Concept explainers
Semisterility in corn, as seen by unfilled ears with gaps due to abortion of approximately half the ovules, is an indication that the strain is a translocation heterozygote. The chromosomes involved in the translocation can be identified by crossing the translocation heterozygote to a strain homozygous recessive for a gene on the chromosome being tested. The ratio of
a. | What types of progeny (fertile or semisterile, green or yellow-green) would you predict from the backcross of the F1 to the homozygous yg mutant if the gene was not on one of the two chromosomes involved in the translocation? |
b. | What types of progeny (fertile or semisterile, green or yellow-green) would you predict from the backcross of the F1 to the homozygous mutant if the yg gene is on one of the two chromosomes involved in the translocation? |
c. | If the yg gene is located on one of the chromosomes involved in the translocation, a few fertile, green progeny and a few semisterile, yellow-green progeny are produced. How could these relatively rare progeny classes arise? What genetic distance could you determine from the frequency of these rare progeny? |
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ND STONY BROOK UNIVERSITY LOOSELEAF GENETICS: FROM GENES TO GENOMES
- 1)se; 12 cM 2)h; 12 cM 3)g; 8 cM 4)se; 8 cMarrow_forwardFemales of wild-type Strain A and males of mutant Strain B, as well as females of mutant Strain B and males of wild-type Strain A, make reciprocal crosses. Explain why reciprocal crosses are needed in genetics experiments involving Drosophila fruit flies.arrow_forwardIn rice, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of rice plants (i.e. the stamen) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile rice plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male-sterile lines. Give the result(s) of the cross and explain the phenotype of the offspring.arrow_forward
- In autotetraploid Chinese primrose (Primula sinensis L.), the gene controlling stigma color is very near the centromere of the chromosome carrying it. The allele G for green stigma is dominant to g for red stigmas. A homozygous green autotetraploid strain is crossed with a homozygous red autotetraploid strain. Each of the F1 GGgg plants would obtain 12 gametes which are 2GG, 8Gg, and 2g. How were these obtained?arrow_forwardIn a cross in Drosophila, a female heterozygous for the autosomallylinked genes a, b, c, d, and e (abcde/ + + + + +) was testcrossedwith a male homozygous for all recessive alleles (abcde/abcde).Even though the distance between each of the loci was at least3 map units, only four phenotypes were recovered, yielding thefollowing data: Phenotype No. of Flies+ + + + + 440a b c d e 460+ + + + e 48a b c d + 52 Total = 1000 Why are many expected crossover phenotypes missing? Can anyof these loci be mapped from the data given here? If so, determinemap distances.arrow_forwardn corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of the corn plants (i.e the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male sterile lines sing the cardboard chips, simulate the crosses indicated below. Give the genotypes and phenotypes of the offsprings in each cross, and properly label the nucleus and the cytoplasm of each individual in the cross Legend male sterile cytoplasm Male fertile cytoplasm FF nucleus Ff nucleus ff nucleus A. Male sterile female x FF male Explain the phenotype of the offspring B. Male sterile female x Ff male Explain the phenotype of the offspringarrow_forward
- In fruit flies, you are mapping three genes in a three point cross. The mutants are hairy body (h), sepia colored eyes (se) and female sterility (g). You cross a heterozygous parent with a homozygous recessive parent and obtain the following results: Type Number h se g. 5 + se + 450 + se g 27 ++g_ h se + + + + h + g. h + + TOTAL is the gene in the middle and the distance in map units between se and g is Oh; 16.4 se; 7.1 Oh; 7.1 70 82 7 327 32 1000 se; 16.4arrow_forwardA yeast geneticist irradiates haploid cells of a strain that is an adenine-requiring auxotrophic mutant, caused by mutation of the gene ade1. Millions of the irradiated cells are plated on minimal medium, and a small number of cells divide and produce prototrophic colonies. These colonies are crossed individually with a wildtype strain. Two types of results are obtained:(1) prototroph × wild type : progeny all prototrophic(2) prototroph × wild type : progeny 75% prototrophic, 25% adenine-requiring auxotrophsa. Explain the difference between these two types of results.b. Write the genotypes of the prototrophs in each case.c. What progeny phenotypes and ratios do you predict from crossing a prototroph of type 2 by the original ade1auxotroph?arrow_forwardInversions are known to affect crossing-over. The following homologs have the indicated order (the filled and open circles indicate centromeres): • (A B C D E) o (A D C B E) c. Diagram the results of a single crossover between homologous genes B and C in the inversion.arrow_forward
- Consider 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_forwardAn individual is heterozygous for a reciprocal translocation, with the following chromosomes: A • B C D E F A • B C V W X R ST • U D E F R ST • U V W X Q. Explain why the fertility of this individual is likely to be less than the fertility of an individual without a translocation.arrow_forwardA PORTION OF THE LINKAGE MAP OF CHROMOSOME 2 IN THE TOMATO IS ILLUSTRATED HERE. ci (compound influorescence) o (oblate) - 15 CM 20 CM p (peach) THE OBLATE PHENOTYPE IS A FLATTENED FRUIT, THE PEACH PHENOTYPE IS HAIRY FRUIT (LIKE A PEACH), AND COMPOUND INFLORESCENCE MEANS CLUSTERED FLOWERS. IGNORE THE PEACH LOCUS. AMONG 1000 GAMETES PRODUCED BY A PLANT OF GENOTYPE O CI /+ +, WHAT TYPES OF GAMETES WOULD BE EXPECTED, AND WHAT NUMBER WOULD BE EXPECTED OF EACH?arrow_forward
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