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- _30*_SP23 - General Biology I (for maj A 1:1 phenotypic ratio will occur when which of the following crosses is done? us page Select one: F1 O a. Ttx Tt O b. TT x tt O c. TT X TT TTx Tt tt x tt tt x Tt GgTt x GgTt GGTT x ggtt GGtt x ggTT O d. e. O f. O g. Oh. Oi. F2 # 80 F3 LA $ 000 000 F4 % F5 MacBook Air < F6 & F71_30*_SP23 - General Biology I (for majors 4 I of evious page A 1:2:1 genotypic ratio will occur when which of the following crosses is done? F1 Select one: O a. GgTt x GgTt O b. TTx TT O c. tt x tt O d. O e. O f. Og. Oh. Oi. TTx Tt tt x Tt GGtt x ggTT GGTT x ggtt TT x tt Tt x Tt F2 80 F3 F4 F5 MacBook Air F6 A F7male affected with haemophilia unaffected male 2 unaffected female Rhesus Rhesus positive negative 30 4 6 7 Rhesus Rhesus Rhesus Rhesus Rhesus positive positive positive positive positive 8 9 10 11 Rhesus Rhesus Rhesus Rhesus negative positive positive positive 12 Rhesus negative The allele for Rhesus positive, R, is dominant to that for Rhesus negative, r. Haemophilia is a sex-linked condition. The allele for haemophilia, h, is recessive to the allele for normal blood clotting, H, and is carried on the X-chromosome. The diagram shows the Rhesus blood group phenotypes in a family tree where some individuals have haemophilia. Use the information in the diagram to give one piece of evidence that the allele for the Rhesus negative condition is recessive. Explain the evidence from the cross between individuals 3 and 4 that the gene controlling Rhesus blood group is not sex-linked
- 30*_SP23 - General Biology I (for majo us page Ö A 3:1 phenotypic ratio will occur when which of the following crosses is done? Select one: O a. Ttx Tt O b. O c. O d. O e. O f. GgTt x GgTt GGtt x ggTT GGTT x ggtt TT x tt tt x Tt tt x tt Oh. TT x Tt Oi. TT x TT g. O g. 000 MacBook Air- 1 attachment If these two individuals with the following genotypes are crossed Parent 1: AAbbCCDDeeFf Parent 2: aaBBccddEEFf What is the expected genotypic ratio of the F, generation? (3 pts) A. ¼ AaBbCcDdEeFf: ½ AaBbCcDdEeFF: ¼ AaBbCcDdEeff B. ½ AaBbCcDdEeFf: ½ AaBbCcDdEeff C. 1 AaBbCcDdEeFF: 1 AaBbCcDdEeFf: 1 AaBbCcDdEeff D. 4 AaBbCcDdEeFF: ½ AaBbCcDdEeFf: 4 AaBbCcDdEeffThe second parental (P2)cross using a Punnett square Male Female Gg gg G GG Gg Because the P1 and P2 crosses consider the inheritance of only one character, seed color, they are called monohybrid crosses. Activity 3. SELF- CHECK 1. Windows peak (S) is dominant over straight hairline (s). Given that the mother has heterozygous gene pair for window's peak and the father has straight hairline. a. What is the genotype of the mother? b. What is the genotype of the father? C. How many kinds of gametes can the mother produce? List them down. d. How many kinds of gametes can the father produce? List them down. e. Perform a cross using a Punnett square. f. Give the genotypic ratio of the cross g. Give the phenotypic ratio of the cross
- Dihybrid crosses I only need the phenotypic ratioThe genotype XXY corresponds to Klinefelter syndrome Turner syndrome Triplo-X Jacob syndromeoter 10 Assignment Saved 10 X linked inheritance Determine the genotypes of the F2 generation in this example of X-linked inheritance of color blindness. ts еВоok Parents Print ceferences X x XBY XBXB Key XB = Normal vision Xb = Color blind оосytes Normal vision XB Xb Color blind XB Phenotypic Ratio Females All Males 1 Y Offspring 080 acer %24 sperm O+
- 1. Study the given alleles. Write the correct phenotype for each genotype. X– normal Genotype XC – Color-blind Phenotype XX XY XCXC www m XX www w ww w 2. Study the given alleles. Write the correct genotype for each phenotype. X- normal Genotype хн- Hemophiliaс Phenotype Hemophiliac female Hemophiliac male Normal female carrier of the gene Normal male Normal femaleW 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.30*_SP23 - General Biology I (for maj s page A 1:1 phenotypic ratio will occur when which of the following crosses is done? Select one: O a. Tt x Tt b. TT x tt O c. TTx TT O d. TTx Tt O e. tt x tt O f. tt x Tt O g. GgTt x GgTt h. GGTT x ggtt GGtt x ggTT O i. MacBook Air