1. F215 5.1.2 Variation By Ms Cullen

2. Some new terminology......... Codominance – two alleles of the same gene are described as codominant if they both appear in the phenotype of a heterozygote. Example: ABO blood group inheritance, the gene for this has 3 alleles I A I B and I O. I O is recessive whilst I A and I B are codominant.

3. Inheritance of blood groups codominance Parent Alleles ABO A AA (A) AB (AB) AO (A) B AB (AB) BB (B) BO (B) O AO (A) BO (B) OO (O) The possible ABO alleles for one parent are in the top row and the alleles of the other are in the left column. Offspring genotypes are shown in white. Phenotypes are red.

4. Another example of codominant alleles is sickle-cell anaemia People with normal haemoglobin are homozygous with a genotype of H N H N and do not have the disease. People who have sickle haemoglobin are homozygous recessive H S H S and suffer from the disease. People with the genotype H N H S are heterozygous. They have some normal haemoglobin and some sickle haemoglobin. The 2 alleles are codominant. Their phentoype is sickle-cell trait. Draw a genetic cross to show the possible offspring of 2 sickle-cell trait parents.

5. Sex Linkage A characteristic is sex-linked if it appears on the X or Y chromosomes which determine sex. The y chromosome is much smaller and therefore carries fewer genes. As a result most of genes on the sex chromosomes are only carried on the X chromosome (X-linked genes). Males often only have 1 allele for sex-linked genes. This means that they usually display this characteristic, even if it is recessive, as they only have 1 copy to express. Examples: colour blindness, haemophilia

7. Haemophilia GenotypePhenotype XHXHXHXH Female normal blood clotting XHXhXHXh Female normal blood clotting (but she is a carrier) XhXhXhXh Lethal X H YMale normal blood clotting X h YMale haemophilia Q Why can’t a man pass haemophilia on to his son?

8. Haemophilia Phenotypes male x female normal clotting Genotypes X H YX H X h Gametes XHXH Y XHXH XhXh XHXH X h X H X H X h YX H YX h Y

9. A family pedigree

11. Dihybrid cross This is a cross between 2 organisms looking at two different genes which are carried by separate chromosomes. As the chromosomes are sorted independently of each other during meiosis, this results in 4 gametes. All the cells in the body have 2 complete sets of each chromosome. This means there are 9 possible genotypes and 4 possible phenotypes.

12. Dihybrid cross An example is Mendel’s peas. Peas can have round (R) or wrinkled (r) seeds and they can be either yellow (G) or green (g) in colour. Which characteristics are dominant? Complete a genetic cross to show wrinkled green seed (rrgg) with round yellow seeds (RRGG).

13. Dihybrid cross Now complete a genetic cross to show the possible offspring (F 2 ) if two of the original round yellow seeds from F 1 generation are crossed.

14. Predicting Phenotypic Ratios Look at the genetic crosses you have completed and try to complete the table: Type of crossParentsPhenotypic ratio in F 1 Phenotypic ratio in F 2 Monohybridhomozygous dominant x homozygous recessive (NN x nn) All heterozygous offspring (Nn) Dihybridhomozygous dominant x homozygous recessive (NNGG x nngg) All heterozygous offspring (NnGg) Codominanthomozygous for 1 allele x homozygous for the other (H N H N x H S H S ) All heterozygous offspring (H N H S )

15. Activity 18: Genetic crosses using Drosophila The male abdomen has a rounded end with a distinctive black tip where the black stripes are very close together. The female has a broad abdomen with a pointed end and the black stripes are well separated so the tip is not black.

16. Activity 18: Genetic crosses using Drosophila Purebred variety will have normal wings Mutant variety will have vestigial wings

17. Activity 18: Genetic crosses using Drosophila Purebred variety will be wild type and have stripy bodies Mutant variety will have ebony bodies

18. Epistatic Genes Sometimes you do not get predicted ratios and this is probably due to epistasis. Epistatic genes are able to mask the effects of other genes. This is usually because 2 genes code for 2 enzymes that control the same metabolic pathway. An example is coat colour and patterning in mammals. This is very complex and can involve as many as 5 interacting genes, genes controlling melanin, genes controlling hair colour etc In an albino mammal the homozygous recessive gene, masks all other genes for colour and patterning.

19. Epistatic Genes An example in humans is the ‘widow’s peak’. This is controlled by one gene and baldness is controlled by another. If a person has the alleles for baldness it doesn’t matter if they also have the alleles for widow’s peak as they will have no hair! The baldness gene is epistatic to the widow’s peak gene.

20. Recessive Epistatic genes Flower pigment is controlled by 2 genes: One gene codes for yellow pigment (Y is the dominant yellow allele). The other gene codes for an enzyme that turns the yellow pigment orange (R is the dominant orange allele) If the flower does not have the Y allele it will not matter if it has the R allele or not, it will be colourless. This is an example of a recessive epistatic allele.

21. Predicting phenotypic ratios for recessive epistatic genes An F 1 cross of flowers YYRR x yyrr will produce all YyRr F 2 offspring. Now cross two of the F 2 offspring and predict the phenotypic ratio.

22. Dominant Epistatic genes Squash colour is controlled by 2 genes: The colour epistatic gene (W/w) and the yellow gene (Y/y). The no-colour white allele (W) is dominant over the coloured allele (w). Therefore WW or Ww will be white and ww will be coloured. The yellow gene has the dominant yellow allele (Y) and the recessive green allele (y). If the plant has at least one W the squash will be white, masking the yellow gene.

23. Predicting phenotypic ratios for dominant epistatic genes An F 1 cross of squash WWYY x wwyy will produce all WwYy F 2 offspring. Now cross two of the F 2 offspring and predict the phenotypic ratio.

24. Autosomal Linkage This refers to 2 or more genes on the same chromosome. These are normally inherited together, as they are not separated during meiosis. Example: in pea plants the gene for colour of seeds (yellow & green) and the gene for flower colour (coloured & white) are both found on chromosome no. 1.

25. The chi-squared (  2 ) In a dihybrid cross we would predict a phenotypic ratio of 9:3:3:1. However we would not necessarily get these results as fertilisation is a random act. We can use a statistical test, the chi-squared (  2 ), to ensure that numbers are offspring are close enough to the expected ratios and to check that nothing unexpected is going on. The test allows us to compare our observed results with the expected results, and decide whether or not there is a statistical difference between the them.

26. The chi-squared (  2 ) 1.Work out expected results E. 2.Record the observed results O. 3.Calculate the difference between them O – E. 4.Square the result of the difference (O – E) 2. 5.Then divide each squared difference by the expected value (O – E) 2 E 6.Then add up all of these answers. The best way to do this is in a table:

27. The chi-squared (  2 ) Mendel’s peas again.... 144 of them! Round yellow seeds Round green seeds Wrinkled yellow seeds Wrinkled green seeds Observed number O 8626248 Expected number E 8127 9 O – E (O – E) 2 E Σ (O – E) 2 E 22

28. The chi-squared (  2 ) Now we have our chi-squared  2 value we can relate this to the probability that the differences between our expected and observed results are down to chance. A probability of 0.05 means we would expect these differences to occur in 5 out of every 100 experiments, or 1 in 20 just by chance. Q: so what does a probability of 0.01 mean?

29. The chi-squared (  2 ) For biological data a probability of 0.05 is the critical measurement. If the probability is 0.05 or larger then we assume the differences are due to chance, not significant. If the probability is smaller than 0.05 then the difference is significant, and would have to consider what occurred during the cross.

30. Table of chi-squared (  2 ) values Degrees of freedom 0.10.050.010.001 12.713.846.6410.83 24.605.999.2113.82 36.257.8211.3416.27 47.789.4913.2818.46 The degrees of freedom takes into account the number of comparisons made. The larger the number of observed results and expected values, the larger  2 is likely to be and we need to compensate for this. To calculate degrees of freedom calculate: number of classes of data – 1 So for our experiment it would be 4 – 1 = 3 Our  2 = 0.79 and at 3 degrees of freedom is well below the 7.82 value for 0.05 probability on the table. This means our differences were due to chance and not significant.

31. The chi-squared (  2 ) In an exam the formula for the chi-squared (  2 ) will be given to you, but you must be able to calculate it!

32. The chi-squared (  2 ) Calculate the chi-squared (  2 ) for the following genetic cross, is the difference significant? In a tomato experiment the predicted and observed ratios were: Purple stem - jagged leaf Purple stem – smooth leaf Green stem – jagged leaf Green stem – smooth leaf Observed number O 12980 Expected number E 9331