To analyze:
a) Considering random segregation of the chromosome, determine the genotype and proportion of the sperm produced by the trisomic male fruit fly with genotype
b) Determine the expected ratio of the eyeless to normal eyed flies produced by the union of the sperm with the egg having genotype eyey producing an eyeless
Introduction:
The segregation of the chromosome during cell division is an important phenomenon. If the chromosome fails to segregate properly, it is called non-disjunction, and it could have resulted in the daughter cell with one extra chromosome (
The cell with improper distribution of the chromosomes may reduce their survival rate as compared to the normal cell. In the germ cells, non-disjunction results in the formation of gametes with extra or missing chromosomes which leads to the production of aneuploidy gametes. Germ cells, egg or sperms, are haploid cells, containing a single pair of chromosome. The condition of the presence of extra chromosome (
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Genetic Analysis: An Integrated Approach (2nd Edition)
- The karyotype of a young girl who is affected with familialDown syndrome revealed a total of 46 chromosomes. Her olderbrother, however, who is phenotypically unaffected, actually had45 chromosomes. Explain how this could happen. What wouldyou expect to be the numbers of chromosomes in the parents ofthese two children?arrow_forwardIn individuals affected by cystic fibrosis, salt crystals may appear afterperspiration dries up. In addition, the disease causes respiratory disorderswhich can be both debilitating and lethal. It occurs in individuals homozygousfor recessive gene. If 2 normal parents had a daughter with the symptoms ofthis disease, and a normal son, what is the probability that he might be acarrier of the recessive gene?Express answer in fraction form.arrow_forwardDuchenne muscular dystrophy (DMD) is caused bya recessive mutant allele of an X-linked gene calleddystrophin. Rarely, females have disease symptomsas severe as those in males hemizygous for therecessive allele. These females are heterozygous forX-autosome reciprocal translocations where the Xchromosome breakage occurred in the middle of thedystrophin gene, breaking it into two pieces.a. If it is equally likely for X chromosome inactivation to spread from either of the X chromosomeinactivation centers (XICs; see Fig. 12.15) in thecells of this patient, what proportion of her cellswould you expect to have normal function of thedystrophin gene?arrow_forward
- A woman with an abnormally long chromosome 13 (and a normalhomolog of chromosome 13) has children with a man with anabnormally short chromosome 11 (and a normal homolog of chromosome 11). What is the probability of producing an offspringthat will have both a long chromosome 13 and a short chromosome11? If such a child is produced, what is the probability that thischild will eventually pass both abnormal chromosomes to one ofhis or her offspring?arrow_forwardA male Drosophila from a wild-type stock is discovered to haveonly seven chromosomes, whereas normally 2n = 8. Closeexamination reveals that one member of chromosome IV (thesmallest chromosome) is attached to (translocated to) the distalend of chromosome II and is missing its centromere, thusaccounting for the reduction in chromosome number. (a) Diagram all members of chromosomes II and IV duringsynapsis in meiosis I.arrow_forwardThe human genome consists of 23 pairs of chro-mosomes (22 pairs of autosomes and one pair of sex chro-mosomes). During meiosis, the maternal and paternal setsof homologs pair, and then are separated into gametes, sothat each contains 23 chromosomes. If you assume thatthe chromosomes in the paired homologs are randomlyassorted to daughter cells, how many potential combina-tions of paternal and maternal homologs can be gener-ated during meiosis? (For the purposes of this calculation,assume that no recombination occurs.)arrow_forward
- A normal female is discovered with 45 chromosomes, one ofwhich exhibits a Robertsonian translocation containing most ofchromosomes 18 and 21. Discuss the possible outcomes in heroffspring when her husband contains a normal karyotype.arrow_forwardFor each of the following types of chromosomalaberrations, tell: (i) whether the chromosomes of anorganism heterozygous for the aberration will formany type of loop during prophase I of meiosis;(ii) whether a chromosomal bridge can be formedduring anaphase I in a heterozygote, and if so, underwhat conditions; (iii) whether an acentric fragmentcan be formed during anaphase I in a heterozygote,and if so, under what conditions; (iv) whether theaberration can suppress meiotic recombination; and(v) whether the two chromosomal breaks responsible for the aberration occur on the same side or on opposite sides of a single centromere, or if the two breaksoccur on different chromosomes.a. Reciprocal translocationb. Paracentric inversionc. Small tandem duplicationd. Robertsonian translocatione. Pericentric inversionf. Large deletionarrow_forwardin the experiment of following chromosomal dna mvement througgh meiosis, why. do you use non-sister chromatids to demobstrate crossing over? what combinatiobns of alleles could result from a crossover between BD and bd chromosomes? Identify two ways that meiosis contributes to genetic recombination. Why is it necessary to reduce the number of chromosomes in gametes, but not in other cell?arrow_forward
- You have spent two decades studying a muscular dystrophy disorder. All previous pedigree data indicate this is an X-linked dominant trait, you notice that in one instance the trait skips one daughter coming from an effected father. If this disease is truly X-linked dominant, what is a likely explanation as to why it was not seen in this single female? O Only environmental conditions actually cause the disease O This disease may not be 100% penetrant for the disease O Skipping generations is common for X-linked dominant traits O The definition for X-linked dominant no longer means what it did when this disease was first discovered O This disease must not actually be X-linked dominantarrow_forwardEVOLUTION CONNECTION Crossing over is thought to beevolutionarily advantageous because it continually shufflesgenetic alleles into novel combinations. Until recently, it wasthought that the genes on the Y chromosome might degenerate because they lack homologous genes on the X chromosomewith which to pair up prior to crossing over. However, when theY chromosome was sequenced, eight large regions were foundto be internally homologous to each other, and quite a few ofthe 78 genes represent duplicates. (Y chromosome researcherDavid Page has called it a “hall of mirrors.”) Explain what mightbe a benefit of these regions.arrow_forward. Chromosome 3 of corn carries three loci (b for plant-color booster, v for virescent, and lg for liguleless). A testcross of triple recessives with F1 plants heterozygous forthe three genes yields progeny having the followinggenotypes: 305 + v lg, 275 b + +, 128 b + lg, 112 + v +,74 + + lg, 66 b v +, 22 + + +, and 18 b v lg. Give the genesequence on the chromosome, the map distances between genes, and the coefficient of coincidence.arrow_forward
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