CAMPBELL BIOLOGY V.2(BHCC>CUSTOM<)
18th Edition
ISBN: 9781323655573
Author: Urry
Publisher: PEARSON C
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Chapter 15, Problem 4TYU
A planet is inhabited by creatures that reproduce with the same hereditary patterns seen in humans. Three
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In a population of martians the hair color (H= purple or h=green) and eye-color (E=red or e=yellow) are very close together on the same chromosome. One heterozygote male, who knows that his dad was homozygous dominant and his mom had green hair and yellow eyes is considering fertilizing a homozygous recessive female. How many green-haired yellow-eyed offspring (out of 16) would they expect to have?
In genetics, the dash symbol (–) is a “wild card” that stands for either the dominant allele or the recessive allele; for example, A– means the individual has either the genotype AA or Aa. Two genes that undergo independent assortment affect coat color in Duroc pigs. Each gene has two alleles, one of which is dominant for coat color. Genotypes of the form A– B– are red, those of the form A– bb and aa B– are sandy, and genotype aa bb is white. What ratio of red:sandy:white is expected from the cross Aa Bb x Aa Bb?
In pre-DNA fingerprint days, blood type was often used in paternity disputes to exclude possible fathers. If the man in question has type B blood, and the child has type A blood, and the mother has type AB blood, can the man be the father of the child?
Group of answer choices
Yes, but only if he is homozygous for the B allele
No
Yes, but only if he is heterozygous for the B allele
Yes, but only if both of his parents were Type AB
Chapter 15 Solutions
CAMPBELL BIOLOGY V.2(BHCC>CUSTOM<)
Ch. 15.1 - Which one of Mendel's laws describes the...Ch. 15.1 - MAKE CONNECTIONS Review the description of...Ch. 15.1 - WHAT IF? Propose a possible reason that the first...Ch. 15.2 - A white-eyed female Drosophila is mated with a...Ch. 15.2 - Neither Tim nor Rhoda has Duchenne muscular...Ch. 15.2 - MAKE CONNECTIONS Consider what you learned about...Ch. 15.3 - When two genes are located on the same chromosome,...Ch. 15.3 - VISUAL SKILLS For each type of offspring of the...Ch. 15.3 - Prob. 3CCCh. 15.4 - Prob. 1CC
Ch. 15.4 - Prob. 2CCCh. 15.4 - Prob. 3CCCh. 15.5 - Gene dosagethe number of copies of a gene that are...Ch. 15.5 - Reciprocal crosses between two primrose varieties,...Ch. 15.5 - WHAT IF? Mitochondrial genes are critical to the...Ch. 15 - What characteristic of the sex chromosomes allowed...Ch. 15 - Why are males affected by X-Iinked disorders much...Ch. 15 - Why are specific alleles of two distant genes more...Ch. 15 - Prob. 15.4CRCh. 15 - Explain how genomic imprinting and inheritance of...Ch. 15 - A man with hemophilia (a recessive, sex-linked...Ch. 15 - Pseudohypertrophic muscular dystrophy is an...Ch. 15 - A wild-type fruit fly (heterozygous for gray body...Ch. 15 - A planet is inhabited by creatures that reproduce...Ch. 15 - Using the information from problem 4, scientists...Ch. 15 - A wild-type fruit fly (heterozygous for gray body...Ch. 15 - Assume that genes, A and B are on the same...Ch. 15 - Two genes of a flower, one Controlling blue (B)...Ch. 15 - You design Drosophila crosses to provide...Ch. 15 - Banana plants, which are triploid, are seedless...Ch. 15 - EVOLUTION CONNECTION Crossing over is thought to...Ch. 15 - SCIENTIFIC INQUIRY DRAW IT Assume you are mapping...Ch. 15 - WRITE ABOUT A THEME: INFORMATION The continuity of...Ch. 15 - SYNTHESIZE YOUR KNOWLEDGE Butter flies have an X-Y...
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- Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?arrow_forwardFrom the pedigree shown here, answer the following questions with regard to individual VII-1. A. Who are the common ancestors of her parents? B. What is the inbreeding coefficient for this individual?arrow_forwardIf 83% of the Elmo Muppet population has a condition that makes them refer to themselves in the third person, and you know that this condition is the result of the homozygous recessive genotype (ee)… You allow the Elmos to reproduce for a while and discover that in the heterozygous condition, the Elmos are capable of referring to themselves in the 3rd person, but only do so upon request. The PBS station puts out a casting call for a locally-produced Sesame Street. This is your chance to make a little cash with all the Elmos you have in your basement. It turns out, however, that the heterozygous Elmos are strongly preferred by the production crew, since they can turn off the third person self-references. So: EE Elmos are NOT acceptable actors, Ee Elmos are preferred actors, and ee Elmos are acceptable, but only when Ee Elmos are unavailable. If having a gig on Sesame Street is critical to Elmo survival: rank the fitness levels of the three genotypes and discuss whether/how long the E…arrow_forward
- On the planet of Caracas, in the Stellar Solorais KaChunka Galaxy, there is a population of Nuggetarians. A phenotype of interest is Peanutus02. There are two alleles for the autosomal gene associated with Peanutus02. A true-breeding cross between a female with the Peanutus02 (disease) phenotype and a male with the wild-type phenotype produces the following offspring counts: Phenotype Number of Males Number of Females Wild-Type 550 562 Peanutus02 0 0 Of the following choices, the most likely mode of inheritance for the Peanutus02 phenotype is: On the planet of Caracas, in the Stellar Solorais KaChunka Galaxy, there is a population of Nuggetarians. A phenotype of interest is Peanutus02. There are two alleles for the autosomal gene associated with Peanutus02. A true-breeding cross between a female with the Peanutus02 (disease) phenotype and a male with the wild-type phenotype produces the following offspring counts: Phenotype Number of Males Number of Females Wild-Type…arrow_forwardTay-Sachs disease is a rare human disease in which toxic substances accumulate in nerve cells. The recessive allele responsible for the disease is inherited in a simple Mendelian manner. For unknown reasons, the allele is more common in populations of Ashkenazi Jews of eastern Europe. A woman is planning to marry her first cousin, but the couple discovers that their shared grandfather’s sister died in infancy of Tay-Sachs disease.a. Draw the relevant parts of the pedigree, and show all the genotypes as completely as possible. b. What is the probability that the cousins’ first child will have Tay-Sachs disease, assuming that all people who marry into the family are homozygous normal?arrow_forwardGenetics, Mating took place between males and females of a fruit fly, and individuals appeared With the following proportions and attributes: 25% Female flies with irregular wings and 25% female flies with irregular wings 25% male flies with irregular wings and 25% male flies with irregular wings . . If a symbol for the adjective of regular wings is symbolized by the symbol B and if a symbol for irregular wings with the symbol b, The genotypes of the parents are? 1. bB, Bb 2. bb,Bb 3. XB Y, Xb Xb 4. XbY, XBXbarrow_forward
- In pedigrees, individuals are usually specified by using a Roman numeral for their generation in the chart and an Arabic number for their position (reading left to right) within that generation. If we use the letter c for the allele that causes cystic fibrosis, what are the genotypes of individuals III-3 and III-4 (the third and fourth individuals shown in generation III) in the pedigree that shows this disease?arrow_forwardStickleback fish that are deep in the ocean can experience a mutation wherein they experience a "loss-of-function" of their spikes. Heterozygotes actually look the same as homozygous wild types at this depth. If a homozygous deep-water wild type stickleback mated with a homozygous shallow-water wild type, what phenotypes would you expect to see?arrow_forwardA planet is inhabited by creatures that reproduce with the same hereditary patterns seen in humans. Three phenotypic characters are height (T = tall, t = dwarf), head appendages (A = antennae, a = no antennae), and nose morphology (S = upturned snout, s = downturned snout). Since the creatures are not “intelligent,” Earth scientists are able to do some controlled breeding experiments using various heterozygotes in testcrosses. For tall heterozygotes with antennae, the offspring are tall antennae, 46; dwarf antennae, 7; dwarf no antennae, 42; tall no antennae, 5. For heterozygotes with antennae and an upturned snout, the offspring are antennae upturned snout, 47; antennae downturned snout, 2; no antennae downturned snout, 48; no antennae upturned snout, 3. Calculate the recombination frequencies for both experiments.arrow_forward
- A genetic engineer is going to cross two watermelon plants to produce seeds for a spring planting. He is breeding for size, and wants to have as many watermelons with the phenotype for long shape as possible. In watermelons, the allele for short shape (R) is dominant to the allele for long shape (r). Would crossing a watermelon homozygous recessive for the trait with a watermelon heterozygous for the trait give the most long watermelons possible? Explain your answer using Punnett Squares.arrow_forwardIn humans, brown eyes are dominant to blue eyes, and the ability to roll your tongue is dominant to not being able to roll your tongue. If a mother who was heterozygous for eye colour and homozygous recessive for tongue rolling had a baby with a father that was heterozygous for both traits, what is the probability that their child would have blue eyes and could roll their tongue?arrow_forwardIn humans, there is a gene called sickle-cell anemia, which produces severe anemia when homozygous recessive. The name of the disease comes from the fact that many red blood cells take on abnormal sickle shape. People homozygous recessive for this trait usually die before adulthood. Heterozygous people appear normal, but when a sample of the blood is held under low oxygen concentration, the red cells take on the sickle shape. A young woman about to be married is concerned about her future children because her brother died of sickle-cell anemia. A sample of her blood sickles under low oxygen concentration, but that red blood of her perspective husband remains normal. What can you say about the woman and the man, and about their future children? HINT: Use Punnett square to help you and receive full creditarrow_forward
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