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)
- Female fruit flies homozygous for the X-linked white-eye alleleare crossed to males with red eyes. On very rare occasions, an offspringof such a cross is a male with red eyes. Assuming these rareoffspring are not due to a new mutation in one of the mother’s Xchromosomes that converted the white-eye allele into a red-eyeallele, explain how a red-eyed male arises.arrow_forwardThe maternal-effect mutation bicoid (bcd) is recessive. Inthe absence of the bicoid protein product, embryogenesis isnot completed. Consider a cross between a female heterozygousfor the bicoid mutation (bcd+/ bcd-) and a homozygousmale(bcd-/ bcd-). How is it possible for a male homozygous for the mutationto exist?arrow_forwardOn rare occasions, an organism may have three copies of achromosome and therefore has three copies of the genes on thatchromosome (instead of the usual number of two copies). Forsuch a rare organism, the alleles for each gene usually segregateso that a gamete will contain one or two copies of the gene. Let’ssuppose that a rare pea plant has three copies of the chromosomethat carries the height gene. Its genotype is TTt. The plant is alsoheterozygous for the seed color gene, Yy, which is found on adifferent chromosome. With regard to both genes, how manytypes of gametes can this plant make, and in what proportions?(Assume that it is equally likely that a gamete will contain oneor two copies of the height gene.)arrow_forward
- In Neurospora, his2 mutants require the amino acidhistidine for growth, and lys4 mutants require theamino acid lysine. The two genes are on the samearm of the same chromosome, in the ordercentromere - his2 - lys4.A his2 mutant is mated with a lys4 mutant. Draw all ofthe possible ordered asci that could result from meioses in which the following events occurred, accountingfor the nutritional requirements for each ascospore.Ascospores without any copy of a chromosome willabort and die, turning white in the process.a. A single crossover between the centromere and his2b. A single crossover between his2 and lys4c. Nondisjunction during the first meiotic divisiond. Nondisjunction during the second meiotic divisione. A single crossover between the centromere andhis2, followed by nondisjunction during the firstmeiotic divisionf. A single crossover between his2 and lys4,followed by nondisjunction during the firstmeiotic divisionarrow_forwardThe 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_forward
- Consider two hypothetical recessive autosomal genesa and b, where a heterozygote is testcrossed to a doublehomozygousmutant. Predict the phenotypic ratios underthe following conditions:(a) a and b are located on separate autosomes.(b) a and b are linked on the same autosome but areso far apart that a crossover always occurs betweenthem.(c) a and b are linked on the same autosome but are soclose together that a crossover almost never occursarrow_forwardA 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_forwardIn Drosophila, the vestigial wings recessive allele,vg, causes the wings to be very small. A geneticistcrossed some true-breeding wild-type males to some vestigial virgin females. The male and female F1 flieswere wild type. He then allowed the F1 flies to matewith one another and found that 1/4 of the male andfemale F2 flies had vestigial wings. He dumped thevestigial F2 flies into a morgue and allowed the wildtype F2 flies to mate and produce an F3 generation.a. Give the genotype and allele frequencies amongthe wild-type F2 flies.b. What will be the frequencies of wild-type andvestigial flies in the F3?c. Assuming the geneticist repeated the selection againstthe vestigial F3 flies (that is, he dumped them in amorgue and allowed the wild-type F3 flies to mate atrandom), what will be the frequency of the wild-typeand mutant alleles in the F4 generation?d. Now the geneticist lets all of the F4 flies mate atrandom (that is, both wild-type and vestigial fliesmate). What will be the…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_forwardDRAW IT Suppose, as in the questionat the bottom of Figure 15.9, the parental(P generation) flies were true-breedingfor gray body with vestigial wings andblack body with normal wings. Draw thechromosomes in each of the four possiblekinds of eggs from an F1 female, andlabel each chromosome as “parental”or “recombinant.”arrow_forwardIn Drosophila, a heterozygous female for the X-linkedrecessive traits a, b, and c was crossed to a male that phenotypically expressed a, b, and c. The offspring occurred inthe following phenotypic ratios.+ b c 460a + + 450a b c 32+ + + 38a + c 11+ b + 9 No other phenotypes were observed.(a) What progeny phenotypes are missing? Why?arrow_forward
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