Homework Answers
recombination frequency= (number of recombinant progenies/total
number of progenies)
100
here the parental genotypes or the arrangement of alleles in the chromosome is not given, to determine which are the recombinant progenies.
( test cross is used to determine the distance between the genes)
if the genes are not recombination frequency is 50%, recombination frequency is the percentage of recombinant progenies if the genes are linked the recombination frequency is less than 50%, that is the percentage of recombinant progenies is less than that of parental type progenies.
here the number of red, oval and yellow, long are less than that of the other phenotypes.
recombination frequency= (( number of red, oval + number of
yellow, long)/ total number of progenies)100
= ((5+5)/100)100
= 10%
recombination frequency in percentage is the distance between the genes in cM. so the distance between the genes=10 cM
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I included everything here , but i only need help with parts
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Given the following data, what is the recombination frequency between these genes for squash color (red, yellow) and squash shape (log, oval)? What is the distance between the two genes in centiMorgans (or map units)? Phenotype Progeny from test cross 46 44 Red, long Yellow, oval Red, oval Yellow, long Total 5 100
4. Given the following data, what is the recombination frequency between these genes for squash color (red, yellow) and squash shape (log, oval)? What is the distance between the two genes in centiMorgans (or map units)? For full credit, you must show your work, i.e. how you made your calculations. (4 points) Phenotype Progeny from test cross 44 Red, long Yellow, oval Red, oval Yellow, long Total 100
Given the following data, what is the recombination frequency between these genes for squash color (red, yellow) and squash shape (log, oval)? What is the distance between the two genes in centiMorgans (or map units)? For full credit, you must show your work, i.e. how you made your calculations. (4 points) 3. Answer this question based on the following information. For full credit, you must clearly show how or explain how you came to this answer. Fur color in Labrador...
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2. Given the following data, what is the recombination frequency? Select an answer below and show how you made your calculaton. Phenotype Progeny from test cross 46 44 Red, long Yellow, oval Red, oval Yellow, long Total a) 45% b) 5% c) 10% d) 90% 3. The genes dumpy (dp), clot (cl), and apterous (ap) are linked on chromosome II of Drosophila. In a series of two-point mapping crosses, the following genetic...
I included everything here , but i only need help with parts
C-F
You are working with a hypothetical fly and have found color and wing mutants. Preliminary work indicates that the mutant traits are recessive and the associated genes are not sex-linked, but beyond that you have no information. You first look at 2 genes, each with two alleles: "Borb" for body color and "Wor w" for wing surface. The red body phenotype is dominant to the yellow body...
4. You are studying three genes in corn and performed a three-point cross. The three genes are green (V) vs Viniscent (V) seedlings, blue (Pr) vs red kernels (pr) and white (B ) or brown (bm) leaf midrib color. The table below list the offspring resulting from a testcross with a female F1 heterozygote. Type of gamete (NR, SCO, DCO Phenotype Progeny Gamete from F1 heterozygote Virescent seedling red 101 kernel Red kernel 175 Red kernel, brown midrib 259 Virescent...
6. (7 points) Dr. O. Sophila is a famous geneticist who studies recombination of genes in fruit flies. She is studying the relationship between a gene vermillion (v) that gives flies purple eyes and is recessive to the wildtype red eye color (vt) and the gene yellow (y) that gives flies a yellow body color compared to the dominant brown body color (y) Dr. Sophila has generated females heterozygous for the two different mutations and performs a testcross with mutant...
5. You started working in a Drospohila lab. You are studying two different autosomal traits, body color and antennae length. Gray body (B) is completely dominant to black body (b) and long antennae length (A) is completely dominant to short antennae length (a). You cross true-breeding gray bodied, long antennae files to true-breeding black bodied, short antennae. The F1 progeny all have gray bodies and long antennae. a. You perform a testcross to the F1 progeny produced from the parent...
1. You want to determine whether three prairie dog genes are linked, and if so what is the order of the genes and how many map units are they away from each other. Capital letters are dominant: L = long necked, I = short-necked, G = gray fur, g = brown fur, R = red eyes, r = pink eyes. True-breeding parentals of opposite phenotypes were bred to produce the F1 hybrids, then the F1 hybrids were test crossed to...
1.
2.
13.
Which statement explains why the recombination frequency between two genes is always less than 50%? Recombination cannot be more than 50% because chromosomes are only 50 map units in length. The genotype of the F, gametes will always be 50% parental gametes and 50% recombinant gametes. Genes with a recombination frequency near 50% are unlinked and have an equal likelihood of being inherited together or separately. F1 gametes always have 50% of their alleles from each parental...
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