The probability that individual VI-1 would have this disease isa..pdf

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The probability that individual VI-1 would have this disease is: a. 1/4 b. 1/9 c. 1/18 d. 1/24 e. 1/36 f. 1/64 ODFOODOHO L11110 Solution Answer is option f. 1/64 .There are 6 generations and circle represents female whereas square represents male. The horizontal line between them represents mating whereas vertical line represents children/offspring. The double line represents consanguineous (related by blood) marriages. So the coloured squares/circles represents diseased/affected individuals. The diseased persons are found in the fourth generation and are offsprings of a consanguineous marriage. Since their parents are unaffected they are unaffected carriers, and the recurrence risk of the disease for each future child is 25%, while the risk of a child becoming a carrier like the parents is 50%. The fourth generation unaffected individuals will have probability of getting disease of 1/4. The fifth generation children mate with other individuals and will carry the risk of inheritance of disease probability as 1/4 x 1/4 = 1/16. Again in the fifth generation there is a consanguineous marriage and the VI 1 individual is found unaffected. So his probability of getting disease is 1/16 x 1/4 = 1/64.

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Please, answers all my question is for genetic class 3. Describe the pattern of inheritance shown in the pedigree below, giving evidence to support your answer. Give the genotypes for individuals II-2, II-4 and II-2. Include a legend for the symbols used. (10 marks) Describe the pattern of inheritance shown in the pedigree below, giving evidence to support your answer. Give the genotypes for individuals II-4, II-7 and III-7. Include a legend for the symbols used. ( 10 marks) 5. A type of hemophilia is due to an X-linked recessive allele. An affected man whose phenotype has been successfully managed now has children with a woman who is a carrier. Providing a legend for your, answer the following questions: a. What are the genotypes of the mother and father? (1 mark) b. What are the probabilities that i. a child will be affected with hemophilia. ii. a child is a carrier. iii. their daughter is a carrier. iv. their male fetus will be affected. (4 marks) 6. Genes for two different recessive disorders are linked on the X chromosome; they are 26 map units apart. A woman is a carrier for both disorder A and disorder B , with the alleles for these genes in cis. Her husband has normal alleles for both of these genes. The woman is pregnant with their first son. What is the probability (in percent) this son is affected with both disorders? Explain, showing the proportions and types of the woman's gametes and identifying them as parental or recombinant. (5 marks) 7. In certain human matings of albino and black-haired parents, the offspring are always blackhaired. When these black-haired offspring of the different families mate with each other (non-consanguineously) and their children are pooled to get significant data, these are the children's phenotypic proportions: 9 black-haired children 3 brown-haired children 4 albino children a. How many genes are involved in this trait? b. How many alleles of each gene are there? c. What type of gene interaction is occurring for this trait? Describe it. d. What are the genotypes of the initial albino and black-haired parents? Provide a legend. e. What are the genotypes of their offspring (all black-haired)? f. What are the genotypes and the specific proportions of the albino children in the third generation? g. What are the genotypes of the brown-haired children? 8. Which type of chromosomal aneuploidy typically has less severe effects on an individual: sex chromosome aneuploidy or autosomal aneuploidy? Explain why. (4 marks) 9. An organism has six chromosomes in its somatic cells. Determine the number of gametes with different chromosome compositions that this organism can potentially create. Explain your answer, including the genetic basis for your calculation. Ignore the allelic variation created by crossing over. ( 4 marks) 10. Compare the inheritance and transmission of Xlinked genes with that of mitochondrial genes. (4 marks) 11. A woman has a normal number of fingers and toes. She has many relatives with ex.

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Consider the following pedigree, which traces the inheritance of a single-gene hereditary disease. Only individuals that are either affected or normal are shown. In other words, potential heterozygotes are NOT indicated. Characterize each of the following modes of inheritance as: impossible, unlikely, or probable. Justify your answers. Autosomal dominant Autosomal recessive X-linked dominant X-Iinked recessive Y linked Solution Single gene herediary diseases are mainly caused by a change of any one gene in a particular DNA.Its divided into diiferemt categories like dominant ,recessive and X linked.Square indicates a male and circle female.According to the diagram affected male is crossed with an unaffected female.The progeny obtained is affected daughters and unaffected sons. Autosomal dominant: in this case, the father — has a 50 percent chance of having an affected child with one mutated gene (dominant gene) and a 50 percent chance of having an unaffected child with two normal genes (recessive genes).But, in our case none of the sons are affected which means its criss cross inheritance that is from father to daughter.Hence , this case is unlikely Autosomal recessive:To have an autosomal recessive disorder, you inherit two mutated genes, one from each parent. These disorders are usually passed on by two carriers.The cross given in the question is not between the two carriers.The male parent is affected for sure hence this case is impossible X linked dominant:As in autosomal dominant inheritance, only one copy of a disease allele on the X chromosome is required for an individual to be susceptible to an X-linked dominant disease.Both males and females can be affected, although males may be more severely affected because they only carry one copy of genes found on the X chromosome. Some X-linked dominant disorders are lethal in males.When a female is affected, each pregnancy will have a one in two (50%) chance for the offspring to inherit the disease allele. When a male is affected, all his daughters will be affected, but none of his sons will be affected.So, there are probable chance for this type of inheritance X linked recessive:For X-linked recessive disorders, an affected father who has a mutation in a gene on the X chromosome can transmit either the X chromosome with this mutation or a Y chromosome to his children. If the mother is not affected or a carrier, none of his sons will be affected since they can only inherit a normal X chromosome from their mother and they inherit a Y chromosome from their father. Each daughter will have a 50% chance of being an unaffected carrier and a 50% chance of both X chromosomes being normal.In our cross all daughters are affected which means this type of inheritance is impossible Y linked:This type of inheritance is impossible in this case as all female children are getting affected.If it would have been Y linked then only sons should have been affected as Y chromosome is present only in males.Hence this.

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If the F1 plants with red flowers were crossed to each other, what progeny would be produced and in what proportions? On occasion, genetically identical humans are produced by a natural cloning process. The result is identical, or monozygotic, twins. These individuals have all 23 pairs of chromosomes identical to those of their twin. Fraternal twins are produced when two eggs are ovulated and fertilized simultaneously, but independently. What is the likelihood that two fraternal twins will be genetically identical? On average, what proportion of their alleles do siblings have in common with each other? On average, what is the probability that any particular X-linked allele carried by a man is also carried by his brother? On average, what proportion of their alleles do children have in common with a parent? You work as a genetics consultant for a local attorney. She is currently working on five cases of disputed paternity. Tests have produced the following information on blood types: Solution 2A. Fraternal twins are formed from the fertilization of two different ovules by two different sperms and are hence also known as \'dizygotic\'. This implies that there is hardly any likelihood of being any more identical than two siblings, which is 50 percent. However, they can be formed either due to the presence of hyperovulation gene in the mother, which makes the mother release more than one egg at a time, or by multiple fathers within a span of few days. In the first case, the likehood of being identical is 50 percent or more, whereas in the second case, it\'s only the mother\'s gene which is similar in the twins. 2B. Each sibling gets half an allele from each parent, so 50% is all they get from the parents. So one extreme can be each sibling getting the alleles from parents and which the other didn\'t get, while the other extreme can be each sibling getting exactly the same allele. Hence the probability ranges from 0 (no similarity) to 1 (exactly same). So on a average, the proportion of alleles they share is (0+1)/2 = 0.5. 2C. For a man and his brother, the X linked genes can only come from their mother since they will recieve Y gene from their father. Hence, the likelihood is 50%. However, to get the same X- linked genes from mother, the likelihood reduces to 25% (Please refer to the table below). X1-M, X-F (25%) (Daughter) X2-M, X-F (25%) (Daughter) X1-M, Y-F (25%) (Son) X2-M, Y-F (25%) (Son) 2D. Since each child receives only half allele from each parent, on an average the have only 50% common with a parent.MotherMotherX1-MX2-MFatherX-F X1-M, X-F (25%) (Daughter) X2-M, X-F (25%) (Daughter)FatherY-F X1-M, Y-F (25%) (Son) X2-M, Y-F (25%) (Son).

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5. Use the following pedigree to determine the probability that IV-1 (child yet to be born) will have albinism. = has cystic fibrosis = = consanguineous marriage (cousins) 6. Using the following pedigrees, mark the genotype of each member and determine the mode of inheritance: B) C) I II III IV 1 2 0. Achondroplasia is an autosomal dominant form of dwarfism. People with achondroplasia are heterozygotes because carrying two copies of the dominant, mutant allele is fatal during embryogenesis. Two people with achondroplasia decide to start a family. A) Fill out a Punnett square for this couple. A) What is the probability that their first child born will be of average stature? B) What is the probability that this couple has two daughters with achondroplasia and one son of average stature in that order? C) What is the probability that this couple has two children with achondroplasia and one of average stature in any order? .

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Study Guide Chapter 15 -Chromosomal Basis of Inheritance-Answers 15.1 Mendelian Inheritance and chromosome theory______________________________ 1. Thomas Hunt Morgan identified the first solid evidence associating a specific___gene_____ on a specific chromosome. 2. Why did Morgan choose Drosophila as his experimental organism? (List 3 reasons) They reproduce quickly, a new generation of adults forms every two weeks! Prolific (single matting = hundreds of offspring) Only 4 chromosome pairs = simple genetics 3A. The normal phenotype for a character (phenotype most common in nature) is called___wild type_____. 3B. Provide at least two examples this phenotype in Drosophila. Red eyes gray body 4A. An alternative phenotype for a character (phenotype not common in nature) is called____mutant____. 4B. Provide at least two examples of this phenotype in Drosophila. White eyes Black body 5. Morgan and his students invented a notation to symbolize alleles in Drosophila that differed from the notation Mendel used to represent alleles. Describe how Morgan’s Drosophila characters are named and the symbol used for the allele type. You can think of Morgan’s wild-type allele as equivalent to the dominant allele in Mendel’s naming system. And you can think of Morgan’s mutant allele as equivalent to Mendel’s recessive allele. So whenever you have one wild-type allele and one mutant allele, that organism will have the wild-type phenotype. Mendel Used the first letter of the dominant character name to represent the dominant allele. This letter was capitalized for the dominant allele, lower case for the recessive allele Ex: purple flower allele dominant to white flower allele: P= dominant allele, p=recessive allele Morgan -Used the first letter of the mutant character name to represent the wild-type allele. The letter for the wild-type and mutant alleles are both lower case. The wild-type allele gets a + sign and the mutant does not. Ex: red eye allele wild-type to white eye mutant: w+ = wild-type allele, w= recessive allele. 6. Morgan and his students invented a notation to symbolize alleles in Drosophila. Which of the following genotypes would produce a fly that is wild-type for eye color (red vs. white eyes)? w+ w+ w+ w w w w+w+ and w+w 7. Morgan performed an experiment that yielded a 3:1 ratio of offspring in the F2 generation; however, only the males of this F2 generation had white eyes. White- eye females DO EXIST, so why were only white-eye males observed in this cross? Complete the following chart and describe why Morgan observed these results and how it allowed him to conclusively determine that the gene for eye color was located on the X chromosome. Morgan’s experiment (see Figure 15.4) Use notation developed by Morgan and his students If the gene for eye color is located on the X chromosome…. P generation Red eye Female genotype for eye color = w+w+ White eye Male genotype for eye color= w F1 generatio.

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