18. The following question refers to the pedigree chart in the figure belowfor a family, some of whose members exhibit the dominant trait, W.Affected individuals are indicated by a dark square or circle. What is thelikelihood that the progeny of IV-3 and IV-4 will have the trait? (LS3-3) *12IwwWw2 3 48II OIII2 34IV0%50%75%100%

Name: 18. The following question refers to the pedigree chart in the figure
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Description: 18. The following question refers to the pedigree chart in the figure belowfor a family, some of whose members exhibit the dominant trait, W.Affected individuals are indicated by a dark square or circle. What is thelikelihood that the progeny of IV-

The process of obtaining the genotype of a dominant offspring is difficult as there is the presence…

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The following question refers to the pedigree chart in the figure b a family, some of whose members exhibit the dominant trait, W. ected individuals are indicated by a dark square or circle. What is elihood that the progeny of IV-3 and IV-4 will have the trait? (LS3-3

The following question refers to the pedigree chart in the figure b a family, some of whose members exhibit the dominant trait, W. ected individuals are indicated by a dark square or circle. What is elihood that the progeny of IV-3 and IV-4 will have the trait? (LS3-3

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter12: Mendel, Genes, And Inheritance

Section: Chapter Questions

Problem 7TYK: In cats, the genotype AA produces tabby fur color; Aa is also a tabby, and aa is black. Another gene...

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A heterozygous individual has a _______ for a trait being studied. a. pair of identical alleles b. pair of nonidentical alleles c. haploid condition, in genetic terms
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
1. A disease is known to be inherited through an ADHL mode of inheritance. For a pairing of two affected individuals, what is the expected proportion of offspring that will be affected AND female? A. 0. B. 0.667. C. 0.333.
1. The pedigree shown below illustrates the inheritance pattern for a trait controlled by a single fully penetrant gene with complete dominance. Based on the pedigree do the following: a. List one mode of inheritance consistent with the pedigree.
We have dealt mainly with only two genes, but the sameprinciples hold for more than two genes. Consider thefollowing cross:A/a ; B/b ; C/c ; D/d ; E/e × a/a ; B/b ; c/c ; D/d ; e/ea. What proportion of progeny will phenotypicallyresemble (1) the first parent, (2) the second parent,(3) either parent, and (4) neither parent?b. What proportion of progeny will be genotypically thesame as (1) the first parent, (2) the second parent,(3) either parent, and (4) neither parent?Assume independent assortment.
1. How many different types of gametes can be produced by an individual with the genotype Yy Ww Pp ee Bb? 2. If an individual with the genotype Yy Ww Pp ee Bb is crossed with another individual with the same genotype, what proportion of offspring will be homozygous recessive at all genes? 3. If an individual with genotype Yy Ww Pp ee Bb is crossed with an individual with the same genotype, what proportion of offspring will have the dominant phenotype for the four heterozygous loci?
1. Construct a pedigree chart of the trait traced in the situation below.2. The pedigree chart must be constructed with the names and genotypes at the bottom of every individual in the chart.If it is autosomal: Always use the first letter of the dominant trait to represent the dominant allele. For example, if the dominant trait is red, so the allele R-red, while r-yellow. For example, if the person is homozygous dominant for curly hair, we write CC as his genotype. If sex-linked, do not forget to write the parents’ chromosomes (XX-female, XY-male), with their corresponding superscripts (dominant or recessive). The superscript must be based on the first letter of the disorder. For example, the mother is a carrier of hemophilia (which is x-linked recessive), so we write XHXh. If the father has hemophilia, we write XhY. Put the pedigree chart using code properly. The situation is given below: Trait: NeurofibromatosisThe dominant form is neurofibromatosis, caused by the production of…
1. Construct a pedigree chart of the trait traced in the situation below.2. The pedigree chart must be constructed with the names and genotypes at the bottom of every individual in the chart.If it is autosomal: Always use the first letter of the dominant trait to represent the dominant allele. For example, if the dominant trait is red, so the allele R-red, while r-yellow. For example, if the person is homozygous dominant for curly hair, we write CC as his genotype. If sex-linked, do not forget to write the parents’ chromosomes (XX-female, XY-male), with their corresponding superscripts (dominant or recessive). The superscript must be based on the first letter of the disorder. For example, the mother is a carrier of hemophilia (which is x-linked recessive), so we write XHXh. If the father has hemophilia, we write XhY. Put the pedigree chart using code properly. The situation is given below: Trait: NeurofibromatosisThe dominant form is neurofibromatosis, caused by the production of…
1. A human female "carrier" who is heterozygous for the recessive, sex-linked trait causing red-green color blindness (or alternatively, hemophilia), marries a normal male. What proportion of their male progeny will have red-green color blindness (or alternatively, will be hemophiliac)? * a. 100% b. 75% c. 50% d. 25% e. 0% 2. A human female "carrier" who is heterozygous for the recessive, sex-linked trait red color blindness, marries a normal male.What proportion of their female progeny will show the trait? * a. All b. ½ c. ¼ d. 0 e. 3/4 3. Women have sex chromosomes of XX, and men have sex chromosomes of XY. Which of a man's grandparents could not be the source of any of the genes on his Y-chromosome? * a. Father's Mother b. Mother's Father c. Father's Father d. Mother's Mother, Mother's Father, and Father's Mother e. Mother's Mother 4. Male-pattern baldness is an example of a sex-influenced trait. The baldness allele, which causes hair loss, is influenced by the hormones…
1) Dr. Thompson is hoping to produce a tester outcome from her cross. Which of the following would depict a tester PHENOTYPE? a) mmnn. b) MmNn. c) mn. d) MN. 2) By decomposing the cross, MmNn X Mmnn, into single gene outcomes, the possibility of obtaining a double recessive outcome for the ‘M’ gene is: a) 50%! b) 100! c) 0%! d) 25%!
. A geneticist mapping the genes A, B, C, D, and E makestwo 3-point testcrosses. The first cross of pure lines isA/A ⋅ B/B ⋅ C/C ⋅ D/D ⋅ E/E × a/a ⋅ b/b ⋅ C/C ⋅ d/d ⋅ E/EThe geneticist crosses the F1 with a recessive tester andclassifies the progeny by the gametic contribution ofthe F1:A ⋅ B ⋅ C ⋅ D ⋅ E 316a ⋅ b ⋅ C ⋅ d ⋅ E 314A ⋅ B ⋅ C ⋅ d ⋅ E 31a ⋅ b ⋅ C ⋅ D ⋅ E 39A ⋅ b ⋅ C ⋅ d ⋅ E 130a ⋅ B ⋅ C ⋅ D ⋅ E 140A ⋅ b ⋅ C ⋅ D ⋅ E 17a ⋅ B ⋅ C ⋅ d ⋅ E 131000The second cross of pure lines is A/A • B/B • C/C • D/D •E/E × a/a • B/B • c/c • D/D • e/e.The geneticist crosses the F1 from this cross with arecessive tester and obtainsA ⋅ B ⋅ C ⋅ D ⋅ E 243a ⋅ B ⋅ c ⋅ D ⋅ e 237A ⋅ B ⋅ c ⋅ D ⋅ e 62a ⋅ B ⋅ C ⋅ D ⋅ E 58A ⋅ B ⋅ C ⋅ D ⋅ e 155a ⋅ B ⋅ c ⋅ D ⋅ E 165a ⋅ B ⋅ C ⋅ D ⋅ e 46A ⋅ B ⋅ c ⋅ D ⋅ E 341000The geneticist also knows that genes D and E assortindependently.a. Draw a map of these genes, showing distances inmap units wherever possible.b. Is there any evidence of interference?
1) By decomposing the cross, MmNn X Mmnn, into single gene outcomes, the possibility of obtaining a double recessive outcome for the ‘N’ gene is: a) 50%! b) 100! c) 0%! d) 25%! 2) A 16-square Punnett is time-consuming to draw out. Dr. Thompson can easily solve this problem by decomposing her dihybrid crosses into separate monohybrid Punnett squares, which would also be quicker than using a forked line diagram. If so, how many individual monohybrid Punnett squares will she need to draw? a) Four Punnett squares. b) Two Punnet squares. c) One Punnett square. d) Sixteen Punnett squares since there are 16 possible outcomes.