BROOKER BIOLOGY
5th Edition
ISBN: 9781307656152
Author: BROOKER
Publisher: MCG/CREATE
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Chapter 17.4, Problem 2EQ
Summary Introduction
To determine:
The results that led Morgan to conclude that eye color in fruit flies is associated with the sex of the individual.
Introduction:
Morgan's experiment explains the process of linkage and recombination. The linkage phenomenon is defined as the transfer of two closely situated or linked genes to pass on to the next generation. Higher the linkage, lesser is the recombination frequency because closely linked genes will always be inherited together.
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estion 2 of 15
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Match each phenotype description to its corresponding sex chromosome genotype in humans.
XX with SRY on X
phenotypically female
but karyotype indicates
presence of both
sex chromosomes
XY with SRY deleted
phenotypically male but
karyotype indicates
presence of only
X chromosomes
hp
XXXX
phenotypically female
with some abnormalities
and overexpression of X
chromosome genes
Answer Bank
A
0
XXY
phenotypically male
with sterility
and hypogonadism
€
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ES
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Attempt 2
Apr 1
XYY
phenotypically male
with an increase in
average stature
3:52
STAY
please help asap
Below you will find a spreadsheet of the "class" data, use this to complete data table 2. Remember, the class is
a total of 24 students, to calculate frequency use the genotype with the highest frequency for each
characteristic and divide that number by 24. Don't forget to answer the question on the bottom of page 107!
Gender
Letters
XY
XX
Sex
X, Y
11
13
XX
Homozygous Heterozygous Homozygous
Recessive
Genotype
highest
frequency
Characteristic
Dominant
Aa
Eyebrows
Eye Shape
Hitchhikers
А, а
R, r
14
10
11
3
Rr
Hh
Н, h
D, d
Е, е
F,f
G, g
В, Ь
15
Earlobes
4
8.
12
dd
Ee
Tongue Roll
Widows Peak
7
10
13
Ff
Face Shape
9.
12
3
Gg
Eye Color
14
Bb
MacBook Air
esc
4)
FI
F2
F4
F5
F6
F8
F9
F10
F1I
F12
@
#
$
&
Chapter 17 Solutions
BROOKER BIOLOGY
Ch. 17.1 - Prob. 1CCCh. 17.1 - Prob. 2CCCh. 17.1 - Prob. 3CCCh. 17.1 - Prob. 4CCCh. 17.1 - Mendels Laws of Inheritance Concept Check: What...Ch. 17.2 - Prob. 1CSCh. 17.2 - Prob. 2CSCh. 17.2 - Prob. 1CCCh. 17.3 - Prob. 1CCCh. 17.3 - Prob. 2CC
Ch. 17.4 - Prob. 1CCCh. 17.4 - Prob. 1EQCh. 17.4 - Prob. 2EQCh. 17.4 - Prob. 3EQCh. 17.5 - Prob. 1CSCh. 17.5 - Prob. 1CCCh. 17.6 - Prob. 1CCCh. 17 - Prob. 1TYCh. 17 - During which phase of nuclear division does the...Ch. 17 - Prob. 3TYCh. 17 - Which of Mendels laws cannot be observed in a...Ch. 17 - During a __________blank, an individual with the...Ch. 17 - Prob. 6TYCh. 17 - Prob. 7TYCh. 17 - A hypothetical flowering plant species produces...Ch. 17 - Genes located on a sex chromosome are said to be...Ch. 17 - Prob. 10TYCh. 17 - Prob. 1CQCh. 17 - A cross is made between individuals having the...Ch. 17 - Core Concept: Systems We can view life as a...Ch. 17 - Discuss the principles of the chromosome theory of...Ch. 17 - When examining a human pedigree, what patterns do...
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- Linkage Mapping Using Dihybrid Testcrosses in Fruit Flies In fruit flies, the normal honey-gray body-color (bl*) is dominant to black (bl), having normal red eyes (pu+) is dominant to purple (pu), and having normal wings (vg+) is dominant to vestigial (vg). The three genes are located close together on the same chromosome. Your task is to use data from three dihybrid testcrosses to (1) calculate recombination frequencies, (2) convert recombination frequencies into map distances, and (3) build a map of the chromosome interval covered by the three genes, indicating their order and distances between them. BLACK PURPLE DIHYBRID CROSS In the parental generation, you mate a pure-breeding wild-type female (bl+/blt;put/pu+) with a pure-breeding black, purple male (bl/bl;pu/pu) to produce an F1 generation that is all wild-type (bl+/bl;pu*/pu). Note that the F1 flies are all dihybrid. Next, you mate several F1 dihybrid females (bl+/bl;pu*/pu) with tester males, which are black, purple…arrow_forwardLinkage Mapping Using a Trihybrid Testcross in Fruit Flies Remember the black, purple, and vestigial genes from Exercise 1? Now you're going to have to construct a map of the same region from trihybrid testcross data. The new map should be similar to the one from Exercise 1, but it won't be identical because we're starting with a new, independent data set. BLACK PURPLE VESTIGIAL TRIHYBRID TESTCROSS In the parental generation, you mate a pure-breeding wild-type female (bl+/bl*pu*/pu+vg+/vg+ with a pure-breeding black, purple, vestigial male (bl/bl;pu/pu vg/vg) to produce an F1 generation that is all wild-type (bl/bliput/pu;vg+/vg). Note that the F1 flies are all trihybrid. Next, you mate several F1 trihybrid females (bl/bl:put/pu;vg+/vg) with tester males, which are black, purple, vestigial (bl/bl;pu/pu;vg/vg). The offspring of this trihybrid testcross are: Phenotype Wild-type Black, purple, vestigial Vestigial Black, purple Purple, vestigial Black Purple Black, vestigial bl pu vg bl pu…arrow_forwardDiscuss Concepts One of the human chromosome pairs carries a gene that influences eye color. In an individual human, one chromosome of this pair has an allele of this gene that contributes to the formation of blue eyes. The other chromosome of the pair has an allele that contributes to brown eye color (other genes also influence eye color in humans). After meiosis in the cells of this individual, what fraction of the nuclei will carry the allele that contributes to blue eyes? To brown eyes?arrow_forward
- Activity 9GENES SEGREGATION BY BEAN-O TYPE Materials: Colored beans to represent the gene for a certain trait of garden pea. (60 pieces of the same size and shape; 30 pieces of which are of different colors from rest) Procedures: 1. Place the 30 pieces beans (one color) to one container and the other 30 pieces to the other container. Assign which color of beans will represent dominant and recessive genes of a specific trait of garden pea 2. Label the paper container as First Parent P1, male and other P1 female 3. At random, segregate the beans on the Lab table. Assume fertilization occurs and First filial generations (F1) are formed 4. Categorize the formed First filial generations 5. Put all the beans in one container. Assume that all the first filial generations undergo self-fertilization 6. Pick up the beans by two from the container without looking at it and collect the second Filial generation (F2) 7. Categorize the formed Second filial generations into two (2) phenotype, and…arrow_forwardCan you help?arrow_forwardasap pleasearrow_forward
- please help?arrow_forwardExplain step by step.arrow_forwardmapping gene The genes for ruby eyes (rb), tan body (t) and cut wings (ct) are all found on the X-chromosome of Drosophila melanogaster. All of these are recessive traits. They map in the order rb, ct, t with 12.5 map units between rb and ct and 7.5 map units between ct and t. Suppose you cross a cut wing male with a homozygous female that is both tan and has ruby eyes. What will the F1 females look like? Draw map of the section of the X chromosomes that has these 3 genes for the F1 females Assume you testcross your F1 females. What progeny classes would you expect? ii. Give approximate numbers for each class based on a total of 2000 progeny. Assuming the i=1 and there are no double crossovers. Assuming the i=0 and there are the expected number of double crossovers.arrow_forward
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