Human Heredity: Principles and Issues (MindTap Course List)
11th Edition
ISBN: 9781305251052
Author: Michael Cummings
Publisher: Cengage Learning
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- What is the correct table in these two given? 1. Take two coins and assume that heads represent the dominantallele (A) and tails represents the recessive allele (a). The genotype for each coin isheterozygous (Aa).2. Assume that each coin represents one parent. When a single coin is flipped, one gameteis formed (through the process of meiosis). If the flipped coin is on heads, then thegamete has the dominant allele (A). When both coins are flipped simultaneously, therewill be two possible gametes that can combine through fertilization to form a zygote. Eachtime you flip both coins, you will record the “genotype” of the offspring.3. Flip the coins 100 times and record your results in the chart belowarrow_forward38 Which cross would produce phenotypic ratios that would best illustrate the Law of Dominance? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. TT X tt TT X Tt tt X tt TTX TT 身arrow_forward25. a)What is “dihybrid cross”? b) Consider two traits color and height. For color, let Y=yellow be dominant over y=green; for height, let T=tall be dominate over t=short. i. Set up the P1 generation ii. enumerate the possible gametes in the sex cells iii. Put the possible gametes along the top and sides of a Punnett square. iv. Hence, or otherwise, find a) the proportion (in percentage) of the F1 are yellow and tall with genotype YyTt b) The probability of obtaining, the genotype YyTtarrow_forward
- please answer question 6.arrow_forwardIncomplete Dominance: 1) On the planet Iota Geminorum IV there exists a small hairy animal known as a tribble. It comes in three colors blue, red, and purple. This trait is controlled by a single locus with incompletre dominance. A homozygous CBCB is blue, a homozygous CRCR is red, and a heterozygous CBCR is purple. a. What would be the genotypes and phenotypes of the offspring if a blue tribble were crossed with a red one? b. If the offspring in #1 were crossed what would the genotypic and phenotypic ratios be of this generation?arrow_forwardIn guinea pigs, the black coat (B) is dominant over the white coat (b), and straight hair (H) is dominant over curly hair (h). Using a Punnett square, complete the cross between a heterozygous black, curly-haired individual and a homozygous straight-haired, white individual. State the parent genotypes and gametes, and the F1 phenotypes and genotypes.arrow_forward
- DIHYBRID CROSS: In rabbits, assume that spotted skin (G) is dominant over non-spotted skin (g) and that wooly hair (B) is dominant over nonwooly hair (b). Cross a marriage between a male rabbit with the genotype GGBB with a female rabbit with the genotype ggbb. Show the P1, F1, P2 and F2 generation on a Punnett Square. Give the genotypic and phenotypic ratio of the offspring and the tally of genotype and phenotype of the offspring.arrow_forward9. Please determine whether the allele responsible for the trait is dominant or recessive. (A) (B) 1 2 10.For a cross between a female color blind carrier and a male color blind person, what is the probability for the female offspring to be color blind? What is the probability for the male offspring to be color blind? Please explain. 11.Consider the genetic map shown below AD Which two alleles have the highest recombination frequency? Which two have the lowest recombination frequency? O Focus ! 72°F DELL F5 F6 PrtScr Insert Hor FZ F8 F9 F10 F11 F12 23 %24 3. 4. 8 E R Tarrow_forwardGiven the karyotype shown at right, is this a male or a female? Normal or abnormal? What would the phenotype of this individual be?arrow_forward
- 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?arrow_forwardAs it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progenyarrow_forwardA true-breeding rabbit with agouti (mottled, grayish brown) fur crossed with a true-breeding rabbit with chinchilla (silver) fur produces all agouti offspring. A true-breeding chinchilla rabbit crossed with a true-breeding Himalayan rabbit (white fur with pigmented nose, ears, tail, and legs) produces all chinchilla offspring. A true-breeding Himalayan rabbit crossed with a true-breeding albino rabbit produces all Himalayan offspring. Explain the inheritance of the fur colors.arrow_forward
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