2. How many genetically different kinds of gametes can an individual with each of the following genotypes produce? Assume each gene is on a separate pair of chromosomes. a. Genotype AA aa Aa AaBB Number of genetically different gametes Genotype AaBb AaBb CC AaBb Cc AaBb CcDd EeFf Number of genetically different gametes 0 Given the information in the table above, what general rule exists for determining the number of different gametes an organism of a given genotype can produce? Explain.

2. How many genetically different kinds of gametes can an individual with each of the following genotypes produce? Assume each gene is on a separate pair of chromosomes. a. Genotype AA aa Aa AaBB Number of genetically different gametes Genotype AaBb AaBb CC AaBb Cc AaBb CcDd EeFf Number of genetically different gametes 0 Given the information in the table above, what general rule exists for determining the number of different gametes an organism of a given genotype can produce? Explain.

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 1SQ: A heterozygous individual has a _______ for a trait being studied. a. pair of identical alleles b....

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9. Make a pedigree for each of the following situations. For each individual, write the individual's genotype (when possible) next to the individual's symbol (e.g. O xty, I Gg): a. Two parents do not have cystic fibrosis and they have a daughter with cystic fibrosis and a son who does not have cystic fibrosis. The daughter grows up and she mates with a male who does not have cystic fibrosis. Their only child is a boy and he has cystic fibrosis. b. A man with hemophilia mates with a female without hemophilia. They have one son and one daughter. The daughter has hemophilia and the son does not have hemophilia. The son grows up, and he marries and mates with a female. Their only child is a boy, and he has hemophilia.
6. A person is simultaneously heterozygous for two autosomal genetic traits. One is a recessive condition foralbinism (alleles A and a); this albinism gene is foundnear the centromere on the long arm of an acrocentricautosome. The other trait is the dominantly inheritedHuntington disease (alleles HD and HD+). TheHuntington gene is located near the telomere of oneof the arms of a metacentric autosome. Draw all copies of the two relevant chromosomes in this person asthey would appear during metaphase of (a) mitosis,(b) meiosis I, and (c) meiosis II. In each figure, labelthe location on every chromatid of the alleles forthese two genes, assuming that no recombinationtakes place.
3) A. How are the following proteins involved in cell division: condensing, cohesin, kinetochore complex, tubulin? B. Why does phenotype not always reveal genotype? C. How can Mendel's postulates of segregation and independent assortment be explained by meiosis? D. Explain the genotypic and phenotypic ratios predicted when performing a monohybrid and a dihybrid cross of heterozygotes? E. How can you use the product rule and sum rule to calculate the probability of a particular genotype resulting from crossing parents of specific genotypes?
9.) In humans, when a dark-skinned person reproduces with a very light-skinned person, their children may range in color from dark to brown. This occurs because skin color is controlled by more than one gene (the exact # of genes is not known). For your convenience, a table is presented below that shows several possible genotypes & phenotypes: Genotypes Phenotypes AABB Very dark skin AABb or AaBB Dark skin AaBb, AAbb, or aaBB Medium brown skin Aabb or aaBb Light skin aabb Very light skin If a man with dark skin whose genotype is AaBB reproduces with a woman who has light skin (aaBb), what are the possible skin colors that their children will have? Develop a Punnett square and list the possible genotypes & phenotypes.
9.) In humans, when a dark-skinned person reproduces with a very light- skinned person, their children may range in color from dark to brown. This occurs because skin color is controlled by more than one gene (the exact # of genes is not known). For your convenience, a table is presented below that shows several possible genotypes & phenotypes: Genotypes Phenotypes Very dark skin Dark skin AABB AABB or AaBB AаBb, AAЬЬ, or aaBB Aabb or aaBb Medium brown skin Light skin Very light skin aabb If a man with dark skin whose genotype is AABB reproduces with a woman who has light skin (aaBb), what are the possible skin colors that their children will have? Develop a Punnett square and list the possible genotypes & phenotypes.
8. Answer the following questions for the two following pedigrees In Pedigree A, is the red highlighted trait dominant or recessive? Why? а. b. What is the genotype of individual C? In Pedigree B, is the red highlighted trait dominant or recessive? Why? а. d. What is the genotype of individual D? Pedigree A Pedigree B mele male : foncle I femole teffected D offected
6). To investigate the genetic control of locomotive behavior in the diploid nematode, Caenorhabditis elegans, eight "wiggly" mutants were isolated, which wiggle in an uncoordinated way rather than smoothly gliding. These eight mutants were crossed together in all possible pairwise combinations, yielding the results summarized in the table above (+ means that all the progeny were wild type, and m means that all the progeny showed the wiggly mutant phenotype). Organize the eight mutations into con lementation groups, each consisting of a group of mutations affecting the same gene. 1 2 3 4 5 6 7 8 1 E N+E 2 3+1 m + m 4 5 + m m E+ + + EE 6 ++E E+ +EE m m + + + + + + + + + m 7 8 ++4 ∞+E +E m m ++ E
1. The pedigree below shows the incidence of rare, autosomal dominant disorder called Ehlers-Danlos disease. The pedigree covers three generations of a particular family and also shows individual genotypes at a potential marker locus (M). a) Indicate the phase of all gen II and III individuals. DdM1M3 ddM2M6 II DDM3M6 ddM4M5 III DdMзM4 DdMЗМ5 DDM3M4 ddM3M5 DDM3M4 ddM5M6 DDM3M4 ddM4M6 ddM5M6 ddM5M6 b) Which, if any, of the gen III individuals are recombinants? c) Calculate the LOD score as a test of physical linkage between the marker (M) and the disease locus. d) What do you conclude about linkage between D and M?
I. Male Drosophila from a true-breeding wild-typestock were irradiated with X-rays and then mated withfemales from a true-breeding stock carrying the following recessive mutations on the X chromosome:yellow body (y), crossveinless wings (cv), cut wings(ct), singed bristles (sn), and miniature wings (m).These markers are known to map in the order:y - cv - ct - sn - mMost of the female progeny of this cross were phenotypically wild type, but one female exhibited ct and snphenotypes. When this exceptional ct sn female wasmated with a male from the true-breeding wild-typestock, twice as many females as males appearedamong the progeny.a. What is the nature of the X-ray-induced mutationpresent in the exceptional female?b. Draw the X chromosomes present in the exceptional ct sn female as they would appear duringpairing in meiosis.c. What phenotypic classes would you expect to seeamong the progeny produced by mating the exceptional ct sn female with a normal male from a truebreeding wild-type…
2. Null mutations are valuable genetic resources becausethey allow a researcher to determine what happens to anorganism in the complete absence of a particular protein. However, it is often not a trivial matter to determinewhether a mutation represents the null state of the gene.a. Geneticists sometimes use the following test forthe nullness of an allele in a diploid organism: If theabnormal phenotype seen in a homozygote for theallele is identical to that seen in a heterozygote(where one chromosome carries the allele in question and the homologous chromosome is known tobe completely deleted for the gene) then the alleleis null. What is the underlying rationale for thistest? What limitations might there be in interpreting such a result?
2. a. A Drosophila male from a true-breeding stockwith scabrous eyes was mated with a female from atrue-breeding stock with javelin bristles. Both scabrous eyes and javelin bristles are autosomal recessive mutant traits. The F1 progeny all had normaleyes and bristles. F1 females from this cross weremated with males with both scabrous eyes andjavelin bristles. Write all the possible phenotypicclasses of the progeny that could be produced from the cross of the F1 females with the scabrous, javelin males, and indicate for each class whether it is arecombinant or parental type.b. The cross in part (a) yielded the following progeny:77 scabrous eyes and normal bristles; 76 wild type(normal eyes and bristles); 74 normal eyes andjavelin bristles; and 73 scabrous eyes and javelinbristles. Are the genes governing these traits likelyto be linked, or do they instead assort independently? Why?c. Suppose you mated the F1 females from the crossin part (a) to wild-type males. Why would thiscross fail…
1. The figure below represents a karyotype performed on a somatic cell of an unknown organism. Answer the following questions using the figure. a. Circle (and label) 3 different sets of homologous chromosomes? b. This individual is heterozygote for gene A (Aa), Homozygote recessive for gene B (bb) and gene homozygote dominant for gene c (CC). Also, gene B and gene C are linked. Using the homologous chromosomes you circled in part a), provide a representation of the location of the A, B and C, genes and alleles on the homologous chromosomes and sister chromatids in such a way that is consistent with the scenario described above in part b). Choose the chromosomes of your liking to show an appropriate representation. TMIID IL MI