1. In the broad bean, a pure-breeding variety with green seeds and black hilums (the point of attachment of the seed to the pod) was crossed with a pure breeding variety with yellow seeds and white hilums. All the F 1 plants had yellow seeds and black hilums. When these were allowed to self-fertilize, the plants of the F 2 generation produced the following seeds: yellow seeds with white hilums31 yellow seeds with black hilums93 green seeds with white hilums 8 green seeds with black hilums28 –Which characteristics do you consider to be dominant and which recessive? (1) –Devise suitable symbols for the alleles of the genes involved (1) –What genetic ratio is suggested by the numerical data in the experiment (1) Construct suitable crossing diagrams to show the phenotypes and genotypes of the plants and their gametes in each generation. Colour code the offspring – one colour for each phenotype. Neat answers please! (7) How many of the offspring are heterozygous for both traits? (1) How many of the offspring if crossed with an individual of an identical genotype would NOT be able to produce offspring that are homozygous dominant for both traits? (1)

2. Chi Squared test in genetics

3. What we will learn today… Use Chi squared test to compare results with those expected and to determine whether the results are significant or non significant.

4. Chi Squared What’s it all mean – simple enough equation but it is looking at how close your results are to what you expected to get. Put simply – you’re rarely going to get exactly what you predict but is the difference due to chance or is it due to something else?

5. Sexing crocodiles 2 crocodiles procreate and the female lays her eggs. What percentage of male and female crocodile hatchlings would you expect? 50/50?

6. WRONG! Would it surprise you to find that they are all either female or male? So your expected value of 50/50 is massively different to what you see (100/0). This difference must be due to something else and not merely coincidence.

7. Just for interest Temperature dependent sex determination The sex of their hatchlings is determined not by genetics, but by the average temperature during the middle third of their incubation period. If the temperature inside the nest is below 31.7 °C or above 34.5 °C the offspring will be female. Males can only be born if the temperature is within that narrow 5-degree range.

8. Another simple example Flipping a coin What would you think if after 100 flips there was 100 heads and no tails?

9. 4 grain phenotypes in the above ear of genetic corn: Purple & Smooth (A), Purple & Shrunken (B), Yellow & Smooth (C) and Yellow & Shrunken (D). They are produced by the following two pairs of heterozygous genes (P & p and S & s) located on two pairs of homologous chromosomes (each gene on a separate chromosome): Dominant Genes Recessive Genes P = Purple p = Yellow S = Smooth s = Shrunken

10. Chi Square Problem: An ear of corn has a total of 381 grains, (216 Purple & Smooth, 79 Purple & Shrunken, 65 Yellow & Smooth, and 21 Yellow & Shrunken).

11. Your Null Hypothesis: There is no significant difference between the expected and observed frequencies of phenotypes. In other words, This ear of corn was produced by a dihybrid cross (PpSs x PpSs) involving two pairs of heterozygous genes resulting in a theoretical (expected) ratio of 9:3:3:1.

12. Phenotype Observed Number ( Expected Ratio Expected Number E [O- E] 2 E Purple & Smooth 2169 381 x 9/16 = 214 4 ÷ 214 = 0.019 Purple & Shrunk en 793381 x 3/16 = 7164/71 = 0.901 Yellow & Smooth 653381 x 3/16 = 7136/71 = 0.507 Yellow & Shrunk en 211381 x 1/16 = 249/24 = 0.375 Total Number : 381---------- Chi Square Value: 1.80

13. Degrees Of Freedom: Number of phenotypes - 1. In this problem the number of phenotypes is four; therefore, the degrees of freedom (df) is three (4 - 1 = 3). In the following Table 3 you need to locate the number in row three that is nearest to your chi square value of 1.80.

14. Probability Value: Get a chi-squared table and look up the value for the appropriate significance level (usually 5%) and the degrees of freedom. In the table 5% is shown as 0.05. If our chi sq value is smaller than the value from the tables, we accept the null hypothesis: There is no significant difference between observed and expected. Think: If

15. In fruit flies, grey is dominant to black and normal wings are dominant to vestigial… Thomas Hunt Morgan conducted a number of breeding experiments with drosophila. The results of a cross between two parents heterozygous for both traits is shown in the table. Does this deviate significantly from the expected results? PhenotypesNumber of flies Grey winged180 Black vestigial52 Grey vestigial14 Black winged12

16. Yes the differ significantly So what’s the explanation?

17. Linkage Genetic linkage occurs when particular genetic loci or alleles for genes are inherited jointly. Genetic loci on the same chromosome are physically connected and tend to stay together during meiosis, and are thus genetically linked. This contravenes Mendel’s law of independent assortment. http://www.ndsu.nodak.edu/ins truct/mcclean/plsc431/linkage/l inkage1.htmhttp://www.ndsu.nodak.edu/ins truct/mcclean/plsc431/linkage/l inkage1.htm

18. Any two genes which occur on the same chromosome are said to be linked. All the genes on a single chromosome form a linkage group. Consider two genes with two alleles each: Gene 1: A is dominant over a Gene 2: B is dominant over b Linkage

19. For non-linked genes We can represent this individual with a diagram showing two homologous pairs: AaBb AAaaBBbb ABabAbaB What gametes can be produced? Possible types of gamete: This fits Mendel’s Law of Independent Assortment: any one of a pair of contrasted characters may combine with any of another pair.

20. Aa B b A B a b ABab For linked genes This individual can be represented with only one homologous pair: What gametes can be produced? Possible types of gamete: It is impossible to get the gametes Ab or aB therefore Mendel’s law does not apply.

21. YY R R In practice On tomatoes the gene for flower colour and fruit colour are on the same chromosome i.e. they are linked genes. What would the ratio be at F 2 if we crossed the pure lines of each phenotype? Y: yellow flowers y: white flowers R: red fruit r: yellow fruit F 1 generation: all have yellow flowers and red fruit yy rr Y R y r meiosis Y R y r fertilisation

22. Y R Self cross F 1 generation What would the ratio be at F 2 if we crossed the pure lines of each phenotype? Y: yellow flowers y: white flowers R: red fruit r: yellow fruit F 2 generation: ratio is 3 yellow flowers + red fruit to 1 white flowers + yellow fruit i.e. 3:1 and NOT 9:3:3:1 as we would expect. y r Y R meiosis Y R y r y r Y R y r Y R y r Y R y r Y R Y R y r y r

23. When the actual cross was performed, however, the following results are typical if 100 F2 plants are produced: –Yellow flowers, red fruit: 68 –Yellow flowers, yellow fruit: 7 –White flowers, red fruit: 7 –White flowers, yellow fruit: 18 What’s the explanation?

24. Linkage means Mendel’s laws do not always apply. However, crossing over can separate linked genes.

25. Y R Crossing over of F 1 generation at prophase 1 y r y r Y R Y R y r y r Y R Y R y r y r Y R Crossing over occurs Recombinants have a new genotype What gametes will be produced?

26. Gametes produced by F 1 generation following crossing over Y R y r y r Y R Y R r Yy R y r YRyrYryR This will produce an F 2 generation with our ratio of 9:3:3:1 (work it out if you don’t believe me)

27. Let’s go back to our example off fruit flies (drosophila). Here’s the original table and PhenotypesNumber of flies Grey winged180 Black vestigial52 Grey vestigial14 Black winged12 PhenotypesNumber of flies Grey winged192 Black vestigial64 Grey vestigial1 Black winged2 What might the data below tell us about the gene loci (compared to the above) if we had found the following data instead?

28. Polygenic Inheritance Polygenic inheritance refers to a single characteristic that is controlled by more than two genes By increasing the number of genes controlling a trait, the number of phenotype combinations also increase, until the number of phenotypes to which an individual can be assigned are no longer discrete, but continuous What type of distribution do you think you would see with such a pattern of inheritance?

31. The colour of human skin is determined by the amount of dark pigment (melanin) it contains At least four (possibly more) genes are involved in melanin production; for each gene one allele codes for melanin production, the other does not The combination of melanin producing alleles determines the degree of pigmentation, leading to continuous variation