Answered: Same genotype, but with close linkage.…

Same genotype, but with close linkage. Dad's Mom's Chromosomes Chromosomes A a Different from previous bl A a В B slide b Genotype of Dad' gametes From Previous Slide Aabb of Genotype Mom's gametes

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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Question

100%

Same genotype, but with close linkage.
Dad's
Mom's
Chromosomes
Chromosomes
A
a Different from previous
bl
A
a
В
B slide
b
Genotype of
Dad' gametes
From Previous
Slide
Aabb
of
Genotype
Mom's
gametes

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18
Draw two replicated, condensed, homologous chromosomes that have the genes E and F on them. This individual is heterozygous for gene E, and homozygous recessive for gene F. Be sure to label your chromosomes for full credit.
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name a disorder karyotype analysis, pick and explain a disorder associated with a variation in chromosome number type of variation in chromosome number. causes of variation. phenotypic symptoms of disorder. name a disorder.
Short Answer 3. The pedigree shown is of a family with an X-linked recessive disease. The sex chromosomes are indicated. Individual II-2 is the result of a non-disjunction event in one of the parents. I || 1 2 XHXH XhY XhY 12 XhXhY A) If the nondisjunction event that gave rise to Il-2 occurred in the father, in which phase of meiosis did it happen? Explain your reasoning. B) Upon further testing, it is discovered that Individual 1-1 has a mild case of the disease. Please give a short explanation as to how that is possible.
Short Answer 4. You identify two recessive mutations in fruit flies, one that results in curly wings (cc) and one in an ebony body (ee). You cross two flies that are heterozygous for both traits (CcEe), and find 352 flies progeny in the next generation with the following phenotypes: wild type - 193 wild type wings, ebony body - 69 You believe the curly and ebony alleles assort independently and are not linked. Perform a chi square analysis for this cross to test the null hypothesis. A) Show at least the following parts of your work: the expected vs. observed numbers for each phenotype, and the final X² value. Observed Expected (o-e)²/e wild type body, curly wings - 64 curly wings, ebony body - 26 Edit View Insert Format Tools 12pt ✓ Paragraph Table X² U AV T² v BIUA V V QV G✓ ⠀
What is the difference between these two. How would the karytotype look like