3 is supposed to have cn in the middle can you help

Biology Today and Tomorrow without Physiology (MindTap Course List)
5th Edition
ISBN:9781305117396
Author:Cecie Starr, Christine Evers, Lisa Starr
Publisher:Cecie Starr, Christine Evers, Lisa Starr
Chapter10: Biotechnology
Section: Chapter Questions
Problem 1VQ
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3 is supposed to have cn in the middle can you help

eet01-xythos.content.blackboardcdn.com/blackboard.learn.xythos.prod/5a3683e2386d2/118193597X-Blackboard-S3-Bucke
1_11_2021
texture gene
B
b
2. At first glance, it appears that only 3 genes are linked, as we have 8 phenotypic classes (of all
4 were linked with recombination taking place, we would have 16 phenotypic classes).
However, note that for genes A and D we have only 2 phenotypes: Ad and aD, whereas the
other 3 genes are present in all possible combinations (e.g. Ab, AB, aB, ab). In this case
genes A and D are so close together that no detectable recombination can take place between
them. All 4 genes are linked, but genes A and D are acting as a single gene.
The linkage map is:
10.3 mu
7 /10 - 107% + @ »
P
14.6 mu
10mu
V
an
3. This experiment is analyzed as in the OnLine_Linkage Ppt, slides 5 & 6. The linkage map of
the genes in the heterozygous parent is:
a/D
21 mu
cn
odor gene
4. & 5. This experiment is analyzed as in the OnLine_Linkage Ppt, slides 5 & 6. The linkage
map is:
13.1 mu
16.7 mu
0
15.4 mu height gene
30 mu
14.7 mu
ct
Sequence and arrangement of genes in the heterozygous parent for 5 a) is
++s/pot.
6. This experiment is analyzed as in the OnLine_Linkage Ppt, slides 5 & 6. The linkage map is:
f 4.8
14 mu
6.4 mu
WX
br
C
CV
7. This experiment is analyzed as in the OnLine_Linkage Ppt, slides 5 & 6. The linkage map
is:
S
7
Transcribed Image Text:eet01-xythos.content.blackboardcdn.com/blackboard.learn.xythos.prod/5a3683e2386d2/118193597X-Blackboard-S3-Bucke 1_11_2021 texture gene B b 2. At first glance, it appears that only 3 genes are linked, as we have 8 phenotypic classes (of all 4 were linked with recombination taking place, we would have 16 phenotypic classes). However, note that for genes A and D we have only 2 phenotypes: Ad and aD, whereas the other 3 genes are present in all possible combinations (e.g. Ab, AB, aB, ab). In this case genes A and D are so close together that no detectable recombination can take place between them. All 4 genes are linked, but genes A and D are acting as a single gene. The linkage map is: 10.3 mu 7 /10 - 107% + @ » P 14.6 mu 10mu V an 3. This experiment is analyzed as in the OnLine_Linkage Ppt, slides 5 & 6. The linkage map of the genes in the heterozygous parent is: a/D 21 mu cn odor gene 4. & 5. This experiment is analyzed as in the OnLine_Linkage Ppt, slides 5 & 6. The linkage map is: 13.1 mu 16.7 mu 0 15.4 mu height gene 30 mu 14.7 mu ct Sequence and arrangement of genes in the heterozygous parent for 5 a) is ++s/pot. 6. This experiment is analyzed as in the OnLine_Linkage Ppt, slides 5 & 6. The linkage map is: f 4.8 14 mu 6.4 mu WX br C CV 7. This experiment is analyzed as in the OnLine_Linkage Ppt, slides 5 & 6. The linkage map is: S 7
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