In Drosophila, a mutation which causes a defect in the cuticle of every segment, A1-A8 (abdominal segments) in the embryonic pattern is called a mutation. Select one: O a. gap O b. embryonic polarity O c segment polarity O d. homeotic (Hox) O e pair-rule
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- 1. If a drosophila contains diploid chromosomes 2n=8, how many chromosomes are there in n? A. 2 B. 4 C. 6 D. 8 B. The process of spermatogenesis where the spermatids undergo morphological changes to assumed the structure of the spermatozoa is called? These changes include: A. Spermatocytogenesis; Heads are larger and tail is absent. B. Spermiogenesis; Acrosome is added to the head cap, the tail is present C. Spermiation; The sperm cells are ready for "swimming" towards the egg D. Spermatogenesis; Heads are bulging and the tail is present1. At which stage in development would a cell's gene expression not be altered between cell types? a) pre-gastrulation b. All of these are correct c. blasula d. post-fertilization e. zygote 2. Dihybrid crosses: In dogs, black coat color(B) is dominant to yellow coat fur (b), and straight fur (F) is dominant to curly fur (f). The coat color gene and the fur texture gene are on different chromosomes, so they assort independently, and are not sex linkied. Determine the ratio of having a yellow dog with straight hair if the parents are: BbFF x Bbff a. What are the phenotype of the parents? b. Complete a punnet square. List all the genotypes of the predicted offsprin of this mating c. Based on your Punnet sqaure, list all the genotypes for offspring that would yield a dog with yellow straigh hair d. Based on the punnet square, what is the predicted phenotypic ratio for those parents having a dog with black curly hair.A mutation occurs in the bicoid gene in a female fly. What will result if her eggs are fertilized? 1.Her embryos will be normal if they are fertilized by a male fly that contributed a wild type allele for bicoid 2.Her embryos will be normal if the male that fertilizes her eggs also has that same bicoid mutation 3.All of her embryos will be disrupted irrespective of the genotype of the father 4.Her embryos will develop normally but the adults will be sterile
- genes relies on the existence of ovoD, a dominant female sterile mutation of the ovo gene, whichis located near the middle of the acrocentricDrosophila X chromosome. Females that are ovoD/ovo+ are sterile; ovoD-containing germ-line cellscannot produce eggs.a. Mutations in gene X are recessive lethals, so homozygotes for these mutations do not develop intoadults. Explain how researchers could use the ovoDmutation in a mitotic recombination experiment todetermine (i) whether or not females might supplythe RNA or protein product of gene X to the eggsthey make in their ovaries, and (ii) whether thismaternally supplied product is needed for properdevelopment of their progeny. Where in the genome would gene X need to be located for thisapproach to work?b. The ovoD mutant gene has been cloned, so genomicDNA for this mutant gene is available. How couldyou use this cloned DNA to determine whether any embryonic lethal mutation located anywhere in thegenome was an allele of a maternal effect…Part 1= You are to discover the chromosomal location (autosomal or X linked) and mode of inheritance for two new fly mutations couchpotato (cp) and happyhour (hh) The homozygous couchpotato (cpcp) phenotype is (sedentary) an extreme lack of activity The homozygous happyhour (hhhh) can drink any other fly under the table so phenotype (drinker) PARENTS Cross Male double homozygous mutants cp hh with Female wild wild CP HH are to be crossed BUT FIRST you don’t know if either gene is on the X chromosome Parental diploid genotypes could be: male cpcphhhh female CPCPHHHH if Both genes autosomal or Male XcpYhhhh female XCPXCPHHHH if cp on the X chromosome and hh autosomal or Male cpcpXhhY female CPCPXHHXHH if hh on the X chromosome and cp autosomal What are the female and male PARENTAL GAMETE GENOTYPES for the above parental crosses if: a) Both genes were autosomal ((REMEMBER BOTH GENDER HAVE ONE GAMETE GENO) b) cp gene on the X chromosome and hh autosomal…1. Genes in different chromosomes ______ during meiosis. Genes that are very close together in the same chromosome are ______. a. do not assort independently; linked b. do not assort independently; unrelated c. assort independently; linked d. assort independently; unrelated 2. Suppose that codons consisted of 4 nucleotides instead of 3 and that there were only 2 different bases. How many amino acids could be encoded by this variant of the genetic code?
- 4. In Drosophila, AAAXX is --while in humans, AAXXXY is-- a. Metamale---female b. Intersex-male С. Metafemale---metamale d. Male-female e. Intersex---female7 I of A OY embryo or egg, where an egg is missing both X chromosomes and is fertilized by a normal sperm: O a. forms an inviable embryo O b. develops normally O c. forms a male with Klinefelter Syndrome O d. forms an individual with Turner Syndrome 2 37Gametogenesis is d. The process of gamete formation which goes through one division of meiosis Oc. The process of gamete formation which goes through one division of mitosis b. The process of gamete formation where the number of chromosomes is halved for 23 chromosomes a. The process of gamete formation where the number of chromosomes is doubled for 46 chromosomes None of the choices are correct Question 3 or False: Meiosis undergoes two sets of divisions.
- Name: 2. Some ladybugs have 10 black spots on their shells and some have 4. When true breeding 10 spot individuals are crossed with true breeding 4 spot individuals, the offspring have 7 spots. a. Propose two distinct explanations for this finding. Explain the nature of spot inheritance in each case. D. rew bacteria erred them figure above would of14N7 then ely 2 moldon omrod abitqaq ratlsmmem sge of delw toiisoibom s s neu not oomod aid to soubnup ogusl onomod odi to slevel ismon oouborg o o consu ud od souboini bluow uoy dairlw yd za00oq sdi mialex b. Propose an experiment that would distinguish between these possibilities. sor proieins57. male sex organs; unusually small testes A. down syndrome B. edward's syndrome C. patau's syndrome D. klinefelter's syndrome 60. In karyotyping, the purpose of culture is to be able to: A. halt cell division in metaphase B. obtain more speciens C. observe mitosis D. generate more cellsAaBbCcDdE 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