10 m.u. In silk moths red eyes (r) and white banded wing (w) are recessive to normal eyes (R) and normal wings (W). A moth homozygous for red eyes and white banded wings is crossed with an individual who is homozygous normal for both traits. The F, progeny are testcrossed with the following progeny produced: Normal eyes, normal wings Red eyes, normal wings 19 Normal eyes, white banded wings Red eyes, white banded wings 426 a. What phenotypic proportions would be expected if the genes for red eyes and white banded wings were located on different chromosomes? b. What is the genetic distance between these two genes? 418 16

10 m.u. In silk moths red eyes (r) and white banded wing (w) are recessive to normal eyes (R) and normal wings (W). A moth homozygous for red eyes and white banded wings is crossed with an individual who is homozygous normal for both traits. The F, progeny are testcrossed with the following progeny produced: Normal eyes, normal wings Red eyes, normal wings 19 Normal eyes, white banded wings Red eyes, white banded wings 426 a. What phenotypic proportions would be expected if the genes for red eyes and white banded wings were located on different chromosomes? b. What is the genetic distance between these two genes? 418 16

Name: In silk moths red eyes (r) and white banded wing (w) are recessive to
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Description: In silk moths red eyes (r) and white banded wing (w) are recessive to normal eyes (R) and 10m.u. aparnormal wings (W). A moth homozygous for red eyes and white banded wings is crossedwith an individual who is homozygous normal for both traits. The F

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 3CS: A couple was referred for genetic counseling because they wanted to know the chances of having a...

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A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What is the chance that this couple will have a child with two copies of the dominant mutant gene? What is the chance that the child will have normal height?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?
S (smooth) is dominant to s (hairy) L (long pod) is dominant to I (short pod) (Ll is medium length) 7. Given this cross: Yy RR Bb SS LI (male) cross with yy RR Bb Ss Ll (female) a. How many different gametes can be formed by the female plant? b. How many different genotypes are possible in the F1 offspring? c. How many different phenotypes are possible in the F1 offspring? d. What percent of the F1 individuals will be: i. green, bitter, and smooth. ii. hairy, medium, and sweet iii. round, bitter, and long. 8. Two babies in a maternity ward have lost their identity bands, and there is some confusion about their footprint records. Baby #1 is type A blood; #2 is type B blood. If you are one of the mothers and your blood type is 0, which one of the following statements applies. Explain your answer. Neither baby could be yours. The type A baby is yours. The type B baby is yours. Either baby could be yours. 9. In cocker spaniels, the following genotypes and phenotypes are found: AABB =…
A- Normal xy Aa XX Carrier ху ху A- XX XX What inheritance pattern is Ад 43. illustrated in the following pedigree? a- Aa
BIU E I = E E 11 Section 2: Codominance 3. In shorthorn cattle, coat color may be red, white, or roan. Roan is an intermediate phenotype expressed as a patches of red and patches of white hairs. The following data are obtained from various crosses: Red x red all red White x white all white Red x white all roan Roan x roan 25 % red, 50% roan, 25% white a. From these results, determine the appropriate allele symbols and indicate which genotypes lead to which phenotypes. b. What are the genotypes of the parents and offspring in each cross above? C. Predict the results of a red x roan cross. d. Predict the results of a white x roan cross. acer
Heterozygous Cp cp chickens express a condition called creeper, in which the leg and wing bones are shorter than normal (cp cp). The dominant Cp allele is lethal when homozygous. Two alleles of an independently segregating gene determine white (W-) versus yellow (ww) skin color. From matings between chickens heterozygous for both of these genes, what phenotypic classes will be represented among the viable progeny and what are their expected relative frequencies?
n mice, the Ay allele of the agouti gene is recessive lethal, but it is dominant to wild type for yellow coat color. For an unlinked gene called C, cc makes no pigment and are albino while C_ mice will make pigment in in their coat. cc is epistatic to A or AY with regard to coat color, but not for viability. What phenotypic ratio would you expect from the progeny produced by two mice of the genotype: AAy Cc? (Note that “agouti” is the wild-type coat color.) 9 agouti : 3 black : 3 white : 1 yellow 3 agouti : 6 yellow : 3 white 12 yellow : 3 black: 1 brown 9 agouti : 3 yellow : 4 white 8 yellow : 3 black: 1 brown
ybrid Cross - Parental (P) vill be observing the F, offspring of the cross shown in nage. The purple colof (P) of the kernel is the result of ment called anthocyanin, which is dominant and not nked. /hat are the genotypes of the F, offspring if both nts shown are homozygous? F1 Mendel's dihybrid cross always showed a the same ratio ONLY for traits that are not linked on the same mosome and inherited together. If the genes are linked the ratio will not follow Mendel's dihybrid ratio. What is atio seen in dihybrid crosses that are not linked? F2 EXPERIMENTAL QUESTION: Are the genes for color (P) and shape (R) linked (on the same chromosome)?
Drosophila females of wild-type appearance but heterozygous for three autosomal genes are mated with malesshowing the three corresponding autosomal recessivetraits: glassy eyes, coal-colored bodies, and striped thoraxes. One thousand (1000) progeny of this cross aredistributed in the following phenotypic classes:Wild type 27Striped thorax 11Coal body 484Glassy eyes, coal body 8Glassy eyes, striped thorax 441Glassy eyes, coal body, striped thorax 29a. Draw a genetic map based on these data.b. Show the arrangement of alleles on the two homologous chromosomes in the parent females.c. Normal-appearing males containing the samechromosomes as the parent females in the precedingcross are mated with females showing glassy eyes,coal-colored bodies, and striped thoraxes. Of 1000progeny produced, indicate the numbers of thevarious phenotypic classes you would expect.
Here are the progeny of this cross: (Note that the categories are not in any particular order.)Fly type # of prog. Phenotype symbols Categorywt eyes black body wt wings 97 grn+ blk crv+Green eyes black body curved wings 709 ParentalGreen eyes wt body wt wings 9Green eyes black body wt wings 162wt eyes wt body wt wings 727wt eyes black body curved wings 12wt eyes wt body curved wings 179Green eyes wt body curved wings 105Total = 2000 9.) Write the phenotype symbols in the right-hand column. The first one has been done for you.10.) Next to that, label all fly categories as parental (NCOs), SCOs, and DCOs. One has been donefor you.11.) After each SCO/DCO label, write which gene got “unlinked” in these offspring.12.) Put these three genes into a genetic map in the proper order.13.) Calculate the genetic distance between the genes and label the map with these distances.14.) Calculate the cross-over interference15.) Return to questions #1-6 above. For question 6, you gave your opinion, but…
Consider the following pedigree. 하 3 10 (5 3 2 (a) What pattern of transmission is most consistent with this pedigree? (1) autosomal recessive, (2) autosomal dominant, (3) X-linked recessive, (4) X-linked dominant. (b) If individual V-2 marries a normal individual, and if the condition has a pene-trance of 85 percent, what is the probability that their second child will express the trait? (c) On the third line, what does the diamond with a 10 in the middle mean?
In cucumbers: warty (T) > smooth (t) dull (D) > glossy (d) warty, dull x smooth, glossy P: TTDD X ttdd F1: warty, dull x smooth, glossy TtDd x ttdd If RF between T and D loci known to be 24%, how many total non- recombinant progeny do you expect out of 200 total progeny?