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Dominance Inheritance Review Incomplete Inheritance Co-dominance Inheritance SBI3U.

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Presentation on theme: "Dominance Inheritance Review Incomplete Inheritance Co-dominance Inheritance SBI3U."— Presentation transcript:

1 Dominance Inheritance Review Incomplete Inheritance Co-dominance Inheritance SBI3U

2 When one (1) allele is stronger (so dominant) than the other allele in heterozygous genotype, resulting in phenotype of the strong (dominant) Examples: a) For hair texture trait  Curly hair is dominant (C)  Straight hair is recessive (c)  Therefore, Cc genotype translates to Curly hair phenotype b) For dimple trait  Having dimple is dominant (D)  Not having dimple is recessive (d)  Therefore, Dd genotype translates to having dimple phenotype c) For hitchhiker thumb trait  Having the hitchhiker thumb is dominant (_____)  Not having the hitchhiker thumb is _______ (___)  Therefore, ______ genotype translates to _______

3 Sample problem: Dwarfism, is a dominant trait. If both parents with dwarfism phenotype have normal looking offspring, a) determine the genotype of the parents b) the percentage of their child looking normal c) the percentage of their grandchildren from the normal offsprings looking dwarfish given that their life partners are also normal Answers: a) P: Dd x Dd (because if either one of the parents has DD, none of their children will look normal) DD = Dd = dd = 142414142414 = 25% (dwarfism) = 50% (dwarfism) = 25% (normal) c) Zero percent (0%)  F 1 = dd x dd b) D d D DD Dd d Dd dd

4 Sample problem: Breast cancer has been tracked to associate with genes BRCA1 & BRCA2, which are tumour suppressors and function to inhibit the growth of tumour (cell cycle genes) It is a recessive trait. That means, if the gene is negatively mutated and the offsprings carry both mutated alleles, she is more likely to get breast cancer. a) Determine the phenotype of the genotype variants for this trait b) If the mother is heterozygous for the trait and the father has normal alleles, how many of their 6 children will i) be carrier of the cancer trait (heterozygous) ii) be homozygous for normal BRCA1 iii) develop breast cancer Answers: a) BRCA1 + BRCA1  normal; least risk for cancer BRCA1 + brca1  carrier of the cancer trait brca1 + brca1  develop cancer b) i) 6 ii) 6 24242424 = 3 children iii) zero BRCA1 BRCA1 brca1 BRCA1 BRCA1 brca1 BRCA1 brca1

5 Also known as Blending Inheritance When two (2) alleles are equally dominant (same strength), they interact to produce a new, third, phenotype  This means, the heterozygous genotype has its own phenotype  Three (3) phenotypes are produced from such cross Although Mendel did not observe such pattern of inheritance with peas, there are many examples found in nature. When the heterozygous genotype produces a different/new phenotype

6 A red hibiscus is crossed with a white hibiscus What are their genotypes? P: (H R H R ) x (H W H W ) F 1 : H R H R = ¼ = 25%  Red H R H W = ½ = 50%  Pink H W H W = ¼ = 25%  White F 2 : Predict the phenotype ratio for self-pollinated hybrid hibiscus.

7 In guinea pigs, colour of coat is determined by at least three alleles.  Yellow coat is determined by the homozygous genotype YY,  White by the homozygous genotype WW, and  Cream by the heterozygous genotype YW. Determine the expected genotype and phenotype ratio of the F1 generation which would result from a cross between: a) two cream coloured guinea pigs; b) a yellow coated and a cream coated animal Your Turn:

8 … Solutions Let G Y G Y represent yellow coat Let G W G W represent white coat Let G Y G W represent cream coat a) P generation phenotypes: genotypes: cream x cream G Y G W x G Y G W GYGY GYGY GWGW G Y G Y G Y G W G W G Y G W G W G W F 1 : genotypes G Y G Y = ¼ = 25% (yellow) G Y G W = ½ = 50% (cream) G W G W = ¼ = 25% (white) Therefore, the expected genotypic ratio is: 1 : 2 : 1 the expected phenotypic ratio is: 1 : 2 : 1 b) P generation phenotypes: genotypes: cream x yellow G Y G W x G Y G Y GYGY GYGY GYGY G Y G Y G W G Y G W G Y G W F 1 : genotypes G Y G Y = ½ = 50% (yellow) G Y G W = ½ = 50% (cream) Therefore, the expected genotypic ratio is: 1 : 1 the expected phenotypic ratio is: 1 : 1

9 1. Howdy! My name is Bob Howard, and I own 20 purebred red cows. Something strange happened several months ago. During a violent storm, all of the fences that separate my cattle from my neighbors cattle blew down. During the time that the fences were down, three bulls, one from each neighbor, had access to my cows. For awhile, I thought that none of the bulls found my cows, but over the months, I have come to the conclusion that all of my cows are expecting calves. One of the bulls is the father. Which bull is it? A local college professor told me to use a little genetics detective work to figure out who the father is. He told me to collect information about each of the bulls, and to read articles about genetics and Gregor Mendel's experiments in genetics. So, I did exactly what he said. I compiled the information. Now, I need your help to make sense of the data and to figure out who the father is. After reading through the information, maybe you can tell me why my red cows had 9 roan calves and 11 red calves. I don’t really understand how this happened. When you have determined which bull is the father, please tell me the answer.

10 This is Rocky. He is a 2,200 pound Red bull. The colour of Rocky&’s calves, if mated with a red cow, can be determined by using a Punnett square. His offspring will also be unique in colour compared to the other two bulls. This is Rufus. He is a 1,920 pound White bull. The colour of Rufus’ calves can also be figured out, if he is mated with a red cow, by using a Punnett square. The colour of his calves will also differ from the other two bulls offspring. This is Ferdinand. He is 2,000 pound Roan bull. The laws of genetics tell us that the offspring he produces will probably be different, in colour, than the other two bulls’ offspring. Using a Punnett square, you can see the gene combinations, for colour, that Ferdinand’s offspring could have if he mates with a red cow. The colour of Ferdinand’s calves has to do with probability. Who did it? … a bonus mark for first correct answer posted on Wiki discussion

11 Co-dominanceCo-dominance

12 When the dominancy (strength) of the allele changes unpredictably  That is, sometimes one allele is dominant; other times it is not  This results in the re-appearance of the both parental phenotypes which causes the offspring phenotype to look patchy! The genotypes are not blended and they still obey Mendel’s law of segregation, resulting in a mixture of the phenotype. Haemoglobin The gene for haemoglobin Hb has two co-dominant alleles: i) Hb A (the normal gene) ii) Hb S (the mutated gene) Heterozygous: Hb A Hb S is good for protecting from Malaria normal red blood cell sickled red blood cell

13 Cat coat: orange (OO) x black (oo) = Tortoiseshell maternal ‘O’ allele paternal ‘o’ allele mosaic adult

14 Phenotype Genotype PictureAntibodiesNotes Type A Type B I A I A or I A i I B I B or I B i B – Antibodies A - Antibodies Type AB IAIBIAIB No antibodiesUniversal Recipient Type Oii A – Antibodies B – Antibodies Universal Donor

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