Introduction to Genetic Analysis
11th Edition
ISBN: 9781464109485
Author: Anthony J.F. Griffiths, Susan R. Wessler, Sean B. Carroll, John Doebley
Publisher: W. H. Freeman
expand_more
expand_more
format_list_bulleted
Question
Chapter 4, Problem 35P
a.
Summary Introduction
To determine: The two examples of the second-division segregation patterns in the cross.
Introduction. The law of independent assortment was given in a dihybrid cross which is the cross between the alleles of two pairs of contrasting characters. The alleles produced by each character are independent to combine with the alleles of the other character.
b.
Summary Introduction
To determine: The data that can be calculated from the 8 percent value of segregation.
Introduction. The chi square value x2 is calculated as the summation of observed values subtracted by the expected values and dividing the resultant by the number of expected values.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
What would justify the following ratio appearing after phenotyping the outcome of a crossing
trial: 8.9: 2.9: 3.2:1?
Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer.
a
b
C
d
Obviously this represents independent assortment based on crossing dihybrid
heterozygotes.
Obviously this represents gene linkage based on test crossing a dihybrid
heterozygote.
Obviously this represents the results of a trihybrid test cross.
Obviously this represents independent assortment based on crossing monohybrid
heterozygotes.
. The Neurospora cross al-2+ × al-2 is made. A linear tetrad analysis reveals that the second-division segregation frequency is 8 percent.a. Draw two examples of second-division segregationpatterns in this cross.b. What can be calculated by using the 8 percent value?
What are the expected phenotypic ratios from the following cross:Tt Rr yy Aa × Tt rr YY Aa, where T = tall, t = dwarf, R = round,r = wrinkled, Y = yellow, y = green, A = axial, a = terminal; T, R,Y, and A are dominant alleles. Note: Consider using the multiplication method in answering this problem
Chapter 4 Solutions
Introduction to Genetic Analysis
Ch. 4 - Prob. 1PCh. 4 - Prob. 5PCh. 4 - Prob. 12PCh. 4 - Prob. 13PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - Prob. 19P
Ch. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 21.1PCh. 4 - Prob. 21.2PCh. 4 - Prob. 21.3PCh. 4 - Prob. 21.4PCh. 4 - Prob. 21.5PCh. 4 - Prob. 21.6PCh. 4 - Prob. 21.7PCh. 4 - Prob. 21.8PCh. 4 - Prob. 21.9PCh. 4 - Prob. 21.10PCh. 4 - Prob. 21.11PCh. 4 - Prob. 21.12PCh. 4 - Prob. 21.13PCh. 4 - Prob. 21.14PCh. 4 - Prob. 21.15PCh. 4 - Prob. 21.16PCh. 4 - Prob. 21.17PCh. 4 - Prob. 21.18PCh. 4 - Prob. 21.19PCh. 4 - Prob. 21.20PCh. 4 - Prob. 21.21PCh. 4 - Prob. 21.22PCh. 4 - Prob. 21.23PCh. 4 - Prob. 21.24PCh. 4 - Prob. 21.25PCh. 4 - Prob. 21.26PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 38.1PCh. 4 - Prob. 38.2PCh. 4 - Prob. 38.3PCh. 4 - Prob. 38.4PCh. 4 - Prob. 38.5PCh. 4 - Prob. 38.6PCh. 4 - Prob. 38.7PCh. 4 - Prob. 38.8PCh. 4 - Prob. 38.9PCh. 4 - Prob. 38.10PCh. 4 - Prob. 38.11PCh. 4 - Prob. 38.12PCh. 4 - Prob. 38.13PCh. 4 - Prob. 38.14PCh. 4 - Prob. 38.15PCh. 4 - Prob. 38.16PCh. 4 - Prob. 38.17PCh. 4 - Prob. 38.18PCh. 4 - Prob. 38.19PCh. 4 - Prob. 38.20PCh. 4 - Prob. 38.21PCh. 4 - Prob. 38.22PCh. 4 - Prob. 38.23PCh. 4 - Prob. 38.24PCh. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 48PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Prob. 53PCh. 4 - Prob. 54PCh. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - Prob. 57PCh. 4 - Prob. 58PCh. 4 - Prob. 59PCh. 4 - Prob. 60PCh. 4 - Prob. 62PCh. 4 - Prob. 63PCh. 4 - Prob. 64PCh. 4 - Prob. 65PCh. 4 - Prob. 66PCh. 4 - Prob. 67PCh. 4 - Prob. 68PCh. 4 - Prob. 69P
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- Consider three genes L, U, and W, for which the count of F2 phenotypes after a 3-point cross is as follows: Phenotype F2 count: L U w 19 L u W 1 l u W 21 L U W 33 l U W 274 l u w 41 l U w 2 L u w 259 Which of the following statements about genes L, U, and W are TRUE? (may be more than one correct ans) A. L, U, and W are each on a different chromosome B. Only U and L are on the same chromosome C. Only U and W are on the same chromosome D. Only W and L are on the same chromosome E. L, U, and W are all on the same chromosomearrow_forwardA 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. Labelclearlyarrow_forwardKernel color in wheat is controlled by 2 pairs of genes (AABB). Determine the color of each offspring with the following genotypes: (Note: 4 alleles – red; 3 – medium red; 2 – intermediate red; 1 – light red; 0 – white). CAPITAL letters only and with spaces when applicable. AABb - AaBb - AABB - aaBb - aabb -arrow_forward
- A series of three-point testcrosses is made to determine the genetic map order of seven linked allele pairs: A/a, B/b, G/g, H/h, Q/q, R/r, and Y/y.From each cross between a triply heterozygous parent listed below, two recombinant classes were noticed as the least frequent among all 8 progeny classes, and are listed at the right in the table. A. For each testcross write the genotype of the F1 heterozygous parent. F1 Parental Phenotype Least frequent F2 Phenotype 1.AHB&ahb AHb & ahB 2.RYh&ryH RYH & ryh 3.BhY&bHy Bhy & bHY 4.qYB&Qyb qYb & QyB 5.AbQ&aBq Abq & aBQ 6.ghR&GHr ghr & GHR B. Write the unified map order of these genes, showing your reasoning.arrow_forwardDetermine whether the 3 phenotypes will be in either a 9:3:4 or 12:3:1 ratio. (Yellow/Red/Purple) Observe the red and yellow kernels (up to 100) and use this to see the ratio and determine the deviation (number of observed minus the number of expected). I am having trouble on how to exactly get this. Please help.arrow_forwardThe pedigree below shows the phenotypes of the ABO blood groups and Rhesus factors [positive (+) and negative (-)] for several members of a family. I (B+ AB- 1 2 3 4 II O- A+ В- B- AB+ A+ 1 2 4 5 6 a. What are the ABO blood group genotypes of individuals I-1 and I-2? b. Which child/ren of individual I-4 can donate blood to him? c. Which individual in the pedigree can donate blood to all the other individuals in the pedigree?arrow_forward
- ) You are performing a dihybrid cross with a strain of flies that is ebony (e) and a strain that is pink (p). The results of the F2 are shown below. Use x? test to help you determine if e and p are linked (Note: refer to Table I for chi-square table). Number 1827 Phenotype + p 424 e + 386 e p 298 Genetic map is the relative positions of genetic markers on a chromosome. Explain the rationale behind genetic mapping.arrow_forwardFor the cross: PpAa x PpAa P = purple flowers (Dominant) p = white flowers A = axial flowers (Dominant) a = terminal flowers a. What are the possible gamete classes that can form from these parents? b. What are the expected offspring genotype classes and ratios/proportions/fractions which will result from the cross? c. What are the expected offspring phenotype classes and ratios/proportions/fractions which will result from the cross? 2. Predict ratios/proportions/fractions of genotypes and phenotypes of the following crosses. T = tall stem t = dwarf stem P = purple flowers p = white flowers G = green pods g = yellow pods A = axial flowers a = terminal flowers R = round peas r = wrinkled seeds A. ttPp x Ttpp B. GgRr x ggRr C. PpGg x ppggarrow_forwardIn a cross between mice the genotypes AB/ab ab/ab, what is the recombination frequency if the progeny numbers are 72 AB/ab, 68 ab/ab, 17 Ab/ab, and 21 aB/ab? The alleles are shown for each chromosome, separated by a slash (/).arrow_forward
- Consider the following dihybrid testcross: B/b • E/e × b/b • e/e For the progeny from this testcross, determine the relative proportions (from 0% to 100%) of each genotype if the two genes: a) are linked (dominant alleles in cis conformation) with no crossing over: Be/be: be/be: BE/be: bE/be: b) assort independently. B/b; E/e: B/b; e/e: b/b; E/e: b/b; e/e: c) are linked (dominant alleles in cis conformation) and 20 map units apart. Be/be: be/be: BE/be: bE/be:arrow_forwardE. W. Lindstrom crossed two corn plants with green seedlings and obtained the following progeny: 3583 green seedlings, 853 virescentwhite seedlings, and 260 yellow seedlings . Q. Explain how color is determined in these seedlings.arrow_forwardWhat will be the results of the following crosses, where N is black, n is brown, L is short hair and l is long hair. Make the tables and the corresponding explanation: 1. Crossing a Nnll woman dog with a Nnll dog. 2. Crossing a NnLl woman dog with a NnLl dog 3. Crossing a woman dog nnLl with a dog NNllarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Biology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage Learning
Biology (MindTap Course List)
Biology
ISBN:9781337392938
Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg
Publisher:Cengage Learning
Introduction to the NIOSH Manual of Analytical Methods Fifth edition; Author: Centers for Disease Control and Prevention (CDC);https://www.youtube.com/watch?v=B5rUrKLMoas;License: Standard Youtube License