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HL Genetics Topic 10.2. Assessment Statements 10.2.1 Calculate and predict the genotypic and phenotypic ratio of offspring of dihybrid crosses involving.

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Presentation on theme: "HL Genetics Topic 10.2. Assessment Statements 10.2.1 Calculate and predict the genotypic and phenotypic ratio of offspring of dihybrid crosses involving."— Presentation transcript:

1 HL Genetics Topic 10.2

2 Assessment Statements Calculate and predict the genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes Distinguish between autosomes and sex chromosomes Explain how crossing over between non-sister chromatids of a homologous pair in prophase I can result in an exchange of alleles Define linkage group Explain an example of a cross between two linked genes Identify which of the offspring are recombinants in a dihybrid cross involving linked genes.

3 Dihybrid crosses Mendel’s Peas Seed shape – some round, others wrinkled (allele for round is dominant) Seed colour – some green, others yellow (allele for yellow is dominant) Mendel crossed true breeding plants with each other One parent: homozygous dominant for both traits (round and yellow seeds) RRYY Other parent: homozygous recessive for both traits (wrinkled and green) rryy

4 When both parents are homozgous – all the F1 offspring are the same genotype and phenotype R = allele for round peas r = allele for wrinkled peas Y = allele for yellow peas y = allele for green peas Parent phenotypes Round yellow Green wrinkled Parent genotypes RRYYrryy Parent gametes RYry F1 genotypes RrYy F1 phenotypes Round yellow

5 Cross the F1 double heterozygous Allowing heterozygous offspring to self-pollinate Phenotype Ratio is: 9 3 1

6 Dihybrid Phenotype Ratios Homozygous parents AABB x aabb All F1 offspring the same Heterozygous parents AaBb x AaBb 9:3:3:1 phenotype ratio 9AB 3Ab 3aB 1ab  A new shuffling of the alleles has created a new combination which does not match either of the parents’ genotypes. Recombinants

7 Autosomes and sex chromosomes Sex chromosomes: X and Y (one pair) Autosomes: any chromosome not X or Y (22 pairs) Sex-linked gene is located on a sex chromosome. Autosomal gene is located on one of the autosomes. On which type of chromosome is the gene for haemophilia found? So, the gene is known as ___. On which type of chromosome is the gene for protein production in the pancreas found? So, the gene is known as ___.

8 Exchange of alleles by crossing over Two non-sister chromatids can swap segments of their DNA during prophase I of meiosis. This increases genetic variety of chromosomes in gametes

9 There are now thought to be 20,500 human genes on 23 chromosomes Source: Broad Institute of MIT and Harvard (2008, January 15). Human Gene Count Tumbles There must be approx 1000 genes on each chromosome.

10 Linkage group Any two genes which are found on the same chromosome are said to be linked to each other. Linked genes are usually passed on to the next generation together. Linkage group - groups of genes on the same chromosome inherited together Linked genes are the exception to Mendel’s law of independent assortment

11 Linked genes Fruit fly gene for body color is in the same linkage group as the gene for wing length Alleles are G – grey body g – black body L – long wings l – short wings Notation of linked genes is G L The two horizontal bars symbolize homologous chromosomes and that the locus of G is on the same chromosome as L G L

12 Offspring of a dihybrid cross A cross between homozygous fruit flies GGLL x ggll Grey body & long wings x Black body, short wings F1 flies will be all heterozygous for both of the traits GgLl Phenotype : Grey body & long wings If these F1 heterozygotes were allowed to reproduce together the ratio of phenotypes produced would be expected to be 9:3:3:1

13 Recombinants After the F2 flies were identified the phenotypes were; Grey long wing Grey short wing Black long wing Black short wing This is far from the expected 9:3:3:1 ratio which should give This is a sign that the genes are on the same chromosome (genes are linked)

14 Offspring of a dihybrid cross using linkage notation A cross between homozygous fruit flies GL gl GL x gl Grey body & long wings x Black body, short wings F1 flies will be all heterozygous for both of the traits GL gl Phenotype : Grey body & long wings If these F1 heterozygotes were allowed to reproduce together the ratio of phenotypes produced would be expected to be 9:3:3:1

15 Offspring of a dihybrid cross using linkage notation If these F1 heterozygotes make gametes The gametes will be GL or gl unless crossing over happens – because the genes are on the same chromosome The F1 will contain more than expected of these genotypes G L GL GL gl gl GL gl

16 Offspring of a dihybrid cross using linkage notation If The F1 will contain more than expected of these genotypes The Grey short wing and black long wing phenotypes Are Recombinants - G L GL GL gl gl GL gl Grey long wing Grey short wing Black long wing Black short wing

17 Topic 10.3

18 Assessment Statements Define polygenic inheritance Explain that polygenic inheritance can contribute to continuous variation using two examples, one of which must be human skin colour.

19 Polygenic Inheritance when two or more genes influence the expression of one trait Eg Skin Colour, Height

20 Continuous and discontinuous variation When an array of possible phenotypes can be produced, it is called continuous variation Examples: skin color, height, body shape, and intelligence These traits are also influenced by environmental conditions When only a number of phenotypes can be produced, it is called discontinuous variation Examples: earlobe attachment, blood group

21 Graphical representation Continuousvariation Discontinuous variation

22 Eye Color Iris is made up of zones, rings, streaks or speckles of different colored pigments with varying intensities What color are your eyes, really? Since there is so much variety, eye color must be influenced by multiple alleles and has continuous variation.

23 Skin color How does the existence of multiple alleles controlling skin colour result in the appearance of many different shades of skin colour in humans?

24 Thoughts How do people of varying degrees of skin color relocated to parts of the world that receive differing amounts of sunlight get vitamin D? How do others fight off the sun? Should there be equal esteem for all humans? Why is human diversity so often used to divide and discriminate, rather than be appreciated, respected, and celebrated?

25 Polygenic inheritance of color in wheat. Kernal color in wheat is determined by two genes. A range of colors occur, from white to dark red, depending on the combinations of alleles. Dark red plants are homozygous AABB and white plants are homozygous aabb.

26 Crossing individuals with the phenotype extremes yield offspring that are a 'blend' of the two parents. When these homozygous phenotypes are crossed AABB x aabb Dark x white the F1 offspring are all double heterozygous AaBb.

27 But what happens when the two double heterozygous genotypes are crossed? Parent Phenotypes: all brown Genotypes: AaBb x AbBb Punnet square: ABAbaBab ABAABBAABbAaBBAaBb AbAABbAAbbAaBbAabb aBAaBBAaBbaaBBaaBb abAaBbAabbaaBbaabb There is no blending in the offspring. Offspring can be more extreme than either parent There are grades of colour – evidence of polygenic inheritance.

28 Skin color How does the existence of multiple alleles controlling skin colour result in the appearance of many different shades of skin colour in humans?

29 Human skin colour is controlled by multiple alleles (and the environment) It is known that at least three genes control skin color, let’s call them genes A, B, and C. Someone who is AABBCC would have very dark skin color and someone who is aabbcc would have very light skin color. If they got married and had children, their children would all be AaBbCc and have mid-brown skin. If two of those people would get married and have children, the Punnett square would look like the one above.


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