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BIO 181 Lecture 9: ch9

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School
NCSU
Department
Biological Sciences
Course
BIO 181
Professor
Lisa M Paciulli

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Related Questions

57. Line 5 (A) for (int j = -i*2; j<= i*2; j++){ (B) for (int j = 0; j< i+2+1; j++){ (C) for (int j = -i; j> i; j++) { (D) for (int j = 0; j<= i*2; j++){ (E) for (int j = i; j<= i+2; j+i) {

55. Line 3 (A) for (int j = height; j< height - i; j+-height) { (B) for (int j = 0; j< 2 + height; j+=height) { (C) for (int j = 0; j< height - i * 2; j++) { (D) for (int j = 0; j< height - i-1; j++) { (E) for (int j = 0; j< height ; j++) {

5.

 Please explain, I understand how todetermine the single and double cross progeny, but not sure whereto to from there.


The following genotypesare a result of a three-point cross between

Offspring

#

z z j j oo

690

z+ z j+j oo

6

z z j+j o+o

119

z+z jj o+o

45

z+z j joo

108

z z j+j oo

49

z z j j o+o

5

z z+ j j+ oo+

675

  1. What are thefrequencies of the parental and the recombinantgenotypes?

  1. Which genotypes arethe result of a double recombination?

z+ z j+j o+ o (6) and z zj j o+ o (5) (is this correct?)

  1. Generate a geneticmap for the three genes z, j, and o.