BackgroundProblem N° 5Objective: To apply classical transmissiongenetics in a trihybrid cross.Flower position, stem length, and seed shape were three of the traitsthat Mendel studied. Each is controlled by an independently assortingBOTANIQUE.Légumineusesgene and has dominant and recessive expression:Solve it using the rules of probability insteadAxial (flowers originate along the stems) is dominant over terminal(flowers on top of the stem).of Punnet squares, bifurcation forks, ormental visualization.Long ("tall") stem is dominant over short stem ("dwarf").Let's use the symbols T, t for genesdetermining axial vs. terminal; Le, le for tallvs. dwarf; K, k for round vs. wrinkled (fromRound seed is dominant over wrinkled.If a plant that is heterozygous for all three characters were allowed toself-fertilize, what proportion of the offspring would be expected to beas follows?the German noun knitter = wrinkle).One final note: the phenotype "tall plant"may be conferred by the genotype Lele ora. homozygous for the three dominant traitsLele; this can be abbreviated as Le-, wherethe dash represents the alternative allele.The homozygous lele can only show therecessive phenotype.b. homozygous for the three recessive traitsSince an organism only shows the recessivephenotype when it is homozygous for thatparticular genotype, a homozygousrecessive is usually referred as the recessive.heterozygous (for all three characteristics)C.POIS binailleveneC dnd e lwaliee al ddn ai la mod l dead d fare ce d griPISUMd. homozygous for axial and tall, heterozygous for seed shapeBackgroundProblem N° 6Objective: To illustrate the differences betweenthe dominant/recessive model vs. partial (orincomplete) dominance model. In incompletedominance, the phenotype is neither dominantnor recessive, but intermediate instead.Explain why a monohybrid cross of snapdragon plants carrying white flowers with red-flowered plantsproduces all pink flowers in the F1 generation. [Hint: What makes flowers red, purple, yellow, etc.?]a.b.What are the genotypic and phenotypic ratios of snapdragon flowers in the F2 generation?True breeding individual = Homozygous foreither trait. BackgroundProblem N° 7Objective: To illustrate the use of testcrosses to establish unknown genotypes.In dogs (Canis familiaris), there is a hereditary deafness caused by the recessive gene d. A kennel owner has amale dog (Fido, of course) that she wants to use for breeding purposes. She comes to you for genetic advice. Thedog can hear, so you know that his genotype is either DD or Dd.For this problem, you are allowed to usePunnett squares or pedigrees.If Fido's genotype were Dd, the owner would not use him for breeding so that the deafness gene will not bepassed on. The dog's genotype can be ascertained via a test-cross with a deaf female (dd). Draw Punnett squaresor pedigrees to illustrate these two possible crosses.A test cross uses an individual whosegenotype we can be certain of, i.e., ahomozygous recessive for a particular trait.Such organism is crossed with one whosegenotype we do not know (remember thehomozygous recessive always express therecessive phenotype). In this case, a dd dogwill be deaf and a hearing dog can be DD orDd (i.e., D- represents both homozygousdominant and heterozygous until thegenotype is determined).O000우The results of a test cross will be eitherdd X DD → All Dd (all hearing), the unkonwnUse fractions or decimals to indicate proportions: In each case, what proportion of the offspringа.genotype is DD;would be expected to be hearing?ordd X Dd → ½ Dd + ½ dd (half hearing,half deaf), the unknowngenotype is Dd.b.What proportion would be expected to be deaf?So, the proportion of hearing dogs willindicate the unknown genotype.C.How could you tell the genotype of the male dog (in parental generation P)?d. Also, using a Punnett square, how would you explain to the kennel owner that two hearing dogscould produce deaf offspring? Indicate to the kennel owner what kind(s) of gametes (ova or sperm)could hearing dogs produce to conceive a deaf puppy

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Problem N° 5 Flower position, stem length, and seed shape were three of the traits that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression: Axial (flowers originate along the stems) is dominant over terminal (flowers on top of the stem). Long ("tall") stem is dominant over short stem ("dwarf"). Round seed is dominant over wrinkled. If a plant that is heterozygous for all three characters were allowed to self-fertilize, what proportion of the offspring would be expected to be as follows? a. homozygous for the three dominant traits b. homozygous for the three recessive traits c. heterozygous (for all three characteristics) d. homozygous for axial and tall, heterozygous for seed shape

Problem N° 5 Flower position, stem length, and seed shape were three of the traits that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression: Axial (flowers originate along the stems) is dominant over terminal (flowers on top of the stem). Long ("tall") stem is dominant over short stem ("dwarf"). Round seed is dominant over wrinkled. If a plant that is heterozygous for all three characters were allowed to self-fertilize, what proportion of the offspring would be expected to be as follows? a. homozygous for the three dominant traits b. homozygous for the three recessive traits c. heterozygous (for all three characteristics) d. homozygous for axial and tall, heterozygous for seed shape

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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Background Problem N° 6 Objective: To illustrate the differences between the dominant/recessive model vs. partial (or incomplete) dominance model. In incomplete dominance, the phenotype is neither dominant nor recessive, but intermediate instead. a. Explain why a monohybrid cross of snapdragon plants carrying white flowers with red-flowered plants produces all pink flowers in the F1 generation. [Hint: What makes flowers red, purple, yellow, etc.?] b. What are the genotypic and phenotypic ratios of snapdragon flowers in the F2 generation? True breeding individual = Homozygous for either trait.
please help with these two problems. They are not hard I just need a better understanding?
Question 3 first worksheet help
needing help with (c) and (d)
Show me how to do this punnet square problem
Please do not give solution in image format thanku
Genetic Problems Read the following genetic problems, and then complete the activities and questions for each problem 1. In the pea plant, the green pod color allele is dominant to the yellow pod color allele. o Set-up a monohybrid cross between a pea plant that is heterozygous and a true-breeding yellow pod pea plant. o What percentage of the offspring produced from this cross would you expect to have yellow pods? 2. Within a mouse population, the black fur allele (B) is dominant to the white fur allele (b) and the short whisker allele (S) is dominant to the ngs long-whisker allele (s). o A heterozygous black-furred short-whiskered mouse is crossed with a homozygous white-furred long-whiskered mouse. o What percentage of the offspring will be black-furred with long whiskers? 3. A physician is examining the blood types of children from a family. o Child 1 has blood type AB, Child 2 has blood type B, Child 3 has blood type B, and Child 4 has blood type A. o Based on the phenotypes of…
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elp Normal ▾ B IU A - Corn Breeding / Genetic Problem A corn breeder is working with 2 inbred corn plants. He is interested in two traits: High yield (produces lots of corn) and drought tolerance (does not need lots of water). High yield (H) is dominant to low yields (h). Drought tolerance (D) is dominant to drought intolerance (d). M BEE Inbred 1 is true breeding for drought tolerance and low yield (DDhh). Inbred 2 is true breeding for drought intolerance and high yield (ddHH). In the parent cross (P) the corn breeder crossed inbred 1 with inbred 2. The resulting F₁hybrid (DdHh) was heterozygous for drought tolerance and high yields. Phenotypically all the corn needs little water to grow and produces lots of corn. This is the corn that is sold to farmers, planted in the Spring and harvested in the Fall. Explain why the farmer does NOT hold back some of the Fall harvested (F₂) kernels and plant them next Spring. You must show your work. (Hint: As a %, how many of the F2 kernels still…