Campbell Biology
12th Edition
ISBN: 9780135188743
Author: Urry
Publisher: PEARSON
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Chapter 16, Problem 16.2CR
DRAW IT Ø Redraw the Punnett Square on The right side of Figure 14.8 as two smaller monohybrid Punnett squares, one for each gene. Below each square. list the fractions of each
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Chapter 16 Solutions
Campbell Biology
Ch. 16.1 - Given a polynucleotide sequence such as GAATTC,...Ch. 16.1 - VISUAL SKILLS Griffith was trying to develop a...Ch. 16.2 - What role does complementary base pairing play in...Ch. 16.2 - Identify two major functions of DNA pol III in DNA...Ch. 16.2 - Prob. 3CCCh. 16.2 - Prob. 4CCCh. 16.3 - Describe the structure of a nucleosome, the basic...Ch. 16.3 - Prob. 2CCCh. 16.3 - MAKE CONNECTIONS Interphase chromosomes appear to...Ch. 16 - What does it mean wheti we say that the two DNA...
Ch. 16 - DRAW IT Redraw the Punnett Square on The right...Ch. 16 - Prob. 16.3CRCh. 16 - In his work with pneumonia-causing bacteria and...Ch. 16 - What is the basis for tlie difference in how the...Ch. 16 - In analyzing the number of different bases in a...Ch. 16 - The elongation of the leading Strand during DNA...Ch. 16 - In a nucleosome, the DNA is wrapped around (A)...Ch. 16 - E. coli cells grown on, 15N medium are transferred...Ch. 16 - A biochemist isolates, purifies, and combines in a...Ch. 16 - The spontaneous loss of amino groups from adenine...Ch. 16 - MAKE CONNECTIONS Although the proteins that cause...Ch. 16 - EVOLUTION CONNECTION Some bacteria may be able to...Ch. 16 - SCIENTIFIC INQUIRY DRAW IT Model building can be...Ch. 16 - Prob. 12TYUCh. 16 - Prob. 13TYU
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- What is the labeled part of these figures?arrow_forwardConsider a Droscophilia fly with a genotype of Nn XqYY. The dominant allele of the sex-linked gene specifies a black body and the recessive a white body. The recessive autosomal allele specifies hairy bristles while the dominant allele specifies smooth bristles. i)What is the ploidy of this fly? ii) What would the sex of this fly be? iii)What would the phenotype of this fly be with respect to these two loci?arrow_forwardIn com, male sterility is controlled by maternal cytoplasmic elements. However, the presence of a nuclear fertility restorer gene (F_) restores fertility to male sterile lines. Using the following cardboard chips, simulate the crosses indicated below. Present your cross to your instructor for checking. After being checked, draw your simulated crosses in the spacesprovided below. Give the genotypes and phenotypes of the offspring in each cross, and properly label the nucleus and the cytoplasm of each individual in the crosses.Big white circular chips Male sterile cytoplasmBig peach circular chips Male fertile cytoplasmSmall peach circular chips FF nucleusSmall half-white-half-peach circular chips Ff nucleusSmall white circular chips ff nucleus a. male sterile female x FF maleExplain your results. b. male sterile female x Ff maleExplain your results.arrow_forward
- Here are the progeny of this cross: (Note that the categories are not in any particular order.)Fly type # of prog. Phenotype symbols Categorywt eyes black body wt wings 97 grn+ blk crv+Green eyes black body curved wings 709 ParentalGreen eyes wt body wt wings 9Green eyes black body wt wings 162wt eyes wt body wt wings 727wt eyes black body curved wings 12wt eyes wt body curved wings 179Green eyes wt body curved wings 105Total = 2000 9.) Write the phenotype symbols in the right-hand column. The first one has been done for you.10.) Next to that, label all fly categories as parental (NCOs), SCOs, and DCOs. One has been donefor you.11.) After each SCO/DCO label, write which gene got “unlinked” in these offspring.12.) Put these three genes into a genetic map in the proper order.13.) Calculate the genetic distance between the genes and label the map with these distances.14.) Calculate the cross-over interference15.) Return to questions #1-6 above. For question 6, you gave your opinion, but…arrow_forwardConsider the selfed offspring of a AaBbCcDd individual: What is the probability that offspring will have the following genotypes: AABBCCDD AaBbCCDd ● • A_B_C_D_ BbCCDd ● ● ● ● ● 1/4 x 1/4 x 1/4 x 1/4= 1/256 1/2 x 1/2 x 1/4 x 1/2= 1/32 3/4 x 3/4 x 3/4 x 3/4= 81/256 1 x 1/2 x 1/4 x 1/2= 1/16 Same genotype as the parent? A.1 B. 1/1/12 C. D.1/8 E.1/16arrow_forwardThe mother of a family with 10 children has blood typeRh+. She also has a very rare condition (elliptocytosis,phenotype E) that causes red blood cells to be oval rather than round in shape but that produces no adverseclinical effects. The father is Rh− (lacks the Rh+ antigen)and has normal red blood cells (phenotype e). The children are 1 Rh+ e, 4 Rh+ E, and 5 Rh− e. Information isavailable on the mother’s parents, who are Rh+ E andRh− e. One of the 10 children (who is Rh+ E) marriessomeone who is Rh+ e, and they have an Rh+ E child.a. Draw the pedigree of this whole family.b. Is the pedigree in agreement with the hypothesisthat the Rh+ allele is dominant and Rh− is recessive?c. What is the mechanism of transmission ofelliptocytosis?d. Could the genes governing the E and Rh phenotypesbe on the same chromosome? If so, estimate the mapdistance between them, and comment on your resultarrow_forward
- Draw a Punnett square for the dihybrid cross described below (it is the same story as given for question 8, above) and use it to fill in the blanks correctly in the text that follows. NOTE: Please type in whole numbers, no symbols. 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…arrow_forwardThe aldose gene is 10 map units from the fructose gene. (A = dominant allele, can produce aldose; a = recessive allele, cannot produce aldose; F = dominant allele, can produce fructose; f = recessive allele, cannot produce fructose). You cross worms with the chromosomes shown below. What is the probability that offspring will be able to produce only aldose (and not fructose)? A f a f ----- X ----- a F a f A) 10% B) 5% C) 90% D) 45% E) 0% F) 25% G) 50% H) 33%arrow_forwardShow a completed punnett square for the following Dihybrid Cross: AaBb x AAbb. A=normal; a= short; B=bright; b=dull. Show your work for full credit. Give the expected genotypic and phenotypic ratios.arrow_forward
- The picture below represents a G1 cell from a newly discovered species that uses the X/Y sex determination system. Alleles for the different autosomal genes of interest are indicated on the chromosomes, and genes R and T are 36 cM apart. Red lines show maternal chromosomes, and blue lines show paternal chromosomes. Answer the following questions about this individual. a) This individual can produce a pool of different gametes. List any TWO potential gametes that will occur at different frequencies. Give the alleles of each gamete contained within a set of brackets, and indicate the expected frequency (up to two decimals) at which that gamete would occur. b) Name any two loci in this individual that can result in recombination in the gametes via the process of independent assortment. c) What is the sex of this individual? Justify your answer. d) Give the term that will describe the morphology of the chromosome that carries the R/T alleles. e) Provide the genotype of the fertilizing male…arrow_forwardA large ear of corn has a total of 433 grains, including 271 purple & smooth, 73 purple & shrunken, 63yellow & smooth, & 26 yellow & shrunken. Conduct a X^2 analysis to determine if this ear of corn was produced by adihybrid cross (PpRr x PpRr) resulting in a phenotypic ratio of 9:3:3:1.arrow_forwardIn Figure 5-33, which is the rarest λ genotype producedin the initial lysate?arrow_forward
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