1. If two genes are non-allelic, then there is no gene which is aligned on a similar locus of a particular character. As m1 and m2 are two nonallelic then it is a chance of epistasis where one gene modifies how another is expressed. But we know that they are recessive non alleles with the mutant phenotype, so when they will cross with each other then we will get m1/m2 genotypes from m1/m1 homozygotes and m2/m2 homozygotes crossing. Hence, option A is correct.
2.If alleles m1 and m2 are nonallelic, then F1 hybrid genotype must be a monohybrid due to only one nonallelic form is considered. Hence, option A is correct.
3. If alleles m1 and m2 are nonallelic, then the F1 hybrid would have a mutant phenotype due to the presence of a mutant gene which suppresses/ masks the phenotype of m1 nonallele. Again m1 and m2 would fail to complement. It may have a chance of F1 hybrid would have a wild phenotype.
Hence, options A,B and C are correct.
answer fully for thumbs up Hi mc woulu womplement 10. If alleles m1 and m2 are...
1) The alternate forms of a gene for the same trait are known as -A)alleles. B)phenotypes. C)genotypes. D)codominants. E)incomplete dominants. 2) Mendel carried out most of his research with A)livestock -B)pea plants. C)guinea pigs. D)fruit flies. E)bacteria. 3) Which of the following is true according to Mendel's law of segregation? A)Each individual contains two alleles for each trait. B)Fertilization restores the presence of two alleles. C)Alleles separate from each other during gamete formation. D)Each gamete contains one copy of each...
Multiple Choice
1. You count 1000 seeds from a monohybrid cross (i.e., single-locus heterozygote crossed with single-locus heterozygote). How many seeds do you expect to display the dominant phenotype? a. 1000 b. 750 c. 500 d. 250 2. Which of the following is among the purposes of a genetic dissection analysis? a. To determine how two alleles at a locus interact with one another. b. To determine the order of intermediaries in a genetic pathway. c. To determine whether a...
Merritt, 2016 BIO 1511 NAME:F. I.) 10-15 Post Lab #10 Post-LabE19: BASIC GENETICS PROBLEMS # Correct Answers: (16) (Last) Lab Day: MWF(circle one) Solve the genetics problems in this handout. For all problems, assume complete dominance, assume one gene codes for one trait, and assume there are only 2 different alleles possible for each gene (dominant or recessive) 1. If a homozygous dominant red flowered plants are crossed with homozy phenotypes and genotypes of the offspring? Show the genotypes of...
Match the following terms with the appropriate description
below:
a. alleles b. autosomes c. dominant allele d. genotype e.
heterozygous f. homozygote g. phenotype h. recessive allele i. sex
chromosomes
1. ________________ genetic make-up
2. ________________ how genetic make-up is expressed
3. ________________ chromosomes that dictate most body
characteristics
4. ________________ alternative forms of the same gene
5. ___________an individual bearing two alleles that are the same
for a particular trait 6. ________________ an allele that is
expressed, whether in...
Help Center 07 Question 38 What phenotypic ratio of offspring would result from a cross between a white-eyed male fruit fly, and a female that is heterozygous for red eye color? (See Chapter 8.3 Figure your e Text) 2 white-eyed females 2 red-eyed males V none of these 2 red-eyed female 1 white-eyed female: 1 red-eyed male: 2 white-eyed male 1 red-eyed female 1 white-eyed female 1 red-eyed male: 1 white-eyed male 2 red-eyed females: 2 white-eyed males Question 39...
Bio215 Problem Questions Hassanaton MULTIPLE CHOICE 1) Pea plants were particularly well suited for use in Mendel's breeding experiments for all of the following reasons exep that ) peas show easily observed variations in a number of characters, such as pea shape and flower color. B) it is possible to completely control matings between different per plants. C) it is possible to obtain large numbers of progeny from any given cross. D) peas have an unusually long generation time. E)...
2. A dominant allele H reduces the number of body bristles that Drosophila flies have, giving rise to a “hairless” phenotype. In the homozygous condition, H is lethal. An independently assorting dominant allele S has no effect on bristle number except in the presence of H, in which case a single dose of S suppresses the hairless phenotype, thus restoring the "hairy" phenotype. However, S also is lethal in the homozygous (S/S) condition. What ratio of hairy to hairless flies...