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
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Chapter 4, Problem 21.6P
Summary Introduction
To determine: The column marked "Progeny
Introduction: Phenotypes are symbolized for each progeny class.
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In this case a family history revealed a genetic basis for the disorder. The pedigree is shown in Fig. 1 Below.
Key
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Female: affected
Female: unaffected
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IV
V
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10
9
10
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Male: affected
Male: unaffected
Deceased
Disease status not given
Dizygotic twins
Monozygotic twins
Fig. 1 Disease pedigree. Five generations I, II, III, IV, V are shown. Females are represented by circles, males by squares, dizygotic (non-identical) twins by
diagonal lines originating from the same point, Monozygotic (identical) twins by diagonal lines originating from the same point and joined symbols and deceased
by a diagonal line through the symbol. Filled symbols indicate that the individual displays the disease phenotype. Unfilled symbols indicate that the individual
does not display the disease phenotype. Carriers of the disease are not indicated. Information on disease status is not known for generation I and is omitted for
the individuals represented by a symbol with an asterisk.…
Is this karyotype male or female?
What kind of error (if any)
Name of
syndrome
Draw a Punnett square for the dihybrid cross described below (it is the same story as given for question 8, above) and use it to fill in
the blanks correctly in the text that follows. NOTE: Please type in whole numbers, no symbols.
There are two known alleles of gene occupying a specific locus in the X chromosome. The gene in question codes for a transcription factor
involved in digit development. The mutant allele is dominant and gives rise to an additional but non-functioning little finger (polydactyly) on
both hands.
A couple have had their DNA sequenced at the region of interest, the male exhibits polydactyly because of the mutation, the female is
homozygous wild type at the same locus and therefore has the wild type phenotype. Both have green eyes. In this story; eye colour shows a
monogenic autosomal inheritance pattern and the allele for brown eyes shows incomplete dominance with that for blue eyes, the
heterozygote phenotype is green eyes.
The genes for eye colour and…
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
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- Using the conventions of Figure 4-15, draw parents andprogeny classes from a crossP M′′′/p M′ × p M′/p M′′′′arrow_forwardA strain of Neurospora with the genotype H ⋅ I is crossedwith a strain with the genotype h ⋅ i. Half the progeny areH ⋅ I, and the other half are h ⋅ i. Explain how this outcome is possible.arrow_forwardAn individual that is heterozygous for an inversion has the following chromosomes(∗ is the centromere):M N O P Q • R S T Um n o t s r • q p u Assume that a crossover occurred between P and Q. Starting with “M” allele, list the remaining genes in order (NO spaces between the letters) of the chromosome resulting from crossing over. You must use upper and lower-case letters correctly and the * symbol for the centromere(s).arrow_forward
- One yeast strain carries the alleles lys+ and arg+, whereas another strain has lys-3 and arg-2. The two strains were crossed toeach other, and an ascus obtained from this cross has four spores with the following genotypes: lys+ arg+, lys+ arg-2, lys-3arg+, and lys-3 arg 2. This ascus has a. a parental ditype.b. a tetratype.c. a nonparental ditype.d. either a tetratype or a nonparental ditype.arrow_forwardThe following is a linkage map of chromosome 5 for three genes in tomato: (see image) The cross between the triple heterozygote (Lf J W/ lf j w) and a triple homozygous recessive produced 500 progeny. Assume that there is no interference in the Lf-W region. Give the expected number of individuals for each of the following progeny types and show complete solutions.a. with crossover in the Lf-J and J-W regionsb. with crossover in the Lf-J regionc. with crossover in the J-W regiond. without crossover in the Lf-W regionarrow_forwardConsider the following two nonhomologous wildtype chromosomes, where letters or numbers represent genes, the "-" represents the centromere of each chromosome, and chromosomes are shown on separate lines. ABCDE-FGHIJK 123-45678 Identify the type of rearrangement shown in each of the following (A-C) and then identify whether it is balanced or unbalanced. Assume that the individual is diploid and heterozygous for the rearrangement. A. ABCDE-FGHIJKGH 123-45678 Rearrangement: [Select] • Balanced or Unbalanced: [Select] B. ABCDGF-EHIJK 123-45678 Rearrangement: [Select] Balanced or Unbalanced: [Select]arrow_forward
- An individual heterozygous for a reciprocal translocation possesses the following chromosomes: A B • C D E F G A B • C D V W X R S • T U E F G R S • T U V W X Q. Give the products that result from alternate, adjacent-1, and adjacent-2 segregation.arrow_forwardNeurospora of genotype a + c are crossed withNeurospora of genotype + b +. (Here, + is shorthandfor the wild-type allele.) The following tetrads areobtained (note that the genotype of the four sporepairs in an ascus are listed, rather than listing alleight spores):a + c a b c + + c + b c a b + a + ca + c a b c a + c a b c a b + a b c+ b + + + + + b + + + + + + c + + ++ b + + + + a b + a + + + + c + b +137 141 26 25 2 3a. In how many cells has meiosis occurred to yieldthese data?b. Give the best genetic map to explain these results.Indicate all relevant genetic distances, both betweengenes and between each gene and the centromere.c. Diagram a meiosis that could give rise to oneof the three tetrads in the class at the far right inthe listarrow_forwardExplain why it is possible for the proband in the following pedigree to have children of blood types A, B, and AB. Considering epistatic genes, what are the possible genotypes of II-2?arrow_forward
- An individual has the following reciprocal translocation: I - SKLM CDE CDE H B B H M e. What is the name of the structure formed during meiosis, as the chromosomes attempt to synapse? f. What would be the outcome of alternate segregation? g. What would be the outcome of adjacent-1 segregation? h. What would be the outcome of adjacent-2 segregation?| (NOTE: A-B--C-D-E or A-B-O-C-D-E are acceptable formats for chromosome structure)arrow_forwardWhat does the expression linear octad analysis mean?arrow_forwardHere are the progeny of this cross: (Note that the categories are not in any particular order.)Fly type # of prog. Phenotype symbols Categorywt eyes black body wt wings 97 grn+ blk crv+Green eyes black body curved wings 709 ParentalGreen eyes wt body wt wings 9Green eyes black body wt wings 162wt eyes wt body wt wings 727wt eyes black body curved wings 12wt eyes wt body curved wings 179Green eyes wt body curved wings 105Total = 2000 9.) Write the phenotype symbols in the right-hand column. The first one has been done for you.10.) Next to that, label all fly categories as parental (NCOs), SCOs, and DCOs. One has been donefor you.11.) After each SCO/DCO label, write which gene got “unlinked” in these offspring.12.) Put these three genes into a genetic map in the proper order.13.) Calculate the genetic distance between the genes and label the map with these distances.14.) Calculate the cross-over interference15.) Return to questions #1-6 above. For question 6, you gave your opinion, but…arrow_forward
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