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- olving Monohybrid Inheritance single gene. crosses. The henotypic traits controlled by a single gene. The problems are to give you practice in problem solving using Mendelian genetics. A dominant gene (W) produces wire-haired texture in dogs; its recessive allele (w) produces smooth hair. A group of heterozygous wire-haired individuals are crossed and their F1 progeny are then test-crossed. Determine the expected genotyplc and phenotypic ratios among the test cross progeny:6 Huntington's disease causes damage to the nervous system. It is an inherited condition caused by a dominant allele (H). Only individuals who are homozygous for the recessive allele (h) are protected from the disease. The diagram shows the inheritance of Huntington's disease in a family. A F G H K M N Male Female Male with Female with Huntington's disease Huntington's disease (a) Complete the table to show how many people in the diagram fit each description. The first one has been done for you. Number of people who fit the description Description male 7 female with Huntington's disease homozygous recessive heterozygous homozygous dominant188IS Add-oIS al text BIUA E E E E Times New. 12 1 4. Purple fur (P) is DOMINANT in monsters. Yellow fur (p) is RECESSIVE. What is the genotype of a PURE PURPLE monster? = What is the GENOTYPE of a HETEROZYGOUS purple monster? What is the GENOTYPE of a YELLOW monster? Using the punnet square above, make a cross between TWO HETEROZYGOUS PURPLE MONSTERS. POSSIBLE OFFSPRING GENOTYPES PHENOTYPES edu gear El
- CO-DOMINANCE/I... % Ø Oll ^ 6- 3. A homozygous red gummy bear is crossed with a homozygous yellow gummy bear (red and yellow mixed together make orange). You predicted that all the offspring will be orange gummy bears. Create a Punnett Square, then explain if your prediction was correct. 13 r y 2 / 3 | - h ? & 71 u * 82 87% i + | © k ( 93 O D § ) 014 Р 11 -½ { [ [ ^ 11 = ¾ 11 8 '{' \For the following crosses in model organisms that produce a lot of progeny per cross: (Read each statement carefully. Select all of the statements below that are true (that you agree with). Leave any statements that are false (that you do not agree with) un- selected.) a two-factor cross of parents that are dominant for both characteristics should result in all progeny being dominant-traited for both characteristics. a one-factor cross of two homozygous parents is expected to have the same phenotypic and genotypiC ratios. a three-factor cross of AaBbDd x aabbdd should result in a 1:1 three-factor phenotypic ratioRolly's APPLE ACK Free stu X O Leaming Lab Varsity Tutors + tutors.com/learning-lab/assessment/2eb9c784-88d7-4224-91d8-87c72f8edeb1 r rr rr Which best describes the parent genotypes in the punnet square provided? A Homozygous dominant and heterozygous B Two heterozygous C| Homozygous recessive and homozygous dominant D Homozygous recessive and heterozygous I don't know Next Question
- Classical Mendelian Genetics, Incomplete Dominance, Codominance, and Multiple Alleles 1. Complete the table given below regarding the phenotype and genotype ratios in completely dominant traits. R and r represent the dominant and recessive allele, respectively. Type of Cross rrx rr RR x rr Rrx rr Rrx Rr Genotype Ratio Phenotype Ratio RR x Rr RR X RR *How would the genotype ratios be affected if the mode o inheritance in incomplete dominance? Codominance? 2. In dogs, barer trait is controlled by a dominant gene D and the silent trait by the recessive gene d. Normal tail is dependent on a dominant gene M and the screw m. Give the probable genotypes of the parents in the following crosses: Phenotype of parents Phenotypes of progeny Genotypes of parents Barker Barker Silent Silent normal screw normal normal a. silent normal x silent normal b. barker normal x silent normal 0 0 6 2 7 2 8 3 c. barker normal x silent screw 4 5 5 3 d. barker screw x silent normal 6 0 0 0 e. barker screw x…Trivla Game Show _Make Your Own Tri ngston.schoology.com/common-assessment-delivery/start/4789189591?action=onresume&submissionld=463322566 Dillon WF g Aa v Done In guinea pigs, black hair (B) is dominant to white hair (b) and rough hair (R) is dominant to smooth hair (r). What are all the possible genotypes of a guinea pig that has black, rough hair? (Select all that apply.) O BBRR BBRr BBrr BBRR BbRr O bbRR O bbRr O bbrr O Black O White O Rough OSmooth O Rough O SmoothO e. Parent 2: Parent 1: Parent 2: ¡Ai QUESTION 9 Bi QUESTION 8 Let's assume that, in dragons, red scales (B) are dominant to green scales (b), and long tongues (S) are dominant to short tongues (s). The genes that determine these characteristics assort independently. A homozygous red, long-tongued dragon is crossed with a homozygous green, short-tongued dragon. If an F1 dragon is crossed to a homozygous green and homozygous short-tongued dragon, what phenotypes and proportions are expected in the offspring? O a. 100% green and long-tongued O b. ½ green and short-tongued & ½ red and long-tongued O c. ½ red and short-tongued & ½ green and long-tongued O d.% red and long-tongued, % red and short-tongued, ½ green and long-tongued, % green and short-tongued O e. 9/16 red, long-tongued, 3/16 green, long-tongued, 3/16 red, short-tongued, 1/16 green, short-tongued Save and Submit to save and submit. Click Save All Answers to save all answers. 000 MacBook Air
- The ff. mothers, (A) through (E) with given phenotypes, each produced one child whose phenotype is described. For each child, select as the father ONE of the five males whose genotypes are given. Genotype of male (1) IA i MN rr XAY (2) IB i MN RR XAY (3) ii NN rr XAY (4) ii MM rr XaY (5) IAIA MN RR XaY Give possible genotypes of all the mothers and children: Mother Maternal Phenotype Phenotype of child Possible father (write number only) A Type A, M, Rh+, normal color vision Type O, M, Rh+ , normal color vision, male B Type B, N, Rh-, colorblind Type O, N, Rh-, colorblind male C Type O, M, Rh-, normal color vision Type A, MN, Rh+, colorblind, female D Type A, N, Rh+, normal color vision Type AB, MN, Rh+, normal color vision, female E Type AB, MN, Rh- , normal color vision Type AB, M, Rh-, colorblind, maleSchool - Dan. AClassroom What's Your Guardi. V Authentic Italian Re YouTube ver sheet ☆回 Add-ons Help Last edit was seconds ago BIUA Times New... - 14 1 .1 2 II| 4 5 Phenotype: % Go To Next Page 4. Cross a Green flower (YB) with another green flower. Y = yellow, B = Blue. Is this a codominant or an incomplete dominance problem? What is the probability of the offspring genotypes and phenotypes? (Use a Punnett Square and make a table as seen in the problems above to give genotypes and phenotypes) Genotype: edu gear lii lilıTopic: Penetrance. Petal number is controlled by a single gene in merigonias. The gene has a completely dominant wild type allele F that makes a plant have five petals and a mutant recessive six petal allele(f). However the six petal trait is only 50% penetrant. You do the cross Ff x Ff. What fraction of the progeny do you have the 6 petals? what is the meaning for 50% penetrant.