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- Individuals of genotype AaBb were mated to individuals of genotype aabb. One thousand offspring were counted, with the following results: 474 Aabb, 480 aaBb, 20 AaBb, and 26 aabb. What type of cross is it? Are these loci linked? What are the two parental classes and the two recombinant classes of offspring? What is the percentage of recombination between these two loci? How many map units apart are they?A genotypically wild-type fruit fly is crossed to a fly with homozygous recessive mutations in two different genes that result in a wingless and an eyeless phenotype. The F1 generation is 100% phenotypically wild type. Two of the phenotypically wild-type F1 flies are mated and the following offspring are observed in the F2 generation: Wild type: 367 Wingless (wg): 119 Eyeless (ey): 134 Eyeless, wingless (ey; wg): 52 Part 1 Assuming independent assortment, how many F2 offspring would be expected to be homozygous for the wingless mutation but have normal eyes? O224 O126 O 112 O 378 Part 2 How many degrees of freedom are there in the results of this cross? O0 O 1 02 03 O4In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct+ is crossed to a sn+ct male. The F1 flies are interbred. The F2 males are distributed as follows: genotype number sn ct 15 sn ct+ 34 sn+ ct 33 sn+ct+ 18 What is the map distance between sn and ct?
- The genes responsible for albinism (A or a) and pink eyes (P or p) in mice are linked and wild-type pigmentation is dominant in both traits. A homozygous wild-type mouse is mated with an albino, pink-eyed mouse. The resulting F1 mice are crossed with albino, pink-eyed mice. The F2 progeny of this testcross are: 22 7 wild-type body color, wild-type eyes albino body color, wild-type eyes wild-type body color, pink eyes albino body color, pink eyes 10.000 3 18 How many cM apart are the albinism and pink-eye genes? O 50 O 40 O 30 O 20 O 10In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn+ and ct+ is crossed to a sn ct male. The F1 flies are interbred. The F2 males are distributed as follows sn ct 36 sn ct+ 13 sn+ ct 12 sn+ ct+ 39 What is the map distance between sn and ct?In Drosophila, the sepia mutation (se, chromosome 3, position 26) results in dark brown eyes, while cinnabar (cn, chromosome 2, position 57.5) results in bright orange-red eyes. True breeding, wild type females are mated with true breeding males homozygous recessive for both traits. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1
- In beetles, an X-linked gene determines body size, with normal size (M)completely dominant to miniature body size (m). Body color is determined by an autosomal gene with two alleles, where B is incompletely dominant to b such that BB beetles are black, Bb beetles are brown and bb beetles are yellow. Male beetles are heterogametic (XY). The following cross is performed: brown, miniature sized body female X brown, normal sized body male Based on this information, which of the following statements is FALSE? Select 3 correct answer(s) Question 3 options: A) 1/8 of the female progeny will have yellow miniature bodies. B) All of the male progeny will have miniature bodies. C) 1/4 of the total progeny will be black. D) 1/4 of the male progeny will have yellow miniature bodies. E) All of the female progeny will have miniature bodies.…A mutant sex-linked trait called “notched” (N) is deadly in Drosophila when homozygous in females. Males who have a single N allele will also die. The heterozygous condition (Nn) causes small notches on the wing. The normal condition in both male and females is represented by the allele n. a) Indicate the phenotypes of the F1 generation from the following cross: XNXn x XnY b) Explain why dead females are never found in the F1 generation no matter which parents are crossed. c) Explain why the mating of female XNXn and a male XNy is unlikely.In onion, male sterility is produced when the nuclear genotype is aa and the mitochondrial gene S (sterile) are present. Any other combination of nuclear genotype and mitochondrial gene (including gene F for fertile) will result in a male fertile plant. Give the genotypic ratio and the phenotypic ratio or the percentage of male sterile and male fertile offspring that will be produced in the following crosses. 1. Aa + S male x aa + F female 2. Reciprocal cross of number 1. (Note that when we do reciprocal cross, we interchange/swap the genotypes of the parents (if there is a nuclear gene involved, you interchange the nuclear genotype as well). 3. Aa + S female x Aa + F male 4. Reciprocal cross of number 3.
- A parental cross between homozygous wild-type red eyed females (X+ X+) were crossed with white eyed males (XwY). The F1 generation consisted of phenotypic X+Xw females and X+Y males. The F2 generation was produced by crossing the F1 generation and consisted of a ratio of 2 red eyed females (X+ X+ and X+ Xw) : 1 red eyed male (X+Y) : 1 white eyed male (XwY). Attached is a picture of a chi square analysis based on the results from the F2 generation. Please discus what conclusions can be made based on the data/findings. What inheritance pattern is this?The recessive allele s causes Drosophila to have small wings, and the s+ allele causes normal wings. This gene is known to be X linked. If a small-winged male is crossed with a homozygous wild-type female, what ratio of normal to small-winged flies can be expected in each sex in the F1? If F1 flies are intercrossed, what F2 progeny ratios are expected? What progeny ratios are predicted if F1 females are backcrossed with their father?Miniature wings (Xm) in Drosophila result from an X-linked allele that is recessive to the allele for long wings (X*). Give the genotypes of the parents in the following cross: Male parent Female parent Male offspring Female offspring Long Miniature 750 miniature 761 long O male: X* / X* and female X™ /x+ O male: X*/Xt and female Xm /xm O male: X*/ Y and female Xm /xm O male: Xm/ Y and female Xm /xm
