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- For the following Chi-square table, how many degrees of freedom should be used when calculating the p-value? Gender Phenotype O E O-E (O-E)2 (O-E)2/E Male Disease 99 101.5 -2.5 6.25 0.061576 WT 105 101.5 3.5 12.25 0.12096 Female Disease 0 0 0 0 0 WT 202 203 -1 1 0.004926 Total 406 406 19.5 0.187192 You do not need to know the degrees of freedom to calculate the p-value. a. 4 b. 5 c. 2 d. 3I need help in table 2 & 3 please Table 1 Phenotype O E O-E (O-E)^2 (O-E)^2 /E Disease, Male 304 433.875 129.875 16789.68 38.69 Disease, Female 267 433.875 166.875 26855.01 61.89 WT, Male 285 144.625 140.375 19705.14 136.25 WT, Female 301 144.625 156.375 24453.15 156.37 Total 1157 1157 393.20 DF 3 p-value 7.84 Expected progencies as per SLR MOI’s = 1:1 for both male and female Phenotype O E O-E (O-E)^2 (O-E)^2 /E Disease, Male 304 289.25 14.75 217.56 0.75 Disease, Female 267 289.25 22.25 495.06 1.85 WT, Male 285 289.25 4.25 18.06 0.06 WT, Female 301 289.25 11.75 138.06 0.46 Total 1157 1157 3.12 DF 3 p-value 7.84 Progenies are following SR MOI’s that parents have genotypes and progenies in ratio 1:1 In Table 2,…AAeres Shident Dashboard MA NCHLS: goformative.com/formatives/60065d5b16a20c0af8319cc9 Launch Meeting-Zoom O Advanced Biology Exam Mendel x t sd net bookmarks MAGC 4. 5. 6. 7. 8. 6. 10 11 12 19 20 21 22 23 24 25 OO0 00-O 3\ 13 14 15 16 17 18 3/16= 19% 1/16= 6% 0/16= 0% Which of Mendel's "rules," stating that genes for different characteristics are inherited separately, was proved by his dihybrid crosses? Rule of dominance Rule of unit factors Law of independent assortment Law of segregation A female Guinea pip is homoszygous dominant for Black fur (BB) and is mated with a homozygous recessive for white fur (bb) In a itter of 8 offspring there would be a probability of. Draw a punnet square
- polonlinemi.instructure.com/courses/25527/assignments/4743471?module_item_id%3D9163916 Due Feb 19 by 11:59pm Polnts 10O Submitting an external tool A person has blood type B. What combination of alleles could she have to give her this blood type? Select all that apply. O i O AB O Pi « PreviousEdio | Calendar X Edio | Student Da x days/1089977/lessons/1533969/variants/2439102/take/10/ A Q TEXT ANSWER The inheritance patterns for some traits in guinea pigs are listed in the table below. 1. Identify the phenotype of a guinea pig with the genotype HhBBrr. 2. Using the allele symbols in the table, identify the genotype of a guinea pig that is recessive for hair length, heterozygous for hair color, and homozygous dominant for hair texture. Trait Dominant Allele Recessive Allele hair short (H) long (h) length hair color black (B) white (b) rough (R) smooth (r) hair texture BIUG X₁ X¹ EEAA H Normal : √x Enter your answer here ŏooooo Questions Answered 中山川 Ω Ο Τ Edio | Calendar X RepostExchange X Practice All Changes Saved ContinueEdio | Course Student Lesson Using the letter T to represent t x + days/1089940/lessons/1533883/variants/2273231/take/14/ O Alert 1 of 1: Instructional Technology Subsidy Úp. DISMISS * Practice MULTIPLE CHOICE Using the letter T to represent the trait, what is the genotype for individual 7? TT TE it is impossible to determine from this picture Atempe t of 2 683838886 0888.o S of & Total Questions Answered A Changes Saund 21°F Sunny
- ab H Q9. Who's gonna take you home tonight? There has been a mix up in the maternity ward. The babies have managed to remove their wrist bands. On their wrist bands are their blood groups and names. and art Billy Smith Blood group:A Sarah Chadwick Blood group:AB Imran Ahmed Blood group:B Johnny Honest Blood group:0 nislax 80 Complete the table to match each parent to their baby by indicating the parental genotypes using the symbols IA, IB, and i, and then the name of the baby that belongs to each set of moo parents. Consider one blood type might have more than one genotype. ever Parental Blood groups Parents Genotype Baby Blood Group O and O si grappo AB and O 196² A and O AB and A Completed forms to be made available for external moderation.enetic Mutations: Mastery Test X E W C f2.app.edmentum.com/assessments-delivery/ua/mt/launch/49372087/858339845/aHR0cHM6Ly9mMi5hcHAUZWRtZW50dW0uY29tL2xlYXJuZXItdWkvc2Vjb25 mentum Learni... My Virtual Academ... 1 v Next → Genetic Mutations: Mastery Test 3 2022 Edmentum. All rights reserved O Type here to search F 1 F3 00 Order details - Walmart.com 4 Select the correct answer from each drop-down menu. The lac operon in E.coli regulates genes that code for enzymes required for breakdown of lactose. The lac operon is operon that is activated in the presence of F4 Z $ F5 OO % 5 € F6 E A X 6 O F7 8 Free Mind - YouTube & 7 glucose lactose sugar k V F8 SOUS * 8 F9 4X Reset F10 + Next acer F11 F12 Scr Lk - PrtSc SysRq + Pause Break v 67°F Sunny Del Ins Home ►/11 SDo you think a patient needs to have the SAME two mutant alleles of CFTR to present with/have cystic fibrosis? Why or why not?
- How do you get the probability of 81% for A2 being lost? And why do you multiply by 1/4?W O () ENG 9:37 am O GENBIO-1ST-SEM-MIDTERN X 9 Schoology G karyotype of a certain huma x 6 BigBlueButton - GNBIO Messenger My Questions | bartleby + A app.schoology.com/common-assessment-delivery/start/5385424680?action=Donresume&submissionld=643190401 The diagram below shows a karyotype of a certain human. 8. 10 11 12 13 14 15 16 17 18 19 21 22 X Y Based on the karyotype, which of the following statements is most likely true? O The individual has a genetic condition caused by a nondisjunction event. The individual has a genetic condition caused by the X and Y chromosomes being different sizes. O The individual has a genetic condition caused by a chromosomal duplication. O The individual has a genetic condition caused chromosomes number one being different sizes. GENBIO-1ST-SEM-.pdf O 245180335_56899...jpg Show all ... TID N DAD × IDD . I ID.Explain why CVAR should always be larger than the corresponding VAR VaR = Value at risk