color Inheritance of color of wheat kernels works in a similar way to human skin color. A wheat plant which is homozygous dominant for both genes is crossed with one which is heterozygous for both genes. What is the predicted ratio of phenotypes in the cross? F 0 homozygous dominant for both genes Genotype: Punnett Grid: AABB Key to alleles: A = add red a = don’t add red B = add red b = don’t add red heterozygous for both genes Aa. Bb gametes F 1 81

Polygenic Inheritance of Wheat Kernel Color Inheritance of color of wheat kernels works in a similar way to human skin color. A wheat plant which is homozygous dominant for both genes is crossed with one which is heterozygous for both genes. What is the predicted ratio of phenotypes in the cross? F 0 homozygous dominant for both genes Genotype: Punnett Grid: heterozygous for both genes Aa. Bb AABB gametes AB Key to alleles: A = add red a = don’t add red B = add red b = don’t add red Ab a. B ab AB AABB AABb Aa. BB Aa. Bb F 1 82

Polygenic Inheritance of Wheat Kernel Color Inheritance of color of wheat kernels works in a similar way to human skin color. A wheat plant which is homozygous dominant for both genes is crossed with one which is heterozygous for both genes. What is the predicted ratio of phenotypes in the cross? F 0 homozygous dominant for both genes Genotype: Punnett Grid: “All AB” F 1 heterozygous for both genes Aa. Bb AABB gametes AB Ab a. B ab AB AABB AABb Aa. BB Aa. Bb Phenotype ratio: 1 very red : Key to alleles: A = add red a = don’t add red B = add red b = don’t add red 2 Red : 1 pink 83

Gene Interaction The expression of one gene is dependent upon the prior expression of another. In the case of guinea pigs, there is gene interaction for fur color. Key to alleles: C = color c = albino A = agouti a = black The first gene, C, determines whether color is present. The second gene, A, is only expressed if C is first expressed. It determines which color will be produced. 10. 2 Dihybrid Crosses & Gene Linkage 84

Gene Interaction The expression of one gene is dependent upon the prior expression of another. In the case of guinea pigs, there is gene interaction for fur color. Key to alleles: C = color c = albino A = agouti a = black The first gene, C, determines whether color is present. The second gene, A, is only expressed if C is first expressed. It determines which color will be produced. Genotypes cc. AA cc. Aa ccaa CCAA Cc. Aa CCaa Ccaa If the genotype ‘cc’ is present, there will be no expression of color. A will also not be expressed. 10. 2 Dihybrid Crosses & Gene Linkage 85

Gene Interaction The expression of one gene is dependent upon the prior expression of another. In the case of guinea pigs, there is gene interaction for fur color. Key to alleles: C = color c = albino A = agouti a = black The first gene, C, determines whether color is present. The second gene, A, is only expressed if C is first expressed. It determines which color will be produced. Phenotype ratios do not fit the normal 9 : 3 : 1 ratio. Genotypes cc. AA cc. Aa ccaa CCAA Cc. Aa CCaa Ccaa If the genotype ‘cc’ is present, there will be no expression of color. A will also not be expressed. gametes CA Ca c. A ca CA CCAa Cc. AA Cc. Aa Ca CCAa Ccaa Cc. Aa Ccaa c. A Cc. Aa cc. AA cc. Aa ca Cc. Aa Ccaa cc. Aa ccaa 9 agouti : 3 black : 4 albino 10. 2 Dihybrid Crosses & Gene Linkage 86

Polygenic Inheritance of Dog Coat Color Dog coat color demonstrates continuous variation from full color to the lack of color due to the inheritance of S, SI, SP and SW alleles for pigment produced by melanocytes. The traits for pigment are all located on the same chromosome as the genes for deafness, so they are considered linked. SW Maggie’s Genotype: W S SW d S = Contribute regular melanin Sw = do not contribute melanin SP = Contribute piebald melanin SI = Contribute to Irish Spotting d d If Maggie were to have puppies with a white, heterozygous hearing dog, what percentage of her puppies would have hearing? (Deafness is also recessive) 87

Polygenic Inheritance of Dog Coat Color Dog coat color demonstrates continuous variation from full color to the lack of color due to the inheritance of S, SI, SP and SW alleles for pigment produced by melanocytes. The traits for pigment are all located on the same chromosome as the genes for deafness, so they are considered linked. SW Maggie’s Genotype: W S SW d d d SW D SWSWdd SWSWDd Phenotype Ratio S = Contribute regular melanin Sw = do not contribute melanin SP = Contribute piebald melanin SI = Contribute to Irish Spotting If Maggie were to have puppies with a white, heterozygous hearing dog, what percentage of her puppies would have hearing? (Deafness is also recessive) % Hearing – 50% 1: 1 88

Mathematical Questions A trait is controlled by two genes, each with two alleles. How many genotypes & phenotypes are possible for this trait? 89

Mathematical Questions A trait is controlled by two genes, each with two alleles. How many genotypes & phenotypes are possible for this trait? Key to alleles: A = contributing a = non-contributing B = contributing b = non-contributing Same! There are 16 combinations of parent alleles, making 9 different genotypes. 90

Mathematical Questions A trait is controlled by two genes, each with two alleles. How many genotypes & phenotypes are possible for this trait? 4 3 2 Key to alleles: A = contributing a = non-contributing B = contributing b = non-contributing 1 0 There are 16 combinations of parent alleles, making 9 different genotypes. These make up 5 different phenotypes. 91