Recessive maternal effect genes are identified in flies (for example)when a phenotypically normal mother cannot produce any normaloffspring. Because all of the offspring are dead, this female fly cannot be used to produce a strain of heterozygous flies that could beused in future studies. How would you identify heterozygous individuals that are carrying a recessive maternal effect allele? Howwould you maintain this strain of flies in a laboratory over manygenerations?
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Recessive maternal effect genes are identified in flies (for example)
when a
offspring. Because all of the offspring are dead, this female fly cannot be used to produce a strain of heterozygous flies that could be
used in future studies. How would you identify heterozygous individuals that are carrying a recessive maternal effect allele? How
would you maintain this strain of flies in a laboratory over many
generations?
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- In Drosophila, the autosomal recessive dp allele ofthe dumpy gene produces short, curved wings, whilethe autosomal recessive allele bw of the brown genecauses brown eyes. In a testcross using femalesheterozygous for both of these genes, the followingresults were obtained:wild-type wings, wild-type eyes 178wild-type wings, brown eyes 185dumpy wings, wild-type eyes 172dumpy wings, brown eyes 181In a testcross using males heterozygous for both ofthese genes, a different set of results was obtained:wild-type wings, wild-type eyes 247dumpy wings, brown eyes 242a. What can you conclude from the first testcross?b. What can you conclude from the second testcross?c. How can you reconcile the data shown in parts (a)and (b)? Can you exploit the difference betweenthese two sets of data to devise a general test forsynteny in Drosophila?d. The genetic distance between dumpy and brown is91.5 m.u. How could this value be measured?Imagine that you caught a female albino mouse inyour kitchen and decided to keep it for a pet. A fewmonths later, while vacationing in Guam, you caughta male albino mouse and decided to take it home forsome interesting genetic experiments. You wonderwhether the two mice are both albino due to mutations in the same gene. What could you do to find outthe answer to this question? Assume that both mutations are recessive.In mice, dwarfism is caused by an X-linked recessive allele, and pink coat is caused by an autosomal dominantallele (coats are normally brownish). If a dwarf femalefrom a pure line is crossed with a pink male from a pureline, what will be the phenotypic ratios in the F1 and F2 ineach sex? (Invent and define your own gene symbols.)
- In individuals affected by cystic fibrosis, salt crystals may appear afterperspiration dries up. In addition, the disease causes respiratory disorderswhich can be both debilitating and lethal. It occurs in individuals homozygousfor recessive gene. If 2 normal parents had a daughter with the symptoms ofthis disease, and a normal son, what is the probability that he might be acarrier of the recessive gene?Express answer in fraction form.Galactosemia is a recessive human disease that istreatable by restricting lactose and glucose in the diet.Susan Smithers and her husband are both heterozygous for the galactosemia gene.a. Susan is pregnant with twins. If she hasfraternal (nonidentical) twins, what is theprobability both of the twins will be girls whohave galactosemia?b. If the twins are identical, what is the probabilitythat both will be girls and have galactosemia?For parts (c–g), assume that none of the children isa twin.c. If Susan and her husband have four children, whatis the probability that none of the four will havegalactosemia?d. If the couple has four children, what is the probability that at least one child will have galactosemia?e. If the couple has four children, what is the probability that the first two will have galactosemia andthe second two will not?f. If the couple has three children, what is the probability that two of the children will have galactosemia and one will not, regardless of order?g. If…For a certain gene in a diploid organism, eight units ofprotein product are needed for normal function. Eachwild-type allele produces five units.a. If a mutation creates a null allele, do you think thisallele will be recessive or dominant?b. What assumptions need to be made to answer part a?
- In the experiment of Figure shown, Stern followed the inheritancepattern in which females carried two abnormal X chromosomesto correlate genetic recombination with the physical exchange ofchromosome pieces. Is it necessary to use a strain carrying twoabnormal chromosomes, or could he have used a strain in whichfemales carried one normal X chromosome and one abnormal Xchromosome with a deletion at one end and an extra piece of the Ychromosome at the other end?Honeybees are unusual in that male bees (drones) have only onecopy of each gene, but female bees have two copies of their genes.This difference arises because drones develop from eggs that havenot been fertilized by sperm cells. In bees, the trait of long wingsis dominant over short wings, and the trait of black eyes is dominant over white eyes. If a drone with short wings and black eyeswas mated to a queen bee that is heterozygous for both genes, what are the predicted genotypes and phenotypes of male and femaleoffspring? What are the phenotypic ratios if we assume an equalnumber of male and female offspring?A maternal effect gene in Drosophila, called torso, occurs as a functionalallele (torso+) and a nonfunctional, recessive allele (torso−)that prevents the correct development of anterior- and posterior-moststructures. A wild-type male (torso+torso+) is crossed to a female ofunknown genotype. This cross produces offspring (larva) that are allmissing their anterior- and posterior-most structures and therefore die during early development. What are the genotype and the phenotypeof the female fly in this cross? What are the genotypes andphenotypes of the female fly’s parents?
- Two different genes control the expression of kernelcolour in Mexican black corn: a black pigment gene (B)and a dotted pigment gene (D). Gene B influences theexpression of gene D. The dotted phenotype appearsonly when gene B is in the homozygous recessivestate. A colourless variation occurs when both genesare homozygous recessive.After pure-breeding black-pigmented plants werecrossed with colourless plants, all offspring wereblack-pigmented. Plants of the F1 generation are suspected of being heterozygous for both genes. A test crossof colourless plants with the heterozygote plants should produce a genotypic ratio in theoffspring ofa. 1:0b. 3:1c. 2:1:1d. 1:1:1:1. In 1932, H. J. Muller suggested a genetic test to determine whether a particular mutation whose phenotypiceffects are recessive to wild type is a null (amorphic)allele or is instead a hypomorphic allele of a gene.Muller’s test was to compare the phenotype of homozygotes for the recessive mutant alleles to the phenotype of a heterozygote in which one chromosomecarries the recessive mutation in question and thehomologous chromosome carries a deletion for aregion including the gene.In a study using Muller’s test, investigators examined two recessive, loss-of-function mutant alleles ofrugose named rg41 and rgγ3. The eye morphologiesdisplayed by flies of several genotypes are indicated inthe following table. Df(1)JC70 is a large deletion thatremoves rugose and several genes to either side of it.Genotype Eye surface Cone cells per ommatidiumwild type smooth 4rg41/rg41 mildly rough 2–3rg41/Df(1)JC70 moderately rough 1–2rg γ3/rg γ3 very rough 0–1rg γ3/Df(1)JC70 very rough 0–1a. Which allele…Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?