¡Descarga Genetics Exam: Blood Group Probabilities and Family Tree Analysis - Prof. Leiva y más Exámenes en PDF de Genética solo en Docsity! General Genetics 2017/18 First Term Exam Page 1 of 6 FIRST TERM EXAM GENERAL GENETICS NAME: GROUP: 1. A woman with blood group B has several children with the same man. Indicate and justify in each case both parents most likely genotype and the probabilities of each possible genotype in the progeny and the probability of this family in particular. Case 1: 2 sons of group A; 1 son of group B, 1 son of group AB and 1 son of group O Case 2: 10 sons of group AB Case 3: 5 sons of group B and 7 sons of group AB Case 4 6 sons of group A and 1 son of group AB 4 points over 10 It is unknown if the parents are homozygous or heterozygous, so we have to take into account all the possibilities: Mother homozygous IBIB o heterozygous IBi In case 1 There are descendants of all blood groups so the only possibility is that the mother is heterozygous form group B and father heterozygosity for Group A. IBi x IAi So the offspring would be: ¼ IBi + ¼ IAi + 1/4 ii + ¼ IAIB 25% of each type The probability of that is this family will be: 5! 2! 1! 1! 1! 1 4 1 4 1 4 1 4 In case 2 Given the very high number of children of the same blood group the most likely is that both the mother and the father are homozygous (the father of Group A) and all the progeny is the same group AB. IBIB x IAIA So the offspring would be: 100% IAIB The probability of that is this family, if you have 10 children, will be 1 since there is no more possible phenotypes. In case 3 The mother has to be homozygous and the father can be two possible genotypes (IAi ó IAIB). Both options would give the same proportion of blood groups in the offspring, although not the same genotypes: IBIB x IAi ½ IAIB + ½ IBi IBIB x IAIB ½ IAIB + ½ IBIB The probability of that is this family will be: 12! 5! 7! 1 2 1 2 It could also be the case that the genotypes were: IBi x IAIB ¼ IAIB + ¼ IBIB + ¼ IBi + ¼ IAi If that was the case, the following son might be from Group A, but the probability that with 12 children none was from the Group A is 0.15%, which makes it quite unlikely. In case 4 It is likely that the mother will be heterozygous (IBi) and the father homozygous (IAIA) IBi x IAIA ½ IAIB + ½ IAi The probability of that is this family will be: General Genetics 2017/18 First Term Exam Page 2 of 6 7! 6! 1! 1 2 1 2 Could also be the case that the father was AB and genotypes were: IBi x IAIB ¼ IAIB + ¼ IBIB + ¼ IBi + ¼ IAi If it were the case, the next child could be Group B, but the likelihood that 7 children none was from the Group B is 0.04%, which makes it quite unlikely. General Genetics 2017/18 First Term Exam Page 5 of 6 3. A series of crosses were performed in rabbits of different coloured fur. Homozygous Brown rabbits crossed with albino rabbits homozygous giving offspring all black. This descent was allowed it to cross each other and showed the next generation: 380 black rabbits, 135 Brown rabbits and 185 albino rabbits a) issue a hypothesis about the mode of inheritance of coat colour and contrast it with the test chi square. b) explicitly indicate the mode of inheritance of coat colour. c) indicate the genotype of the parent of the crosses made. Data: χ2 with 1 degree freedom at 95%= 3.84 χ2 with 2 degree freedom at 95%= 5.99 χ2 with 3 degree freedom at 95%= 7.82 χ2 with 4 degree freedom at 95%= 9.49 3 points over 10 The most likely hypothesis is that the colour is controlled by two independent loci with two alleles each: the first gene (I, i) controls if the rabbit has or no colour and the second gene (B, b) controls the colour of the fur (Brown and black). In the first gene would have a relationship of dominance I > i, if rabbits are homozygous ii then rabbits will be albino (a possible biochemical explanation would be that it is not synthesized a necessary compound that will synthesize the pigment that gives colour to the hair), if the rabbit is dominant homozygote or heterozygote then the coat may have a colour that will determine the second gene. The second gene relationship between alleles is dominance B > b rabbits will be black if they are BB or Bb and Brown if they are bb. Cross will be: Therefore, at the first cross, we would expect that all the descendants were black (which it is). And in the second cross appear all the possible genotypes giving a segregation 9 black 3 Brown and 4 albino (a simple recessive epistasis). Phenotype Genotype Rate Observed Expected Black I_B_ 9 380 393,75 Brown I_bb 3 135 131,25 Albino iiB_ 4 185 175 16 700 700 IIbb iiBB x iB Ib IiBb IiBb x IiBb Cross 1 Cross 2 9 I_B_ negros 3 I_bb marrones 3 iiB_ albinos 1 iibb albinos General Genetics 2017/18 First Term Exam Page 6 of 6 380 393,75 393,75 135 131,25 131,25 185 175 175 0,48 0,11 0,57 1,16 The value of the chi test since is lower than the limit value for two degrees of freedom (5.99) we have no reason to reject the hypothesis. The mode of inheritance of the fur from the rabbit would be controlled by two genes or loci autosomal with two alleles each one of them: the first would determine the existence or not of colour (I_ gives colour and ii gives albinism) and the second gene type of colour featuring the rabbit (B_ colour black and bb brown colour) always if in the first locus is the allele I present. If not rabbits will be albino regardless of the genotype of the second locus. Both genes segregated independently. Parental rabbits will be: Brown rabbits IIbb and Albino rabbits iiBB.
