Answered: Based on the complementation data…

Based on the complementation data below,how many complementation groups are represented in this table? (+ = complementation,- = no complementation) mutation number 1 2 3 4 6 1 3 4 + 6 7

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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The DNA of every individual in the pedigree shown in image B (below) has been sequenced at the causative locus, all the non- shaded individuals are wild type apart from III.1 and III.6. III.1 and III.6 have both been proven to have the causative allele for the condition but they do not exhibit any of the phenotypic signs or symptoms. Based on this pedigree, what is the level of penetrance for the condition? Please give your answer as a percentage to one decimal place, give the number only, no percentage symbol. ANSWER: Given the information above I calculate the level of penetrance seen in image B to be Blank 1 percent. A KEY Homozygous Homozygous Heterozygous Heterozygous Wild Type Male Female Male Female Male Note: Completely red symbol denotes an individual exhibiting the phenotype of interest CI || III IV V 1/4 1/2 1/2 1/2 1/2 Wild Type Female 1/4 1/2 Affected Known carrier Affected female Normal female Affected male Normal male D ●●●
The DNA of every individual in the pedigree shown in image B (below) has been sequenced at the causative locus, all the non-shaded individuals are wild type apart from III.1 and III.6. III.1 and III.6 have both been proven to have the causative allele for the condition but they do not exhibit any of the phenotypic signs or symptoms. Based on this pedigree, what is the level of penetrance for the condition? Please give your answer as a percentage to one decimal place, give the number only, no percentage symbol. Given the information above I calculate the level of penetrance seen in image B to be "Blank" 1 percent.
Could someone please help me with this grade 11 bio dihybrid cross problem in detail and how to solve this question, using a strategy? Assume that curly hair (C) is dominant to straight hair. Albinism (P ) is recessive to normal skin pigmentation. A woman who is heterozygous for curly hair and albinism has a child. The father is homozygous dominant for curly hair and has albinism. (a) Determine the possible phenotypes for their child. (double-crossing, Puneet square)(b) Calculate the four different probabilities of a child beingboth a male and of each phenotype.(c) What is the probability that the child will expressalbinism and have curly hair like his father?
What genetic event does the DSB homologous recombination model account for that the Holiday model does not?
Determine the gene order (which gene is in the middle?), as you were shown in exercises for Tutorial 6, showing all your work. Construct a genetic map for the three genes, including the map distances between them. Clearly indicate the logic you followed and show all your calculations. Include the full distance calculations for the two most distanced genes (do not just add the other 2 distances).
Species I has 2 n = 16 chromosomes. How many chromosomes will be found per cell in the following mutants in this species? Q. Trisomic
Can you explain genetic map with two point cross? And show me a example with the cross
In a three-point mapping experiment for the genes a-w-ec, the following percentages of events are observed: NCO events: 57%; SCO events between a and w: 18%; SCO events between w and ec: 22%; DCO events: 2% Assuming that w is in the middle, what is the map distance between w and ec? What is the map distance between a and w? Draw the map.
Fill in the blanks: The distances within a ["pair of linked gene", linkage group ", "physical map"] can add up to more than 50cM, but the distance between a ["pair of linked gene","linkage group", "physical map"] must be less than 50cM.
Cystic fibrosis (CF) is an autosomal recessive trait. A three-generation pedigree is shown below for a family that carries the mutant allele for cystic fibrosis. Note that carriers are not colored in to allow you to figure out their genotypes. Normal allele = F CF mutant allele = f What is the genotype of individual #13? A) ff B) FF C) Ff D) it is impossible to tell
a) Are the genes linked? Please explain SPECIFICALLY how you can distinguish between linked and unlinked genes in this instance. b) If the two genes are linked, calculate the % recombination between ser and thr. Show the formula used, as well as all of your calculations. c) Draw a single map illustrating the arrangement of the two genes on the chromosome with respect to each other and to the centromere of the chromosome. Make sure to map ALL three distances.
Now, figure out the complementation groups. This shows how many genes are involved in this pathway. Some of the mutant classes in the previous problem must contain more than one complementation group (gene) For example class 2 is defined by mutants 1,3,4,6, and 7. These mutants may, or may not, all be in the same gene. Complementation testing will help you know if they are, or are not, in the same gene. So there are actually additional steps between some of the letters. For example, maybe instead of XàY, it is really XàFàGàY.