1) (15 point question) Panel A shows the initial crosses and representative chro
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Question
1) (15 point question) Panel A shows the initial crosses and representative chromosomes of animals in a QTL mapping project. The black regions represent chromosome material from the largest purebred line and the white regions represent the other pure bred line used in the experiment. Panels B, C, D and E show the karyotypes representative of the smallest and darkest animals from populations of mice that were created for QTL mapping.
A) (2 points) Panels B to D representative chromosomes from some animals used for a QTL mapping experiment. Which panels are from recombinant inbred lines. Explain how you recognise these animals.
B) (2 points) Assuming that the genes for the size and color traits are equal and additive without significant linkage. What will be the phenotype of the F1 mice? Explain your reasoning.
Pure line A Pure line B RR RR BRExplanation / Answer
A. Panels B to D represent recombinant in bred lines with different scores for each parameter (i.e color and size). The chromosome from 1 strain when crossed with the other strain will produce recombinants based on the level of penetrance of that gene into the crossed breed. Apart from that,the phenotypic trait is a product of two or more genes along with the environment (polygenic in nature). Different levels of QTLs score will be given for each chromosome. Based on the final QTL score, one can apporximately tell the necessary phenotypic traits for these recombinants. The recombinants will have both black and white colored patch (in pictorial representation) in varying levels , each color representing the phenotypic trait. Panels show that these genes are additive in nature and quantitatively as well as physically linked in the chromsome
B. If the size and color traits of the genes are equal and additive, the phenotype will contain equal number of color and size based traits. For eg: If 1 crosses strain 1 to strain 2 and they have 6 babies (called pinky), each of the 6 babies will have 100% of phenotypic trait of strain 1 or of strain 2 as there is no significant linkage. Thus there are chances of equal number of strain 1 and strain 2 in f1 progeny as these genes will be passed onto individually. Also there is a chance of both genes coming into the chromosome provided both of them are in sepearate chromsomes. These all are chance and probabilistic events.
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