Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN: 9780134580999
Author: Elaine N. Marieb, Katja N. Hoehn
Publisher: PEARSON
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Transcribed Image Text:Consider this hypothetical example. Bluegills can either have green eyes or brown eyes. The trait
is controlled by a single gene. Bluegills can either have long or short fins. This trait is controlled by
a single gene that is on a different chromosome from the eye color gene. Having green eyes is
dominant (G), while having brown eyes is recessive (g). Having long fins is dominant (L), while
having short fins is recessive (1). Your not-so-famous instructor will mate a male and female that
both are heterozygous for eye color and heterozygous for fin length. If these 2 bluegills produced
1000 offspring, about how many would have green eyes and short fins?
O 63
O 125
188
O 250
O 563
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- To analyze: According to the given information, wild-type allele of a gene has an A-T base pair at a particular location in its sequence, and a mutant allele of the same gene has a G-C base pair at the same location. Apart from this change, the sequences of the two alleles are same. Based on above information, a question is asked as whether this information tells us anything about the dominance relationship of the allele. If yes, we need to explain why. If no, we need to explain why not.arrow_forwardChoose one answer.arrow_forwardLabrador retrievers are commonly black, brown, or yellow. Two different genes determine coat color in Labrador retrievers. The B gene determines pigment color (B is black, bb is brown). The E gene determines whether the pigment can be deposited in the hair shaft: ee prevents any pigment from being deposited, resulting in yellow dogs. You want to breed Labrador retrievers, and you cross a true-breeding brown Labrador retriever to a true-breeding yellow Labrador retriever. All 15 of the F1 puppies are black. After the puppies grow up, you cross two of the F1 black dogs together. In the F2 generation you get 8 black puppies, 4 yellow puppies, and 3 brown puppies. Question 1 Part 1 What is/are the possible genotype(s) of the true-breeding brown lab in the P generation? bbee bbee Could be either answer bbEE or bbEe Could be either answer bbEE or bbEe bbEE bbEE bbEe bbEe Brown Brown Question 2 Part 2 What is/are the possible…arrow_forward
- Let's continue with these two genes that control the color and texture of tomatoes. Allele "R" generates a red coloration, and is dominant. Allele "r" generates a green coloration and is recessive. Allele "S" generates a smooth texture, and is dominant. Allele "s" generates a rough texture and is recessive. You now decide to cross your dihybrid RrSs but this one with a tomato with Rrss. RrSs x Rrss What is the probability of getting Red tomatoes with Rough texture in this new cross? Assume complete dominance and 100% penetrance and expressivity. 9/16 3/8 1/4 1/16 3/16 1/8 x 1/2arrow_forwardTable 3. Common Single Gene Characteristics in Humans For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 Characteristic Tongue rolling (T). Not able to roll = t Free earlobes (E) attached earlobes = e Dimples (D). No dimples = d Widow's peak (W) straight hairline = w Freckles (F). No freckles = f Interlocking fingers (I). When you clasp your hands together, having the left thumb on top = 1 Phenotype Genotypearrow_forwardSuppose that in a species of mouse, the gene for brown coat color is dominant over the color black. Another gene (C) is responsible for depositing color on the hair. A single dominant allele (C) is sufficient for the color (black or brown) to be deposited on the hairs; those without a copy of the dominant allele are albino. A female mouse homozygous recessive for both characters is crossed with a male mouse heterozygous for both characters. A) Give the phenotypic ratio of the progeny of F1. (Punnet grid if possible) B) What proportion of the offspring of the F1 will have the same genotype as either of its parents?arrow_forward
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- Hello Heliodors! (cont.) Trait A For Heliodors, having four arm-like extensions is dominant over having two. In a family of Heliodors, Hayden has four arm-like extensions like his mother Helen. His father, Harry, and sister, Holly both have two. What are the genotypes of everyone in this family of Heliodors? What is Hayden's genotype? O Not enough information to determine. AA (homozygous dominant) Aa (heterozygous) aa (homozygous recessive)arrow_forwardSuppose a geneticist is using a three-point test cross to map three linked rabbit morphology and behavioral mutations called si, sf, and H. The gene si is associated with the silky fur phenotype, and sf is associated with the short-footed phenotype. Both si and sf are recessive mutations with respect to wild type. H is a dominant mutation that confers the hyper phenotype. The geneticist first crosses true-breeding hyper rabbits to true-breeding silky fur, short-footed rabbits. Next, the geneticist backcrosses the F₁ progeny to the silky fur, short-footed parents, and obtains the results reported in the table. Phenotype hyper silky fur, short-footed short-footed silky fur, hyper silky fur short-footed, hyper silky fur, short-footed, hyper wild type Place the genes in the correct order in the chromosome. LLIIN HEL H Number 815 807 175 169 4 5 27 29 Answer Bank si sfarrow_forwardHello Heliodors! (cont.) Trait A For Heliodors, having four arm-like extensions is dominant over having two. In a family of Heliodors, Hayden has four arm-like extensions like his mother Helen. His father, Harry, and sister, Holly both have two. What are the genotypes of everyone in this family of Heliodors? What is Helen's genotype? O Not enough information to determine aa (homozygous recessive) Aa (heterozygous) O AA (homozygous dominant)arrow_forward
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