Concept explainers
In Drosophilia, seven partial deletion (
Mutation | |||||||
Deletion | a | b | c | d | e | f | g |
1 | + | m | + | m | + | + | + |
2 | m | + | + | + | + | m | + |
3 | m | + | + | + | + | + | m |
4 | m | + | + | m | + | m | m |
5 | + | m | + | m | m | + | + |
6 | m | m | m | m | + | m | m |
7 | m | + | + | + | + | + | + |
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GENETIC ANALYSIS: AN INTEG. APP. W/MAS
- A 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_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 melanogaster, red eyes are dominant over white and the variation for this characteristic is on the X chromosome. Vestigial wings (v) are recessive to normal (V) for an autosomal gene. Predict the appearance of offspring of the following crosses: XW/XwV/v×Xw/Y v/v, Xw/XwV/v×XW/Y V/v.arrow_forward
- Rice has a chromosome number of 2n = 24. If different euploids and aneuploids are available in rice, identify the chromosomal mutation, chromosome configuration, chromosome number, and/or type of gametes in each of the mutations below. Example: For 10B, the answer is “10 II + 2 I”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_forwardAn inversion heterozygote has the following inversion chromosome: What would be the products if a crossover occurred betweengenes H and I on the inverted chromosome and a normalchromosome?arrow_forward
- In 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 Using 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_forwardIn Drosophila,, the curled mutation (cu, chromosome 3, position 50.0) results in wings that curl up, while ebony (e, chromosome 3, position 70.7) results in a dark body. True breeding, wild type females are mated with true breeding males with curled wings and ebony bodies. Considering Drosophila notation, which of the following correctly diagrams the P1 cross? X X ++ e + + + O+ X + X + ■ + X + + + 3+ X X X X + + Y Y cu cu cu + cu cu J e e e e e (D e + cu cu (Darrow_forwardIn corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of 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. Using the following color-coded circles, simulate the crosses indicated below. Put the illustrations of crosses in the spaces provided. Be sure to include in the labels the genotypes and phenotypes of the offspring in each cross. Big light green circle - male-sterile cytoplasm Big orange circle - male-fertile cytoplasm Small orange circle - FF nucleus Small half-light green-half-orange circle - Ff nucleus Small light-green circle - ff nucleusarrow_forward
- In Drosophila,, the curled mutation (cu, chromosome 3, position 50.0) results in wings that curl up, while ebony (e, chromosome 3, position 70.7) results in a dark body. True breeding, wild type females are mated with true breeding males with curled wings and ebony bodies. Considering Drosophila notation, which of the following correctly diagrams the F1 cross? X X 3+ cu e + X X e + + + + + cu e + O + ■ 3+ X X X X Y Y + + ■ cu cu cu ' + ■ cu ■ ' + e + e e e e e + cu +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_forwardDraw 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.arrow_forward
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