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tables attached to fill in the punnett square
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- Incomplete Dominance/Codominance By Tara Chavez. Due on May 7th, 4:00pm Incomplete Dominance r Соpy link 三 MULTIPLE CHOICE QUESTION R RR Rr If you cross a red snapdragon with a white snapdragon what phenotype should you expect in 100% of the offspring if snapdragon flowers are expressed via incomplete dominance? Rr r Pink Red YouTube Rr White Lenovo !!!97 Assume that the trihybrid cross MM SS tt X mm ss TT is made in a plant species in which Mand S are dom- inant but there is no dominance between T and t. Consider the F2 progeny from this cross, and assume independent assortment. (a) How many phenotypic classes are expected? (b) What is the probability of genotype MM SS t? obib (c) What proportion would be expected to be 15 2nod bolleo homozygous for all genes? ablla Insninmobo.(omert 1 Draw Design Layout View Help Times New Roman -12 E-5 B I U M abc X₂ x Y Font 15 Paragraph 4 E 1 + E 3. 1 BE 4 BOBER 5 PCB 3063-Sec 799 Homework 3 Question 3: In Chapter 3, we discuss various mechanisms of sex determination, including the XX/XY system of placental mammals, and the ZZ/ZW system of birds. Monotremes, the most primitive mammals, have a very intriguing sex determination scheme that shares elements of both of these systems. Please use the internet to learn more about how sex in monotremes is determined. Then write a short essay (300-400 words) describing this system and comparing/contrasting it to the XX/XY and ZZ/ZW systems. You need to include at least two references. References A A Mailings Aa aly Y 2... Ap A M Review Acrobat 23 Tel 15
- PunnettSquaresPractice (1).pdf /Downloads/PunnettSquaresPractice%20(1).pdf D Page view A Read aloud V Draw Highlight ge 2 of 3 In dogs, there is a hereditary type of deafness caused by a recessive gene. Two dogs who carry the gene for deafness but have normal hearing are mated. What are the possible genotypes and phenotypes of their offspring and the percent chance for each? D D Fill in the table: Genotype Phenotype Chance In guinea pigs, short hair is dominant over long hair. If a short haired SS guinea pig is crossed with a long haired ss guinea pig, what are the possible genotypes and phenotypes of their offspring and the percent 3.Im pP x PP 6/6 wa s hatarosygous genotypa? Dark eyes Dark eyes 11. What percentage of the offspring will have a heterozygous genotype? 12. What will the ratio of flies with dark eyes to white eyes be? Dd x x Dd 13. What percentage of the offspring will be short? Tall Tall 14. What percentage of the offspring will have a homozygous dominant genotype? TT х Tt 15. What percentage of the offspring will Long beak Long beak have long beaks?VHO Has typ D VIO0a ana a Wom What would be the expected phenotypes (in percent) of their children? In fruit flies, a gene that controls eye color is X-linked. Red eyes are dominant to the allele for white eyes. What 3) are the expected offspring phenotypes (in percent) if a heterozygous red-eyed female is crossed with a white- eyed male? d man with normal 11 1 woman m
- Trivla Game Show _Make Your Own Tri ngston.schoology.com/common-assessment-delivery/start/4789189591?action=onresume&submissionld=463322566 Dillon WF g Aa v Done In guinea pigs, black hair (B) is dominant to white hair (b) and rough hair (R) is dominant to smooth hair (r). What are all the possible genotypes of a guinea pig that has black, rough hair? (Select all that apply.) O BBRR BBRr BBrr BBRR BbRr O bbRR O bbRr O bbrr O Black O White O Rough OSmooth O Rough O Smoothdd-ons Help в I U A Calibri 12 三 三1 |:三 6. Consider a guinea pig with a homozygous genotype and a white fur color phenotype. a. What is the probability this parent will produce a gamete with the dominant allele? b. What is the probability this parent will produce a gamete with the recessive allele? C. If 31 sperm cells are collected from this guinea pig, how many would you expect to have the recessive allele (as determined by sequencing the gene)? !!!Search the menus (Alt+/) A, ? 100% Normal text Calibri + в I UА 11 I 1 2 | 3 5 | 6 7 8 9 10 11. In cockroaches, red exoskeletons (bodies) are recessive to black exoskeletons. A red male cockroach is crossed with a heterozygous black female cockroach. Twenty-four "baby" cockroaches are hatched from the egg. How many of these do you expect to be black? How many do you expects to be red? Allele and Phenotype All Genotype and Phenotype Possibilities Parent Genotypes Punnett Square Answer Letter Phenotype Genotype Phenotype Dad Mom
- Help O Editing A A A E E v 12 x Aa A- AaBbCc AaBbCc No Spacing AaBbCc 三 Normal Heading 1 Font Paragraph Styles Edi Some dogs bark when trailing, others are silent. Barking while trailing (B) is dominant to the silent trailer (b). Erect ears (E) are dominant to drooping ears (e). What kinds of pups would be expected from a heterozygous, erect-eared barker mated to a droop-eared silent trailer? 8. Genotype of the heterozygous, erect-eared barker? Genotype of the droop-eared, silent trailer? Genotypes Phenotypes 80% 127O e. Parent 2: Parent 1: Parent 2: ¡Ai QUESTION 9 Bi QUESTION 8 Let's assume that, in dragons, red scales (B) are dominant to green scales (b), and long tongues (S) are dominant to short tongues (s). The genes that determine these characteristics assort independently. A homozygous red, long-tongued dragon is crossed with a homozygous green, short-tongued dragon. If an F1 dragon is crossed to a homozygous green and homozygous short-tongued dragon, what phenotypes and proportions are expected in the offspring? O a. 100% green and long-tongued O b. ½ green and short-tongued & ½ red and long-tongued O c. ½ red and short-tongued & ½ green and long-tongued O d.% red and long-tongued, % red and short-tongued, ½ green and long-tongued, % green and short-tongued O e. 9/16 red, long-tongued, 3/16 green, long-tongued, 3/16 red, short-tongued, 1/16 green, short-tongued Save and Submit to save and submit. Click Save All Answers to save all answers. 000 MacBook AirYou are to discover the chromosomal location (autosomal or X linked) and mode of inheritance for two new fly mutations coughrotato (cp) and happyhour, (bb) The homozygous couchpotato (cpcp) phenotype is (sedentary) an extreme lack of activity The homozygous happyhour (hhhh) can drink any other fly under the table so phenotype (drinker) PARENTS Cross Male double homozygous mutants cp hh with Female wild wild CP HH are to be crossed BUT FIRST you don't know if either gene is on the X chromosome Parental diploid genotypes could be: male çpcphbbh female CPCPHHHH if Both genes autosomal or Male XcpYhbhh female XCPXCPHHHH if cp on the X chromosome and bh autosomal or Male cpcpXhhY female CPCPXHHXHH if hh on the X chromosome and cp autosomal What are the female and male PARENTAL GAMETE GENOTYPES for the above parental crosses if: a) Both genes were autosomal ((REMEMBER BOTH GENDER HAVE ONE GAMETE GENO) b) cp gene on the X chromosome and hh autosomal (REMEMBER MALES WITH HAVE 2 GAMETE GENO) c)…