Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Textbook Question
Chapter 10, Problem 24P
A boy with Down syndrome (trisomy
Explain how this is possible.
How many chromosomes do you expect to see in karyotypes of the parents?
What term best describes this kind of chromosome abnormality?
What is the probability the next child of this couple will have a normal phenotype and have
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Part of the karyotype of a diploid individual who is heterozygous for one chromosomal
rearrangement is shown in the diagram. The chromosomes involved in the rearrangement, and their
homologous pair, are shown. The location of each gene is labeled using horizontal lines and the
name of each gene is labeled using letters or numbers. Answer the following questions about the
diagram.
A. What rearrangement is shown? Be as specific as possible.
B. Describe a mutation scenario that could cause this rearrangement to be formed.
A
B
D
E
F
G
H
IXI
A
B
с
D
E
IIXDD
1
2
3
4
5
6
F
G
H
D
1
2
34
5
6
X
In the fruit fly Drosophila melanogaster, a recessive condition called eyeless (ey) significantly interferes with normal eye development. Eyes are either very small or absent. The ey gene is found on chromosome 4, which is the smallest of the four chromosomes in the organism. Individuals with a single copy or three copies of chromosome 4 are viable and can reproduce.
A trisomic wild-type male is crossed with a disomic wild-type female: see attached image
Determine the chromosome constitution and phenotypic ratios in the offspring. Assume random segregation of chromosomes into gametes.
Draw a Punnett square for the dihybrid cross.
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 polydactyly show no linkage.
Chapter 10 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 10 - 10.1 Give descriptions for the following...Ch. 10 - 10.2 The human genome contains contains base...Ch. 10 - In eukaryotic DNA, Where are you most likely to...Ch. 10 - 10.4 Describe the importance of light and dark G...Ch. 10 - Human late prophase karyotypes have about 2000...Ch. 10 - Prob. 6PCh. 10 - From the following list, identify the types...Ch. 10 - 10.8 If the haploid number for a plant species is...Ch. 10 - Mating between a male donkey (2n=64) and a female...Ch. 10 - A researcher interested in studying a human gene...
Ch. 10 - 10.11 In what way does position effect variegation...Ch. 10 - 10.12 A pair of homologous chromosomes in...Ch. 10 - 10.13 An animal heterozygous for a reciprocal...Ch. 10 - Dr. Ara B. Dopsis has an idea he thinks will be a...Ch. 10 - A normal chromosome and its homolog carrying a...Ch. 10 - The accompanying chromosome diagram represents a...Ch. 10 - 10.17 Histone protein isolated from pea plants...Ch. 10 - 10.18 A survey of organisms living deep in the...Ch. 10 - In humans that XX/XO mosaics, the phenotype is...Ch. 10 - 10.20 A plant breeder would like to develop the...Ch. 10 - In Drosophilia, seven partial deletion (1to7)...Ch. 10 - Two experimental varieties of strawberry are...Ch. 10 - 10.23 In the tomato, Solanum esculentum, tall ()...Ch. 10 - A boy with Down syndrome (trisomy 21) has 46...Ch. 10 - Experimental evidence demonstrates that the...Ch. 10 - Prob. 26PCh. 10 - Genomic DNA from the nematode worm...Ch. 10 - 10.28 A small population of deer living on an...Ch. 10 - A eukaryote with a diploid number of 2n=6 carries...Ch. 10 - Prob. 30PCh. 10 - For the following crosses, determine as accurately...Ch. 10 - A healthy couple with a history of three previous...
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- Given the karyotype shown at right, is this a male or a female? Normal or abnormal? What would the phenotype of this individual be?arrow_forwardGiven that a human normally contains 46 chromosomes, give the chromosome number for each of the following conditions: Turner syndrome (female, no Barr bodies): Klinefelter syndrome (male, one Barr body): Triploid: Down syndrome (trisomic): Trisomy 13:arrow_forwardGiven the following genetic map, predict the chance an offspring would inherit the exact same haplotypes that the parents had. Assume interference in this area of the chromosome is 5%. Please report your value as a percent (%). A |-- 3 mu -- |-- 12 mu -- | Parent 1 Haplotypes: A1 B6 C14/A5 B1 C3 Parent 2 Haplotypes: A8 B3 C2 / A7 B2 C5arrow_forward
- Assume that color blindness is a recessive character on the X chromosome. A man and woman with normal vision have the following offspring: a daughter with normal vision who has one color-blind son and one normal son; a daughter with normal vision who has six normal sons; and a color-blind son who has a daughter with normal vision. What are the probable genotypes of all individuals?arrow_forwardA woman, seeking genetic counseling, is found to be heterozygous for a chromosomal rearrangement between the 2nd and 3rd chromosomes. Her chromosomes, compared to those in a normal karyotype, are diagrammed here: A E A E B F B LL F 0 0 G C D D H H CHROMO.: 2 2/3 2/3 3 a. What kind of chromosomal aberration is shown? b. Using a drawing, demonstrate how these chromosomes would pair during meiosis. c. This woman has had two miscarriages. She has come to you, an established genetic counselor. She raises the ff. questions: Provide an informed response to her concerns: Is there a genetic explanation of her frequent miscarriages? • Should she abandon her attempts to have a child of her own? If not, what is the chance that she could have a normal child?arrow_forwardThe following image shows a Robertsonian translocation between chromosomes 13 and 21. This has the potential to cause Down Syndrome in offspring. The couple had 6 pregnancies. What is the probable cause for the two miscarriages you see in the pedigree? See the legend for more information. Briefly explain your answer. male female miscarriage chromosome 13 chromosome 21 Robertsonian translocation Carrier Normal der(13;21)(q10;q10) Down syndrome Carrier Down syndrome Normal O DOarrow_forward
- In the pedigree below, male II-1 has Klinefelter syndrome, which is the result of an XXY karyotype. On the X chromosome, a gene called G6PD has two codominant alleles, G6PDA and G6PDB. In this pedigree, A, B, and AB refer to the phenotypes associated with the alleles of this gene. (Note: In this family, no individuals have the AB version of the phenotype.) A A B Based on the information in the pedigree, when could nondisjunction have occurred? Select all correct answers. In Il-1's father, during meiosis I In II-1's mother, during meiosis I In II-1's mother, during meiosis II In Il-1's father, during meiosis IIarrow_forwardA phenotypically abnormal individual has a phenotypically normalfather with an inversion on one copy of chromosome 7 and a phenotypicallynormal mother without any changes in chromosomestructure. The orders of genes along the two copies of chromosome7 in the father are as follows: R T D M centromere P U X Z C (normal chromosome 7)R T D U P centromere M X Z C (inverted chromosome 7) The phenotypically abnormal offspring has a chromosome 7 withthe following order of genes: R T D M centromere P U D T R Using a sketch, explain how this chromosome was formed. In youranswer, explain where the crossover occurred (i.e., between whichtwo genes).arrow_forwardIn Drosophila, the white gene located on the X chromosome affects eye color; an autosomal gene, wingless, is on an autosomal chromosome. Use the following allele symbols: Xw+ _ , Xw+Y = wild type red eyes; X-linked dominant allele Xw Xw , XwY = white eyes; X-linked recessive allele Y = Y sex chromosome vg+ = wild type wings; autosomal dominant vg = wingless; autosomal recessive Predict ratios/proportions of genotypes and phenotypes of offspring from the following cross, of a white-eyed male with wild type wings and a wild type red eyed female with wild type wings: indicate sex of offspring along with phenotypes. XwY vg+ vg x Xw+Xw vg+vgarrow_forward
- You have set up a fly cross between red-eyed male flies and white-eyed female flies. Several days later, you look at the F1 progeny to examine the eye phenotypes. You notice that all of the F1 males have white eyes and all of the F1 females have red eyes. These observations tell you that the gene for red eye color is most likely: Group of answer choices not involved in the observed phenotypes sex-linked autosomal located on chromosome IIIarrow_forwardRed–green color blindness is a human X-linked recessive disorder. A young man with a 47,XXY karyotype (Klinefelter syndrome) is color blind. His 46,XY brother is also color blind. Both parents have normal color vision. Where did the nondisjunction that gave rise to the young man with Klinefelter syndrome take place? Assume that no crossing over took place in prophase I of meiosis.arrow_forwardThe gene for nose shape is found on the X chromosome. Round nose is dominant to pointed nose. Human individuals with XXY (an additional X chromosome) are male. Individuals with XO (only one X chromosome) are female. For the following family, identify the possible nondisjunction events (rare mistakes during meiosis) that could explain the phenotype of the child. A man with a pointed nose and a woman with a round nose have a daughter with a round nose. This daughter has Turner Syndrome (genotype XO: one X chromosome). Nondisjunction must have occurred in melosis 2 of the mother More than one of the answers is correct Nondisjunction must have occurred in meiosis 1 of the father Nondisjunction must have occurred in meiosis 1 of the mother Nondisjunction must have occurred in meiosis 2 of the fatherarrow_forward
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Mitochondrial mutations; Author: Useful Genetics;https://www.youtube.com/watch?v=GvgXe-3RJeU;License: CC-BY