Mainly Q6e DGametes $BSb←SBGametes SBSbAssessment criteria 2.3.2.4Q6. Guinea pigs, which were homozygous for long, black hair were crossed with ones whichwere homozygous for short, white hair. All the F1 offspring had short, black hair.$BVsb.(a) Using suitable symbols, draw a genetic diagram showing parents genotypes, to explainthis result in F1 offspring.Parents genotype is SSBB (long black hair) and SSbb (short white hair)Black hair - BShort hair - Swhite hair - bLong hair - s(b) Complete the Punnett square to show the results of interbreeding the F1 offspring.sb.SsBb(Short black hair)Ssbb.(Short white hair)20SsBb (Short black hair)SSBB(Short black hair)SSBB(Short black hair)SSBB(Short black hair)SsBb(Short black hair)TPhenotypeShort black hairShort white hairLong black hairLong white hairAccessibility: InvestigateSbAaBbCcDdEe AaBbCeDdEeNo SpacingSSBb(Short black hair)SSbb(Short white hair)SsBb(Short black hair)Ssbb.(Short white hair)RatioNormal9:7(c) Complete the following table to show the different phenotypes you would expect in theF2 and their ratio.3:133:131:15AaBbCcDHeading 1SB.SSBB(Short black hair)SsBb(Short black hair)$$.B.B.(Long black hair)ssBb.(Long black hair)(d) State the ratio of phenotypes expected in the F2 offspring.The ratio of phenotype in F2 offspring is 9:3:3:1AaBbCcD.Heading 2ssBb(Long black hair)ssbb(Long white hair)(e) Mendel's Second Law of inheritance states that, "during gamete formation thesegregation of each gene pair is independent of other pairs." How do the diagramsabove prove this law? (AC 2.4).

Name: Mainly Q6e DGametes $BSb←SBGametes SBSbAssessment criteria 2.3.2.4Q6.
Uploaded: 2024-03-20T06:42:55.000Z
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Description: Mainly Q6e DGametes $BSb←SBGametes SBSbAssessment criteria 2.3.2.4Q6. Guinea pigs, which were homozygous for long, black hair were crossed with ones whichwere homozygous for short, white hair. All the F1 offspring had short, black hair.$BVsb.(a) Usin