FIGURE 4-9 Achromosome region containing three linked genes. Because map distances are additive, calculation of A-B and A-C distances leaves us with Map distances are generally additive B Map based on A-B recombination the two possibilities shown for the B-C distance. 5 m.u. Map based on A-C recombination 3 m.u. A 3 m.u. 5 m.u. Possible combined maps 8 m.u. A 3 m.u. -2 m.u. 5 m.u.

FIGURE 4-9 Achromosome region containing three linked genes. Because map distances are additive, calculation of A-B and A-C distances leaves us with Map distances are generally additive B Map based on A-B recombination the two possibilities shown for the B-C distance. 5 m.u. Map based on A-C recombination 3 m.u. A 3 m.u. 5 m.u. Possible combined maps 8 m.u. A 3 m.u. -2 m.u. 5 m.u.

Name: In Figure 4-9, why is only locus A shown in a constantposition? FIGURE
Uploaded: 2024-03-20T06:43:04.000Z
Channel: Bartleby
Description: In Figure 4-9, why is only locus A shown in a constantposition? FIGURE 4-9 Achromosome regioncontaining three linked genes. Becausemap distances are additive, calculation ofA-B and A-C distances leaves us withMap distances are generally additiveBMap

BIOLOGY:CONCEPTS+APPL.(LOOSELEAF)

10th Edition

ISBN:9781305967359

Author:STARR

Publisher:STARR

Chapter15: Biotechnology

Section: Chapter Questions

Problem 2CT: The results of a paternity test are shown in the table below. Numbers indicate the number of short...

See similar textbooks

Similar questions

A molecular biologist is investigating homologous recombination. One aim of this study is to reconstitute stages of the process in vitro. (a) Draw diagrams to show how the four synthetic oligonucleotides below could base-pair to form a stable model Holliday junction. W 5’ GATCGCATTGTAGCCGTAGGTCCACTGTAA 3’ X 5’ GTCCCATACGTAGCCGTAGGACATGTACCG 3’ Y 5’ CGGTACATGTCCTACGGCTACAATGCGATC 3’ Z 5’ TTACAGTGGACCTACGGCTACGTATGGGAC 3’ (b) What is branch migration? (c) What is the name of the enzyme that resolves a Holliday junction into two separate DNA duplexes? (d) On your diagram, indicate how the Holliday junction can be resolved in two different ways and draw the structures of the products.
A Neurospora cross was made between a strain that carried the mating-type allele A and the mutant allele arg-1and another strain that carried the mating-type allele aand the wild-type allele for arg-1 (+). Four hundred linear octads were isolated, and they fell into the sevenclasses given in the table below. (For simplicity, they areshown as tetrads.)a. Deduce the linkage arrangement of the mating-typelocus and the arg-1 locus. Include the centromere orcentromeres on any map that you draw. Label all intervalsin map units.b. Diagram the meiotic divisions that led to class 6. Labelclearly
No need to explain 1.) The phenomenon in which gene present on the same chromosome remain together as gametes are produced is knows as A.) non-disjunction B.) cross-over C.) synapsis D.) linkage 2.) Queen Victoria was a carrier of hemophilia, but there is no record of the disease in any of her ancestors. It is reasonable to suggest that she became a carrier of hemophilia as a result of A.) recombination B.) a mutation C.) inbreeding D.) crossing over 3.) How many primary spermatocytes and primary Oo Yates are required to produced 100 sperm and 100 eggs? A.) 25 primary spermatocytes and 25 primary oocytes B.) 100 primary spermatocytes and 100 primary oocytes C.) 25 primary spermatocytes and 100 primary oocytes D.) 100 primary spermatocytes and 25 primary oocytes
he chances of an individual having a particular group of variable number tandem repeats (VNTRs) can be calculated. If 3 VNTRs are analyzed, one of which has 40 known variations while the other 2 have thirty known variations each, what is the chance of an individual having any specific banding pattern? a 36,000 b 100 c 1/100 d 1/36,000
A molecular biologist is investigating homologous recombination. One aim of this study is to reconstitute stages of the process in vitro. Draw diagrams to show how the four synthetic oligonucleotides below could base-pair to form a stable model Holliday junction. W 5’ GATCGCATTGTAGCCGTAGGTCCACTGTAA 3’ X 5’ GTCCCATACGTAGCCGTAGGACATGTACCG 3’ Y 5’ CGGTACATGTCCTACGGCTACAATGCGATC 3’ Z 5’ TTACAGTGGACCTACGGCTACGTATGGGAC 3’
From a cross between e+ f+ g+ and e− f − g− strains ofNeurospora, recombination between these linkedgenes resulted in a few octads containing the followingordered set of spores:e+ f+ g+e+ f+ g+e+ f − g+e+ f − g+e− f − g−e− f − g−e− f − g−e− f − g−a. Where was recombination initiated?b. Did crossing-over occur between genes e and g?Explain.c. Why do you end up with 2 f+ : 6 f − but 4 e+: 4 e−and 4g+: 4g−?d. Could you characterize these unusual octads as MIor MII for any of the three genes involved?Explain.
. An RFLP analysis of two pure lines A/A • B/B and a/a • b/bshowed that the former was homozygous for a long RFLPallele (l) and the latter for a short allele (s). The two werecrossed to form an F1, which was then backcrossed to thesecond pure line. A thousand progeny were scored asfollows:a. What do these results tell us about linkage?b. Draw a map if appropriate.c. Incorporate the RFLP fragments into your map.
When stained with Giemsa (G-Banding staining), the centromere will be stained: Question 28 options: more robustly (dark band) less robustly (light band)
The recombination frequency in a cross can never exceed 50%. Explain why is this limit?
A pair of paralogous repeats, A and B, have 96% sequence similarity and therefore can promote non-allelic homologous recombination (NAHR). They exist in four possible arrangements in a genome, illustrated below as arrangements 1 – 4. What is the result of NAHR between repeats A and B in arrangement 1? A.Translocation between chromosomes 1 and 2 resulting in monocentric chromosomes B.Deletion or duplication of the region between A and B C.Translocation between chromosomes 1 and 2 resulting in acentric and dicentric chromosomes D.Inversion of the region between A and B
Consider two maize plants:a. Genotype C/cm ; Ac/Ac+, where cm is an unstableallele caused by a Ds insertionb. Genotype C/cm, where cm is an unstable allele causedby Ac insertionWhat phenotypes would be produced and in whatproportions when (1) each plant is crossed with a basepair-substitution mutant c/c and (2) the plant in part a iscrossed with the plant in part b? Assume that Ac and care unlinked, that the chromosome-breakage frequencyis negligible, and that mutant c /C is Ac+.
The results of a paternity test are shown in the table below. Numbers indicate the number of short tandem repeats for loci tested. Whos the daddy? How sure are you?