A B, and Cgenes are linked with Bin the middle. AB are 12 cM and BC are 20 cM apart. If ABdabC is testcrossed to abcabc, what is the expected number of individuals of Aa bb Ccgenotype if 1000 progeny result from this testcross with a coefficient of coincidence (9 of 0.5? Oa. 42 Ob. 48 Oc 47 Od. 60 Oe. 54

A B, and Cgenes are linked with Bin the middle. AB are 12 cM and BC are 20 cM apart. If ABdabC is testcrossed to abcabc, what is the expected number of individuals of Aa bb Ccgenotype if 1000 progeny result from this testcross with a coefficient of coincidence (9 of 0.5? Oa. 42 Ob. 48 Oc 47 Od. 60 Oe. 54

Name: A B, and Cgenes are linked with Bin the middle. AB are 12 cM and BCar
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Description: A B, and Cgenes are linked with Bin the middle. AB are 12 cM and BCare 20 cM apart. If ABdabCis testcrossed to abcabc, what is theexpected number of individuals of Aa bb Cc genotype if 1000 progeny result from this testcross with a coefficient of co

Biology: The Unity and Diversity of Life (MindTap Course List)

14th Edition

ISBN:9781305073951

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter13: Observing Patterns In Inherited Traits

Section: Chapter Questions

Problem 11GP: A single allele gives rise to the Hbs form of hemoglobin. Individuals who are homozygous for the...

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Genotype Phenotype FFEE 3 Feelers, thick okin FFTT 2 feclers, then skin %3D offspring Ft ffFt ffft ffFE ffft 50% 2 feelers with ff thin skin heteorozyjous for both TT FLTT FETT FETT FETT FT FFTE ff Tt FF TE Ff Tt 中TFTEIC TE FF TE FPてヒ 50% 3 ¢eelers 1with thick skin hemozyyous domirant for fe ders Cnd hetero us for skin 2) Volantemn lumbricus have feelers which helps them sense their environment. 2 feelers (F) is dominant over 3 feelers (f). They can also have different skin thickness, thick skin (T) is dominant over thin (t). An individual who is heterozygous for 2 feelers with thin skin is crossed with an individual that has 3 feelers and is homozygous for thick skin. Assess the following answer to the above scenario, identify and explain mistakes that were made.
Po es) 20 F1 KEY CROSS- RR X IT RESULTS- Rr GENOTYPE- RR IT Ơ 天 2 T 1. To determine the results of this experiment, complete the Punnett squares below by writing the correct allele(s) in the space provided P Generation PHENOTYPE- Round seeds x Wrinkled seeds F4 R 1, a r Rr F5 Rr RR Rr RR R X Rr F6 Rr Rr AaBbCcDdE AaBbCcDdE Aa BbCcD Aa BbCcD Heading 1 Normal No Spacing IT IT I F1 F2 F2 KEY CROSS-RrxRr RESULTS ace provid GENOTYPE 2. hat phenotypes are present in the F₁ generation? What gen'otypes are present? F7 3. What phenotypes are present in the F2 generation? In what ratio are they present? PHENOTYPE 4. Does your analysis support or refute Mendel's hypothesis of dominant and recessive inheritance? Explain. DII F8 Title F9 Aa BbCcDdEe Focus Subtitle E E 1
. A plant of genotypea bA Bis testcrossed witha ba bIf the two loci are 10 m.u. apart, what proportion of progenywill be AB/ab?
Tall, Axialnuot ert tna apnibnit alsbneM aneg s ni tisTI s to 2ovitsnietis e1s aelellA Then, perform a cross to determine the genotype and phenotype of the offspring. B. Determine the genotype of the gametes that will come from parent 1 and parent 2. Parents (2) TtAa the tr (1) TtAa Phenotype Tall, Axial D deirw efoug for the Gamete genotype s awon ai selelle emse erti liw ensp A vidual wi resa it. alells ins1sib riliw eneg oakns oRhters. vrld ertt of elei isi miel e ai sqyioner9 JBubivibni ns -X 30 tw Ned st o enmol Jer cenimaob isi 2elellelo isa brtt oi aete egytonea beea9 sewoda eepibeg A fassni to at s pniwollot vileuau nentons ot nolisonag Genotype of Offspring: Phenotype:
3.5 25 An individual of genotype AA Bb Cc DD Ee is test- crossed. Assuming that the loci undergo indepen- mETg dent assortment, what fraction of the progeny are expected to have the genotype Aa Bb Cc Dd Ee? oldizz0g S 16 J60 016 160
feeve uld eved lliw en isnebi l Hiwi shi Seeye euld eved lliw erde 1o erd ,lemetst ai niwi nerth ino AaBbC X AaBbCc, niwt1er would be the probability of offspring with: 4. Based on the mating a.) either the phenotype A-bbC- or the genotype aaBBCC? b.) the genotype AABBCC or the genotype AaBbCc?
Albino rabbits (lacking pigment) are homozygous forthe recessive c allele (C allows pigment formation).Rabbits homozygous for the recessive b allele makebrown pigment, while those with at least one copy ofB make black pigment. True-breeding brown rabbitswere crossed to albinos, which were also BB. F1 rabbits, which were all black, were crossed to the doublerecessive (bb cc). The progeny obtained were 34black, 66 brown, and 100 albino.a. What phenotypic proportions would have beenexpected if the b and c loci were unlinked?b. How far apart are the two loci?
A single allele gives rise to the Hbs form of hemoglobin. Individuals who are homozygous for the allele (HbS/HbS) develop sickle-cell anemia (Section 9.6). Heterozygous individuals (HbA/HbS) have few symptoms. A couple who are both heterozygous for the HbS allele plan to have children. For each of the pregnancies, state the probability that they will have a child who is: a. homozygous for the HbS allele b. homozygous for the normal allele (HbA) c. heterozygous: HbA/HbS
oibai) s moil to to bludy 2019mnes or to in aldiz 212019m nool botibni arrow 100 muzes, nisgn esqyjonsg or to oily r R RE Parent 1 Question 5. In a related species of dragons, the fire breathy and flying loci are present on the same chromosome. Non-dividing diploid cells from two dragons are shown above. 22019 is most flest tedi non si to od to be in baribong A. What is the genotype of parent 1? (Use chromosomal notation if appropriate) non mowied OF Rr/11 B. What is the genotype of parent 2? (Use chromosomal notation if appropriate) L Rr/L LI C. Draw the what a cell from parent 2 would like in PROPHASE of MEIOSIS I. For the purpose of this question, ASSUME NO CROSSOVER between the fire breathing and flying loci. Make sure you label alleles on each DNA molecule. xRixr Parent 2 D. Draw what the two daugher cells would look like after the cell you drew in (C) has undergone meiosis I (the cells are ready to undergo meiosis II; assume no crossover between the indicated loci in meiosis I).…
To determine the recombination frequency of two linked genes, what genotypes of organisms are crossed? O Amy cross between two individuals will yield the same results O A homarygous dominant individual with a homozvgos recesshve O A heterozygous individual with a homozypous dominant individual O A heterozygous individual with a homozygous recessive individual O Two heterorygous individuals
non OBYTOT SEI deutet bos nuol 3. Based on the total number of kernels of corn you counted, calculate the expected number for each phenotype if the parents of this corn were both RrTt. Show how you calculate these numbers below and en transfer them to the observation table (155 Kernels) then INO 4 4. Using chi-square analysis to compare the actual phenotype and expected phenotype numbers determine if you can statistically accept that both parents have the genotype RrTt. Your chi square data will be recorded to helow. nb number) which allows for
Date Name Punnett Square Practice Worksheet 1) For each of the genotypes (AA, Aa or aa) below determine what the phenotype would be. Purple flowers are dominant to white flowers. PP Pp PP Hairy knuckies are dominant to non-hairy knuckles in humans. Hh hh HH Bobtails in cats are recessive. Normal tails are dominant. Tt tt TT 2) For each of the following write whether it is homozygous dominant, heterozygous or homozygous recessive. AA gg Pp Ii TT Use the following information for questions 3-5: In dogs, the gene for fur color has two alleles. The dominant allele (F) codes for grey fur and the recessive allele (f) codes for black fur. 3) The female dog is heterozygous. The male dog is homozygous recessive. Figure out the percentage or ratio of possible phenotypes and genotypes of their puppies by using a Punnett Square. Plae the oileles for the mule parent helm One alele n e h lie: % of possible Genotypes: FF Plar the alfele f he emle p r ti side One elleir on ch tine Ff: % of possible…