1. Determine the sequence of genes A, B, and C on a chromosome. Which two genes are most likely going to be inherited together? Genes Crossover Frequency A & B 24% А& C 11% B & C 13%
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- 6. Suppose a particular gene is required for early development and also later for development of a particulartissue, such as the adult nervous system. By generating a homozygous mutant clone in that tissue of a heterozygote, researchers can circumvent the lethalitythat would result if the entire animal is homozygousfor a loss-of-function mutation in that gene.A technique called MARCM (Mosaic Analysiswith a Repressible Cell Marker) was developed to enable Drosophila geneticists to generate homozygousmutant cell clones that are marked by the presence of areporter protein such as GFP. Marker expression enables the investigator to observe clearly the mutantphenotype within a clone of mutant cells. This technique relies on a yeast protein called Gal80 that is anegative regulator of the Gal4 protein described previously in Solved Problem II. Gal80 binds to Gal4 andprevents it from activating transcription. The idea ofMARCM is that Gal4/UASG-driven GFP expression isblocked by Gal80 throughout…3. In 1988, neurologists in Australia reported the existence of identical twins who had developed myoclonic epilepsy in their teens. One twin remainedonly mildly affected by this condition, but the othertwin later developed other symptoms of full-blownMERRF, including deafness, ragged red fibers, andataxia (loss of the ability to control muscles).Explain the phenotypic dissimilarity in theseidentical twins.1. A monogenic disease is a disease caused by a mutation in a single gene. For instance, sickle-cell anemia is caused by a mutation in the HBB gene, which codes for the B- globin chain of hemoglobin. The beginning of HBB is shown here: 5'-ATGGTGCACCTGACTCCTGAGGAGAAGTCTGCCGTTACT...-3' A. Translate this HBB sequence into an amino acid sequence. B. In terms of amino acids, what is the result of the sickle cell mutation, wherein the bolded red A is changed to a T? This single mutation causes hemoglobin to aggregate, causing red blood cells to deform into a sickle-like shape rather than the normal “biconcave disk" shape. C. What would happen if the bolded blue A were mutated to at T? (This is hypothetical; it's not a mutation found in sickle-cell disease.)
- . The production of pigment in the outer layer of seedsof corn requires each of the three independently assorting genes A, C, and R to be represented by at leastone dominant allele, as specified in Problem 64. Thedominant allele Pr of a fourth independently assortinggene is required to convert the biochemical precursorinto a purple pigment, and its recessive allele pr makesthe pigment red. Plants that do not produce pigmenthave yellow seeds. Consider a cross of a strain of genotype A/A ; C/C ; R/R ; pr/pr with a strain of genotypea/a ; c/c ; r/r ; Pr/Pr.a. What are the phenotypes of the parents?b. What will be the phenotype of the F1?c. What phenotypes, and in what proportions, willappear in the progeny of a selfed F1?d. What progeny proportions do you predict from thetestcross of an F1?1. In corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of corn plants (i.e. the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male- fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male-sterile lines. Using the following color-coded circles, simulate the crosses indicated below. Put the illustrations of crosses in the spaces provided. Be sure to include in the labels the genotypes and phenotypes of the offspring in each cross. Big light green circle cytoplasm Big orange circle cytoplasm Small orange circle Small half-light green-half-orange circle - Ff nucleus Small light-green circle - male-sterile - male-fertile - FF nucleus - ff nucleus a. Male-sterile female x FF male (the term male-sterile is an adjective that describes the female) b. Male-sterile female x Ff male (the term male-sterile is an adjective…AaBbCcDdE AaBbCcDc AaBbC Normal No Spacing Heading 1 Head Activity 1. 3. Based on the following data which of the following would be considered: a. Recessive? b. Dominant? C. A weak loss-of-function mutation? d. A null mutation? e. A haploinsufficiency ? Data - Gene "B" is required for formation of body segments. A wild type fly had 3 body segments. BB (Homozygous dominant) Wild Type Bb (Heterozygous) bb (Homozygous recessive) Lacks all body segments 1 body segment Mutant 1 Wild Type Mutant 2 Wild Type 2 body segments Mutant 3 Wild Type Wild Type 2 body segments Activity 4 tv Ps X W 000 DII DD II
- 2. In corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of corn plants (i.e. the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male-fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male-sterile lines. Using the following color-coded circles, simulate the crosses indicated below. Put the illustrations of crosses in the spaces provided. Be sure to include in the labels the genotypes and phenotypes of the offspring in each cross. Big light green circle Big orange circle Small orange circle Small half-light green-half-orange circle - Ff nucleus Small light-green circle - male-sterile cytoplasm - male-fertile cytoplasm - FF nucleus - ff nucleus a. Male-sterile female x FF male (the term male-sterile is an adjective that describes the female) b. Male-sterile female x Ff male (the term male-sterile is an adjective…1. Choose the phrase from the right column that best fitsthe term in the left column.a. cytoplasmicsegregation1. transmission of genes through maternalgamete onlyb. heteroplasmic 2. cell that has mtDNAs or cpDNAs all of onegenotypec. homoplasmic 3. having gametes of similar sized. maternalinheritance4. a cell with a mixture of different mtDNAsgenerates a daughter cell with only one kinde. uniparentalinheritance5. a specific fraction of wild-type organellarDNAs is required for a wild-type phenotypef. isogamous 6. cell with mtDNAs or cpDNAs with differentgenotypesg. threshold effect 7. transmission of genes through either amaternal or a paternal gamete, but not both2. Uniparental disomy is a rare phenomenon in whichonly one of the parents of a child with a recessivedisorder is a carrier for that trait; the other parent ishomozygous normal. By analyzing DNA polymorphisms, it is clear that the child received both mutantalleles from the carrier parent but did not receive anycopy of the gene from the other parent.a. Diagram at least two ways in which uniparentaldisomy could arise. (Hint: These mechanismsall require more than one error in cell division,explaining why uniparental disomy is so rare.)Is there any way to distinguish between thesemechanisms to explain any particular case ofuniparental disomy?
- 2. Uniparental disomy is a rare phenomenon in whichonly one of the parents of a child with a recessivedisorder is a carrier for that trait; the other parent ishomozygous normal. By analyzing DNA polymorphisms, it is clear that the child received both mutantalleles from the carrier parent but did not receive anycopy of the gene from the other parent.a. Diagram at least two ways in which uniparentaldisomy could arise. (Hint: These mechanismsall require more than one error in cell division,explaining why uniparental disomy is so rare.)Is there any way to distinguish between thesemechanisms to explain any particular case ofuniparental disomy?b. How might the phenomenon of uniparental disomyexplain rare cases in which girls are affected withrare X-linked recessive disorders but have unaffectedfathers, or other cases in which an X-linked recessive disorder is transmitted from father to son?c. If you were a human geneticist and believed oneof your patients had a disease syndrome caused…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?. Among adults with Turner syndrome, it has beenfound that a very high proportion are genetic mosaics.These are of two types: In some individuals, themajority of cells are XO, but a minority of cellsare XX. In other Turner individuals, the majorityof cells are XO, but a minority of cells are XY.Explain how these two patterns of somatic mosaicscould arise.