e last two joints of the idde inger are shortened. In the otheringers, the last two joints are fused into one and are also4119122---ETee C. BrachydactylyOurPedigree C. Brachvcactyy2 3458.10358OL11 121314TV24578 910FIGURE 4Formulating Generalizations1. In pedigree A, why do parents 13 and 14 in generation III havefour deaf children?2. In pedigree A, how can parents 1 and 2 have a deaf child eventhough neither parent is deaf?3. In pedigree A, why don't parents 8 and 9 and 1l and 12 have atleast one deaf child?4. List by number the persons in pedigree A for which youcannot determine a genotype. For example-II-1.5. Why can't you determine the genotypes of these pèople?6. In pedigree B, how can unaffected parents have an affectedchild?7. What is the genotype of the mothers of all affected females inpedigree B? Explain.8. What characteristic do all affected females in pedigree B havein common?9. List by number the persons in pedigree B for which youcannot determine a genotype. Explain why a genorypecannot be determined.10. Is the brachydactyly trait in pedigree C dominant orrecessive? Explain,6. One form of deafness in humans is a 17pe in which the individualinherits not only deafness but also the ability to tilk. Use symbolsBrach.*Pedigree A. Deaf MutismD for dominant, d for recessive.edigree A. Deaf Mutism78543.1314111291074.14151617181213119.10IV234FIGURE 2o* Pedigree B. HemophiliaHemophilia is a disease of the circulatory system. t is sometimescalled bleeder's disease because persons with the distase have a verylong clotting time when injured. An affected person could bleed todeath from minor wounds or intemal injuries. Use symbols H fordominant, h for recessive.HHitHh= nomel nn= hem@e.HH tHh=ノJu%3DPedigree B. Hemophilia12111.34III7.36.7111213 141516 179.1018IV1 2 3 45 67 8GURE 32.

Name: e last two joints of the idde inger are shortened. In the otheringers
Uploaded: 2024-03-20T06:42:55.000Z
Channel: Bartleby
Description: e last two joints of the idde inger are shortened. In the otheringers, the last two joints are fused into one and are also4119122---ETee C. BrachydactylyOurPedigree C. Brachvcactyy2 3458.10358OL11 121314TV24578 910FIGURE 4Formulating Generalizations

Pedigree A is showing autosomal recessive inheritance meaning the trait will only be expressed when…

Science

Biology

Use symbols 3 for dominant, b for recēssivé. dactyly is a condizicain which fingers are abnomally short. ers, the last Ewo joints are fused into one and are also Ewo joints of the iddle änger are shortened. In the other e C. Brachydactyly E Brachvca 4 9 10 5 6 7 8 9 10 11 12 13 14 2 3 4 8 9 10 FIGURE 4 ng Generalizations igree A, why do parents 13 and 14 in generation II have eaf children? igree A, how can parents 1 and 2 have a deaf child even a neither parent is deaf? gree A, why don't parents 8 and9 and 11 and 12 have at one deaf child? - number the persons in pedigree A for which you determine a genotype. For example-II-1. an't you determine the genotypes of these pèople? gree B, how can unaffected parents have an affected s the genotype of the mothers of all affected females in e B? Explain. haracteristic do all affected females in pedigree B have mon? number the persons in pedigree B for which you determine a genotype. Explain why a genorype be determined. brachydactyly trait in pedigree C dominant or

Use symbols 3 for dominant, b for recēssivé. dactyly is a condizicain which fingers are abnomally short. ers, the last Ewo joints are fused into one and are also Ewo joints of the iddle änger are shortened. In the other e C. Brachydactyly E Brachvca 4 9 10 5 6 7 8 9 10 11 12 13 14 2 3 4 8 9 10 FIGURE 4 ng Generalizations igree A, why do parents 13 and 14 in generation II have eaf children? igree A, how can parents 1 and 2 have a deaf child even a neither parent is deaf? gree A, why don't parents 8 and9 and 11 and 12 have at one deaf child? - number the persons in pedigree A for which you determine a genotype. For example-II-1. an't you determine the genotypes of these pèople? gree B, how can unaffected parents have an affected s the genotype of the mothers of all affected females in e B? Explain. haracteristic do all affected females in pedigree B have mon? number the persons in pedigree B for which you determine a genotype. Explain why a genorype be determined. brachydactyly trait in pedigree C dominant or

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter5: The Inheritance Of Complex Traits

Section: Chapter Questions

Problem 8QP: Clubfoot is a common congenital birth defect. This defect is caused by a number of genes but appears...

See similar textbooks

Similar questions

A corn geneticist wants to obtain a corn plant that hasthe three dominant phenotypes: anthocyanin (A), longtassels (L), and dwarf plant (D). In her collection ofpure lines, the only lines that bear these alleles are AALL dd and aa ll DD. She also has the fully recessive lineaa ll dd. She decides to intercross the first two and testcross the resulting hybrid to obtain in the progeny aplant of the desired phenotype (which would have to beAa Ll Dd in this case). She knows that the three genesare linked in the order written, that the distance between the A/a and the L/l loci is 16 m.u., and that thedistance between the L/l and the D/d loci is 24 m.u.a. Draw a diagram of the chromosomes of the parents,the hybrid, and the tester.b. Draw a diagram of the crossover(s) necessary toproduce the desired genotype.c. What percentage of the testcross progeny will be ofthe phenotype that she needs?d. What assumptions did you make (if any)?
HITCHHIKER'S THUMB: The distal (last) joint of the thumb can be bent back to forma 45 degree angle with the proximal joint (attached to your hand) if you arehomozygous recessive (hh) for this trait.Michael has a well-developed hitchhikers thumb, but neither of his parents show thetrait. If his parents have another child, what is the probability that he/she will alsohave a hitchhikers thumb?
A. C. Stevenson and E. A. Cheeseman studied deafness in a family inNorthern Ireland and recorded the following pedigree (A. C. Stevensonand E. A. Cheeseman. 1956. Annals of Human Genetics 20:177–231).
Pedigree attached shows an autosomal recessive genetic disease. G is the normal allele and g is the disease-causing allele. Individual 1’s father is heterozygous (*) and his mother is homozygous dominant. Other individuals in the pedigree may be carriers, but are not marked. The question mark (?) indicates that you do not yet know anything about this individual’s phenotype with regard to the disease. part a) What is the probability that individuals 1 and 2 will have a child (5) who is a boy with the disease (the child is unborn and the sex is not yet known)? a)1/8 b)1/4 c)0 d)1/16 part b) What is the probability that the daughter (6) that individual 3 and 4 just had will have the disease? a)1/8 b)1/6 c)1/4 d)1/12
Hi, I'm having trouble with my study guide for my upcoming genetics exam. If someone could please help with work shown and an explanation it would help so much! Thank you!! 2a. The pedigree below represents inheritance of rare condition. What pattern of inheritance is most consistent with the data? Assign alleles to all individuals to support your answer. If an allele is unknown, assign it a ? symbol. NOTE: Individuals whose phenotype or genotype cannot be determined are assumed to be unaffected and homozygous, unless otherwise indicated. 2b. In addition to the alleles you’ve indicated, describe 2 overall features of the pedigree that make it consistent with your chosen form of inheritance. 2c. Based on your mode of inheritance, what is the probability that the child of couple IV-4 x IV-5 will be affected? Show your work. attached is the pedigree
Tay-Sachs is a recessive lethal disease in which thereis neurological deterioration early in life. This diseaseis rare in the population overall but is found at relatively high frequency in Ashkenazi Jews from EasternEurope. A woman whose maternal uncle had the disease is trying to determine the probability that sheand her husband could have an affected child. Her father does not come from a high-risk population. Herhusband’s sister died of the disease at an early age.a. Draw the pedigree of the individuals described.Include the genotypes where possible.b. Determine the probability that the couple’s firstchild will be affected.
Albinism is an autosomal recessive condition where little or no melanin pigment is produced. Awoman (call her W) had an affected brother (B). This woman's daughter (M) is planning to have achild. The intended father (F) in this mating had a sister (S) who has albinism.a. Draw the pedigree of the family. Use defined symbols to indicate genotypes.
A woman who is heterozygous, Bb, has brown eyes; B (brown) isthe dominant allele, and b (blue) is recessive. One of her eyes,however, has a patch of blue color. Give three different explanationsfor how this might have occurred?
1. You have a line of rice that is true breeding for the recessive traits fo white rice (r) and dwarfisim (d). You cross a member of this line with an individual that is heterozygous at both loci and shows the dominant phenotypes of red grains (R) and tall height (D) yeilding the following progeny. 417 white grains-dwarf 403 red grains-tall 95 white grains-tall 85 red grains-dwarf a) what is the allelic configuration of the heterozygote? b) how far are the genes controlling grain color and plant height in (cM) c) what are the non cross over gametes d) what are the cross over gametes e) what genotype will result in white grains and tall heights (multplie answers) f) if you cross a heterozygote with the same allelic configuration as above, what is the probability that the offspring will have white grain and trall height? Give your answer to number to THREE decimal places not as a percentage (e.g 0.044
In peas, yellow seed (Y) is dominant over green seed (y). Inthe F2 generation of a monohybrid cross that begins when adominant homozygote is crossed with a recessive homozygote,you would expecta. three plants with yellow seeds to every plant with greenseeds.b. plants with one yellow seed for every green seed.c. only plants with the genotype Yy.d. only plants that produce yellow seeds.e. Both c and d are correct.
. A true-breeding strain of Virginia tobacco has dominantalleles determining leaf morphology (M), leaf color(C), and leaf size (S). A Carolina strain is homozygousfor the recessive alleles of these three genes. Thesegenes are found on the same chromosome as follows:M C S6 m.u. 17 m.u.An F1 hybrid between the two strains is now backcrossedto the Carolina strain. Assuming no interference:a. What proportion of the backcross progeny willresemble the Virginia strain for all three traits?b. What proportion of the backcross progeny willresemble the Carolina strain for all three traits?c. What proportion of the backcross progeny will havethe leaf morphology and leaf size of the Virginiastrain but the leaf color of the Carolina strain?d. What proportion of the backcross progeny will havethe leaf morphology and leaf color of the Virginiastrain but the leaf size of the Carolina strain?
E. W. Lindstrom crossed two corn plants with green seedlings and obtained the following progeny: 3583 green seedlings, 853 virescentwhite seedlings, and 260 yellow seedlings (E. W. Lindstrom. 1921. Genetics 6:91–110). a. Give the genotypes for the green, virescent-white, and yellow progeny. b. Explain how color is determined in these seedlings. c. Is there epistasis among the genes that determine color in the corn seedlings? If so, which gene is epistatic and which is hypostatic?