BROOKER BIOLOGY
5th Edition
ISBN: 9781307656152
Author: BROOKER
Publisher: MCG/CREATE
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Chapter 20, Problem 2COQ
Is it possible for a phenotypically normal female fly to be homozygous for a loss-of-function allele in the bicoid gene? What would be the
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In C. elegans, lon-2 and unc-2 are recessive mutations that are 8 map units apart on the X chromosome. An hermaphrodite who is Lon and Unc is mated to a wild-type male. An F1 hermaphrodite is mated to a wild-type male. What are the expected percentages of the different phenotypes among the male progeny?
In fruit flies, the allele for normal wings (V) is dominant to the allele for short wings (v). Suppose two fruit flies heterozygous for the trait are mated.
What is the male fruit fly’s genotype and the female fruit fly’s genotype?
What is the male fruit fly’s phenotype and the female fruit fly’s phenotype?
What will be the genotypic ratio of the F1 generation?
What will be the phenotypic ration of the F1 generation?
Termites have a haplo-diploid sex determination system where females develop from a fertilized egg (they are diploid, having one allele from the
female cueen and one allele from the male), and males develop from unfertilized eggs (they are haploid, having only one allele from the queen).
Assuming that the female parent (queen) is heterozygous for a particular gene, what is the probability that a female offspring will inherit the
recessive allele from her mother? What is the probability that a male offspring will inherit a recessive allele from his mother?
O A. The probability that a daughter will inherit a recessive allele from her mother is 50%; the probability that a son will inherit a recessive allele
from his mather is 50%.
OB. Ihe probobility that a daughter will inherit a recessive allele from her mother is 0%; the probability that a son will inherit a recessive allele
from his mother is 100%.
O C. The probability that a daughter will inherit a recessive allele from her mother is…
Chapter 20 Solutions
BROOKER BIOLOGY
Ch. 20.1 - 20.1 General Themes in Development Concept Check:...Ch. 20.1 - Prob. 2CCCh. 20.1 - Prob. 1CSCh. 20.2 - Development in Animals I: Pattern Formation...Ch. 20.2 - Development in Animals I: Pattern Formation...Ch. 20.2 - Prob. 3CCCh. 20.2 - Core Skill: Modeling The goal of this modeling...Ch. 20.2 - Development in Animals I: Pattern Formation Core...Ch. 20.3 - Development in Animals II: Cell Differentiation...Ch. 20.3 - Development in Animals II: Cell Differentiation...
Ch. 20.3 - Development in Animals II: Cell Differentiation...Ch. 20.3 - Development in Animals II: Cell Differentiation...Ch. 20.3 - Development in Animals II: Cell Differentiation...Ch. 20.4 - Development in Plants Concept Check: Where are...Ch. 20.4 - Prob. 2CCCh. 20 - The process whereby a cells morphology and...Ch. 20 - Prob. 2TYCh. 20 - Positional information is important in determining...Ch. 20 - Morphogens are a. molecules that disrupt normal...Ch. 20 - Prob. 5TYCh. 20 - Arrange the following phases of pattern formation...Ch. 20 - Prob. 7TYCh. 20 - Which of the following genes do not play a role in...Ch. 20 - An embryonic stem cell that can give rise to any...Ch. 20 - Prob. 10TYCh. 20 - Prob. 1CQCh. 20 - The MyoD gene in mammals plays a role in muscle...Ch. 20 - Prob. 3CQCh. 20 - Prob. 1COQCh. 20 - Is it possible for a phenotypically normal female...
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- Human sex chromosomes are XX for females and XY for males. a. With respect to an X-linked gene, how many different types of gametes can a male produce? b. If a female is homozygous for an X-linked allele, how many different types of gametes can she produce with respect to this allele? c. If a female is heterozygous for an X-linked allele, how many different types of gametes can she produce with respect to this allele?arrow_forwardOne reason that worker bees forgo their own reproduction to help their sister (the queen) reproduce is that female bees are more closely related to their sisters than they are to their own offspring. This quirk of genetics results from the fact that bees have haplodiploid sex determination, in which females are diploid, with a mother and a father, but males are haploid, developing from unfertilized eggs. Because males are haploid, they produce sperm by mitosis. Explainwhy haplodiploid sex determination causes females to be more closely related to their sisters than to their offspring.arrow_forwardBees have a haplodiploid sex determination. The recessive allele ivory causes bees to have white eyes. A white-eyed female is mated to a wild-type male. An F1 female is mated to a white-eyed male and some of her eggs are fertilized. She lays eggs after the mating. What are possible phenotypes of the offspring? i) wildtype males ii) wildtype females iii) white-eyed males iv) white-eyed females Group of answer choices A. only i and iii are possible oucomes B. only i and ii are possible oucomes C. all four (i, ii, iii, and iv) are possible phenotypes D. only ii and iii are possible oucomes E. only i,ii, and iii are possible oucomesarrow_forward
- In Drosophila, vermilion eye color is due to a recessive allele (v) located on the X chromosome. Curved wings are due to a recessive allele (cu) located on one autosome, and ebony body is due to a recessive allele (e) located on another autosome. A vermilion male is mated to a curved, ebony female, and the F1 males are phenotypically wild-type. If these males were backcrossed to curved, ebony females, what proportion of the F2 offspring will be wild-type males?arrow_forwardIn fruit flies red (A) eyes are dominant to apricot (a) eyes, and normal (P) wings and dominant to pointed (p) wings. Based on the information above, what are the possible gametes that could be produced by a fly that is homozygous for red eyes and heterozygous for normal wings? Group of answer choices: A) AA, Pp B) AP, Ap, aP, ap C) AP, Ap D) AAPparrow_forwardWhat phenotypic proportions would be expected if the genes for red eyes and for white-banded wings were located on different chromosomes?arrow_forward
- PURPLE VESTIGIAL DIHYBRID CROSS In the parental generation, you mate a pure-breeding wild-type female (put/pu+;vg+/vg+) with a pure-breeding purple, vestigial (pu/pu;vg/vg) to produce an F1 generation that is all wild-type (pu*/pu;vg+/vg). Note that the F1 flies are all dihybrid. Next, you mate several F1 dihybrid females (pu*/pu;vg+/vg) with tester males, which are purple, vestigial (pu/pu;vg/vg). The offspring of this dihybrid testcross are: Phenotype Genotype Tester Gamete Dihybrid Gamete Number Wild-type 437 417 77 59 Purple, vestigial Vestigial Purple Copy the table into your notes and derive the dihybrid gametes following the example in the first section. The columns in blue (phenotypes and numbers of offspring) are what you can see and count. The genotypes of the testcross offspring (orange) must be deduced from the phenotypes and knowing that the tester contributed pu vg gametes. Finally, you can deduce the dihybrid gametes (green) by subtracting the tester gamete contribution…arrow_forwardMales of many diploid species (like us) have X and Y sex chromosomes. They are hemizygous for most X- linked genes. Thus, males express most X-linked alleles, whether they are dominant or recessive in females. In the fruit fly Drosophila, it is common to achieve the equivalent of a test cross of X-linked genes in females by assessing the readily observed phenotypes of their male progeny. Since males do not receive X-linked genes from their father, sires of these crosses can be normal or wild-type flies. In fly genetics, it is conventional to name a gene after the mutant phenotype that enabled its discovery. Your challenge is to establish gene order and map distances between three X-linked genes in Drosophila. Each gene is represented by recessive mutant alleles that express rather distinctive phenotypes relative to their dominant wild-type alternative alleles. Flies expressing fruitless (f) are bisexual, lush (1) have a heightened responses to ethanol, and ken&barbie (kb) lack external…arrow_forwardIn fruit flies, eye color is an X-linked trait. Red eyes (XB) are dominant to maroon eyes (Xb). A male with red eyes mates with a female with maroon eyes. What will be the eye color of the flies in the F1 generation? A) Males: Maroon, Females: Red B) Males: Maroon, Females: Maroon C) Males: Red, Females: Maroon D) Males: Red, Females: Red E) Males: Maroon, Females: Red or Maroonarrow_forward
- In humans, the mic2 gene (involved in antibody production) is found on sex chromosomes and has pseudoautosomal inheritance. If a female that is a mic2a is mated to a phenotypically mic2b male then what would be resulting ratios of the offspring? (note; the mic2a and 2b alleles are codominant) O 50% mic2a and 50% mic2b for both sexes O 100% mic2a/2b for both males and females O all females are mic2a/2b whereas males are all mic2a all females are mic2b whereas males are all mic2aarrow_forwardLet’s suppose that two different X-linked genes exist in mice,designated with the letters N and L. Gene N exists in a dominant,normal allele and in a recessive allele, n, that is lethal. Similarly,gene L exists in a dominant, normal allele and in a recessive allele,l, that is lethal. Heterozygous females are normal, but males thatcarry either recessive allele are born dead. Explain whether or notit would be possible to map the distance between these two genesby making crosses and analyzing the number of living and deadoffspring. You may assume that you have strains of mice in whichfemales are heterozygous for one or both genes.arrow_forwardIn Drosophila, a fully heterozygous female with the X-linked recessive genes a, b, and c (not necessarily in that order on the chromosome) was mated to a male that was genetically a, b, c (not necessarily in that order on the chromosome). The offspring occurred in the following phenotypic ratios: Phenotypes: Numbers: What is the cis/trans arrangement in the heterozygous parent? Wild 426 а, с, b 428 Which gene is in the middle? a 23 c, b 22 If you added 23, 22, 3, and 2, it would give you the map distance between genes C 49 b, a 46 What calculation would you make to determine if interference was occurring? (you don't have to complete the calculation) b. C, a Total 1000 3.arrow_forward
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