For a single Mendelian trait, a homozygous dominant individual is cr.pdf

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Answer the following questions: 1. Given the following matings, what are the predicted genotypes and genotypic ratios of the offspring? a) Aa X aa b) Aa X Aa c) AA X Aa 2. Wet ear wax (W) is dominant over dry ear wax (w) a) A 3:1 phenotypic ratio of F1 progeny indicates that the parents are of what genotype? b) A 1:1 phenotypic ratio of F1 progeny indicates that the parents are of what genotype? 3. A woman is heterozygous for two genes. How many different types of gametes can she produce, and in what proportions? 4. The following diagram shows a hypothetical diploid cell: The recessing allele for albinism is represented by a, and d represents the recessive allele for deafness. The normal alleles are represented by A and D, respectively. a) According to the principle of segregation, what is segregating in these cells? b) According to Mendel’s principle of independent assortment, what is independently assorting in these cells? c) How many chromatids are in this cell? d) Write a genotype of the individual from which this cell was taken. e) What is the phenotype of this individual? f) What stage of cell division (mitosis or meiosis) is represented by this cell? g) After meiosis is complete, how many chromatids and chromosomes will be present in one of the four progeny cells? 5. Draw the following simple pedigree. A man and a woman have three children: a daughter, then two sons. The daughter marries and has monozygotic twin girls (identical twins). The youngest son in generation II is affected with albinism. (Use proper pedigree symbols; draw, scan and add picture to document; do not submit picture independently). 6. A man with blood type A and a women with blood type B have three children. A daughter with AB and two sons, one with type B and one with type O blood. What are the genotypes of the parents? ...

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Hypothesis What is your hypothesis for the genetic characteristics (e.g. dominant/recessive, autosomal/sex-linked, etc.) of the traits you have chosen to study? What observations lead you to make this hypothesis? What phenotypes and phenotypic ratio would you predict for the offspring (the first generation, or F1 generation) of this cross? What genotypes and genotypic ratio would you predict? How did you come to this prediction? Write your explanation as a hypothesis to the experiment you are doing. Enter your answers here. Solution Hypothesis: A cross between pure breeding dominant and recessive genotypes obtain all dominant phenotypes in the F1 generation. Presence of criss cross inheritance confirms the sex linkage of the trait while same proportion of affected son and daughter is the indicator of autosomal inheritance. Dominant trait is expressed in both homozygous and heterozygous genotype while the recessive trait is expressed in homozygous genotype only. Sex linked traits exhibit criss cross inheritance (father to daughter and mother to son). The cross( for dominant/recesive)= TT x tt= Tt Phenotype- all tall Genotype= all Heterozygous (Tt) The cross (for autosomal/sex linked)= XcY (affected father) x XX (normal mother)= XcX (normal but carrier daughters): XY (normal sons)..

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Biology 103 Laboratory Exercise – Genetic Problems Introduction Although the science of genetics has become a highly sophisticated discipline dealing with the interactions of hereditary factors at the molecular level, it has its roots in the basic laws of heredity initially discovered and presented by Gregor Mendel more than one hundred years ago. Mendel's success in discovering these laws was due largely to his application of the simple rules of mathematical probability - the laws of chance - to his observations concerning the inheritance of certain characteristics in the garden pea plant. Reginald Punnett and the Punnett Square The Punnett square is a diagram used by biologists to determine genotypic probability within the offspring from a particular genetic cross. The Punnett square shows every possible genotypic combination of maternal alleles with the paternal alleles for a genetic cross. Punnett squares only give probabilities for genotypes, not phenotypes. The square diagram was designed by the British geneticist, Reginald Punnett (1865-1967) and first presented to the science community in 1905. Punnett’s Mendelism (1905) is considered the first popular science book to introduce genetics to the public. Solving Genetic Problems R R' R RR RR' R' RR' R'R' Maternal alleles A A a Aa Aa Paternal Alleles a Aa Aa The first step in solving a genetic problem is to establish the genetic symbols you will use in your problem solution. Stay consistent by using these same symbols throughout the problem solving process. Represent dominant and recessive alleles (different forms of a gene) using traditional genetic symbols. Dominant alleles should be represented with the capital version of an alphabetic letter while using the lower case version to show recessiveness. For example: B = black color, b = white color. Each individual gene or trait is diploid (2n) in nature and therefore, must be represented with two alleles. Continuing with the alleles mentioned previously, an individual may have the genetic makeup BB, Bb, or bb when using those alleles. Remember that gametes (sperm and egg) are haploid (n) and can only provide one allele per trait. For example: B or b An individual’s genotype contains the possible gametes that can be expected to be produced by that individual. Much of genetics revolves around the probability of the makeup of gametes. If the individual is homozygous, all of the gametes produced will possess the same kind of allele. For example, an individual with the genotype BB would be expected to produce only B gametes and individuals with genotype bb would produce only b gametes. If the individual is heterozygous, that is the individual’s genotype contains one dominant allele and one recessive allele (Bb), the gametes produced will possess one or the other of the two forms of the gene – B or b. ...

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12. Analyzing Coat Color Mutants - Wild type gerbils from Arizona, Montana, indiana and Califomia are black. - You have obtained pure-breeding mutant varieties of gerbils that are white or pink from these different places. - You know that each gerbil strain has a mutation in a single autosomal gene. - The table below shows E1 proseny obtained when crossing purebreeding WT and mutants to each other. - Bolded boxes at the top and side indicate the purebreeding parent strains you crossed to each other. - The intersection of these boxes represents the phenotype of the F 1 offspring of a particular cross. For each cross, you can assume 100% of the F1 progeny are of the color indicated. a) What is the purpose of crossing the mutant to wild type (black) and observing F1 progeny phenotypes? b) What is the purpose of crossing the mutants to each other and observing F1 progeny phenotypes? c) Which of the strains cannot be used for the analysis you described in (b)? Why? d) When you cross pure-breeding white from Arizona and indiana to each other, you get black in the F 1 . What does this indicate about the mutations that give rise to white in these 2 strains? e) To further explain your answer in (d), assign genotypes for the cross between the 2 purebreeding strains from Indiana and Arizona and their resulting F 1 . Use the letter A if only alleles of one gene are segregating in the cross, letters A and B if 2 genes are involved, etc. Denote dominant alleles with a capital letter and recessive alleles with a small letter. .

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If the characteristic followed in the pedigree is x-linked recessive allele, what is III-1\'s genotype? a. Hemizygous for a dominant allele b. Hemizygous for a recessive allele c. Definitely heterozygous d. dominant e. Either heterozygous or homozygous dominant Solution 1st generation parents 1,2 though are normal but must be heterozygotes (Pp) since their children are affected i.e. 2nd generation males 1,3. Since 1st generation male 3 is affected (pp) and 4 is normal but regardless if 4 in 1st generation is PP or Pp their children will be either 4 Pp i.e. heterozygote or 2 affected pp and 2 heterozygotes Pp. But the 2nd generation 4, 5 female are normal but heterozygotes Pp. So the cross between a 2nd generation affected male 3(pp) and a normal but heterozygote female 4(Pp) would have a normal male child only if hemizygous for a dominant allele condition exists. Hemizygous is a condition in which only one gene copy/DNA sequence exists in diploid cells. Males are hemizygous for most genes on sex chromosomes since they have only one X and one Y chromosome unlike two X for females. A recessive X-linked gene requires two mutant alleles to affect the female phenotype. Rules followed here are.

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