小鼠Y染色体上的S基因决定雄性性别的发生,在X染色体上无等位基因,带有S基因的染色体片段可转接到X染...

Created on: February 4, 2025

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小鼠Y染色体上的S基因决定雄性性别的发生,在X染色体上无等位基因,带有S基因的染色体片段可转接到X染色体上。已知配子形成不受S基因位置和数量的影响,染色体能正常联会、分离,产生的配子均具有受精能力;含S基因的受精卵均发育为雄性,不含S基因的均发育为雌性,但含有两个Y染色体的受精卵不发育。一个基因型为XYS的受精卵中的S基因丢失,由该受精卵发育成能产生可育雌配子的小鼠。若该小鼠与一只体细胞中含两条性染色体但基因型未知的雄鼠杂交得F1,F1小鼠雌雄间随机杂交得F2,则F2小鼠中雌雄比例可能为() A. 4: 3 B. 3: 4 C. 8: 3 D. 7: 8

Thought

Examining genetics

I’m looking into how an XYS genotype results in a female mouse once the S gene is lost, leading to the production of viable female gametes.

Assessing genetic patterns

I’m pinpointing the female's genotype, with suggestions pointing to XY. This female, due to the lack of the S gene, is phenotypically female and typically can't produce male offspring.

Mapping out the genes

Tracing the female's XY genotype and unseen male genotype to spot male-to-female ratios, considering factors like the S gene on the X chromosome.

Assessing the father's genotype

OK, let's think it through: the father likely has at least one S gene. This leads to considering possibilities like XY^S, X^S X, or X^S Y^S, while ruling out certain combinations.

Mapping out male patterns

I’m thinking through the possibilities for male genotypes: XY^S, X^S X, and X^S Y^S, steering clear of combinations like Y^S Y^S.

Focusing on sex chromosomes

I'm pulling together the father’s genotype, considering combinations like XY^S and X^S X, while mindful of 2 Y chromosomes being unviable.

Weighing possibilities

I'm collecting info on genotypes and the father: XYS and X^S Y look promising. X^S Y^S is a non-starter due to its lethal implications.

Debunking the theory

OK, let's see. The X^S Y^S genotype works since only one Y chromosome is present, making fatal complications unlikely. Other conditions like Y^S Y^S and X^S X^S are avoided.

Taking a closer look

I'm starting to see that X^S X^S might be plausible if the two X chromosomes don't cause inviability. Other genotypes like X^S Y and Y^S X are also potentially viable.

Piecing together

I’m thinking the mother’s genotype is X Y, with the S gene lost. Normally, XY is male with S, so this female development is intriguing.

Examining the Y chromosome

I’m curious about the mother’s Y chromosome without the S gene, which still lets her produce viable female gametes, as viable gametes emerge as X or Y.

Piecing it together

Shifting my focus to the mother’s ability to produce both X and Y gametes, which is unusual yet follows the problem's logic. The father must carry at least one S chromosome.

Mapping the options

I’m listing each crossing possibility between the mother and father, aiming to pinpoint the correct male to female ratio among the F2 mice. It’s interesting to see where this leads.

Analyzing fertilization

I’m thinking through how the mother’s and father’s gametes combine, considering viability by removing any potential deformations or inviable combinations, and then examining the resulting F1 and F2 generations.

Tracking gene impact

I’m suggesting a simpler method: analyzing the S gene distribution in the population and examining how the mother’s Y gametes and the father’s Y^S gametes influence the F2 sex ratio.

Assessing genotype outcomes

I’m mapping out the distribution of genotypes and sex determination for F1 and F2 generations, using a standard XY^S father and a mother with X Y genotypes.

Analyzing gamete combinations

I’m examining how the mother’s X and Y gametes combine with the father’s X and Y^S gametes, resulting in viable male and female F1 zygotes.

Ratios among viable F1

I’m thinking through the survival rate of F1 zygotes, excluding the lethal combination, and finding the genotypic ratio among the viable ones to be 1:1:1.

Examining gender ratio

I’m noting the gender ratio of F1 progeny from a XY^S father: 2/3 female and 1/3 male, considering the viable genotypes XX, X Y^S, and X Y.

Analyzing F1 mating

Considering female F1 genotypes (XX and XY) and male F1 genotype (XY^S). Female F1 split equally into XX and XY, each 1/3.

Assessing mating dynamics

Females (XX and XY) mate with males (XY^S). With a 2:1 female to male ratio, each male pairs with 2 females on average.

Assessing F2 outcomes

First, half of F2 zygotes are female (XX), the other half male (XY^S). This trend might continue, but let me see further.

Evaluating gamete possibilities

Crossing XY females with XY^S males reveals three viable F2 zygote outcomes: XX (female), XY^S (male), and XY (female), each with a 1/4 chance.

Tracking outcomes

OK, focusing on the critical 3/4 probability for survival, observing a 2:1 female to male ratio among the viable offspring. This makes me think about selection impacts.

Analyzing mating dynamics

I’m weighing the likelihood of XX and XY^S males for Cross 1, then XY females and XY^S males for Cross 2. Random mating ratios could tilt the scales.

Mapping mating dynamics

OK, let’s look at N F1 individuals where 1/3 are XX females, 1/3 are XY females, and 1/3 are XY^S males. Considering random pairing within these groups.

Calculating mating odds

Interestingly enough, with a 1/2 chance of picking XX or XY females, each cross is equally likely. The sex ratio from Cross 1 remains 1:1.

Balancing ratios

Taking a closer look at Cross 1 and Cross 2, each contributing 2k offspring, the combined sex ratio averages out to a balanced 1:1. Progressing through the calculations reveals more clarity.

Breaking down ratios

OK, I’m thinking through the ratio of 2:1 in Cross 2. This leads me to conclude that among 2k offspring, 2/3 are female and 1/3 are male.

Weighing the odds

I'm thinking through the fractions of XX and XY females from a 1/3 ratio, comparing this with the total weighting, and noting the absence of 7:5 in the answer choices.

Mapping out mating odds

OK, let me see. I’m calculating the probability of different pairs within the population and aligning the resulting sex ratios with the provided options to check for congruence.

Refining details

Let's refine the sex ratio, considering cross 2's 1/4 lethal chance, leading to a 3/4 survival and revealing the struggle to match a finite ratio from discrete outcomes.

Mapping out offspring

I'm calculating the total number of litters and their composition by considering the ratio of XX and XY females to XY^S males, keeping in mind the likelihood of each offspring type.

Mapping out the pairing process

I'm detailing the pairing of XX and XY females with XY^S males, ensuring equal mating capacity and averaging offspring per mating to simplify the calculation.

Evaluating potential outcomes

I'm thinking through the expected number of surviving offspring per cross, defined as 1 for simplicity, and exploring the probability of each sex ratio outcome from the offspring.

Piecing it together

I’m examining expected surviving offspring and their sex ratios for both cross types. Cross 1 has a 1:1 ratio, while Cross 2 has a 2:1 ratio with 3/4 surviving children.

Calculating total offspring

First, I’m figuring out the N/3 type 1 matings producing N/6 females and N/6 males. Then, adding type 2 matings, I get N/6 females and N/12 males from N/4 offspring.

Summing up offspring

OK, let me see. I’m balancing the male-to-female ratio, which is 4:3, aligning with the provided answer choices for the scenario. It's interesting to see the ratio match.

Reasoned for 10m 19s

Answer

分析思路概要:

题目给出的关键信息可以归纳为:

  1. S 基因决定雄性

    • 只要合子中含有 S 基因(不管是在 X 还是 Y 上),该个体发育为雄性;
    • 如果不含 S 基因,则发育为雌性;
    • 若合子含有两个 Y 染色体(无论有没有 S 基因),则不发育。
  2. 母鼠的基因型

    • 题目叙述:本来受精卵是 X Y^S,但发育过程中 S 基因丢失,最终长成一个“无 S 基因”的雌鼠,且能产生可育的雌配子。
    • 由于该雌鼠最初带有 X 和 Y 染色体,只不过 Y 染色体的 S 基因丢失,故可将其表述为 X Y^-(Y^- 表示失去 S 基因的 Y)。
    • 这只雌鼠产生配子时,既可以产生 X 型配子,也可以产生 Y^- 型配子,且题意说明这些配子都可受精发育(只要最终合子不出现“两条 Y”的情况即可)。
  3. 父鼠的基因型仅知“含两条性染色体且能表现为雄性”

    • 父鼠必然带有 S 基因(否则不能发育成雄性)。
    • 同时,题目也说“两条 Y 的合子不发育”,所以父鼠不可能是 “Y + Y^S”。
    • 常见且简单的一种情况就是 X Y^S(传统意义上“正常”的雄性:Y 染色体上有 S)。
    • 当然理论上也可能是别的带 S 的组合(如 X^S X、X^S Y^- 等),但只要能证明 X Y^S 这一常见型就能得到题目所需的 F_2 雌雄比,则可判定选项。

下面先以父鼠 = X Y^S为例,详细推导 F_2 的雌雄比例。若能得到与选项对应的结果,则可直接判断答案。


第一代 (F_1) 的产生

  • 母方配子(X Y^- 雌鼠):

    • 可能产生的配子:1/2 概率为 X,1/2 概率为 Y^-。
  • 父方配子(X Y^S 雄鼠):

    • 可能产生的配子:1/2 概率为 X,1/2 概率为 Y^S。

将这两方配子两两组合,得到的合子类型及其命运如下表:

母方\父方X (1/2)Y^S (1/2)
X (1/2)XX(雌)X Y^S(雄)
Y^- (1/2)X Y^-(雌)Y^- Y^S(不发育)
  • 其中,Y^- + Y^S 由于含有 两条 Y,不发育。
  • 其余三种情况均可发育:
    1. XX → 雌
    2. X Y^- → 雌
    3. X Y^S → 雄

因此,F_1 能正常出生的后代中,雌雄比例 = 2 : 1,具体为

  • 2/3 雌(XX 或 X Y^-),
  • 1/3 雄(X Y^S)。

第二代 (F_2) 的产生

F_1 中能存活的个体里:

  • 1/3 为雄鼠 (X Y^S),
  • 2/3 为雌鼠,且这 2/3 的雌鼠里又平均分为:
    • 1/3 总体 = XX 雌鼠,
    • 1/3 总体 = X Y^- 雌鼠。

F_2 是让这些 F_1 的雌雄随机交配后所生。由于 F_1 只有一种雄鼠基因型(X Y^S),雌鼠则有两种 (XX 或 X Y^-)。我们分两类交配来讨论:


1. XX 雌鼠 × X Y^S 雄鼠

  • 母方 (XX) 产生的配子:只会是 X;
  • 父方 (X Y^S) 产生的配子:1/2 X,1/2 Y^S。

于是合子类型:

  1. X(母) + X(父) = XX → 雌 (无 S)
  2. X(母) + Y^S(父) = X Y^S → 雄 (有 S)

该类交配的后代雌雄比 = 1 : 1 (且都能存活)。


2. X Y^- 雌鼠 × X Y^S 雄鼠

  • 母方 (X Y^-) 产生的配子:1/2 X,1/2 Y^-;
  • 父方 (X Y^S) 产生的配子:1/2 X,1/2 Y^S。

组合如下表:

母方\父方X (1/2)Y^S (1/2)
X (1/2)XX(雌)X Y^S(雄)
Y^- (1/2)X Y^-(雌)Y^- Y^S(不发育)
  • 同理,Y^- Y^S 不发育(双 Y)。
  • 其他三种合子都能正常发育:XX(雌)、X Y^-(雌)、X Y^S(雄)。
  • 在这 4 种配子组合中,有 1/4 不发育,剩余 3/4 可以发育且存活。存活下来的 3/4 后代中,2 个雌 (XX、X Y^-),1 个雄 (X Y^S),故该类交配存活后代的雌雄比 = 2 : 1

将两类交配按概率合并

在 F_1 群体里:

  • 一半的雌鼠是 XX,另一半是 X Y^-。
  • 雄鼠都为 X Y^S。
  • 假设每种雌鼠与雄鼠的交配机会相等(即雌鼠里 50% 是第一种,50% 是第二种),则我们把两种交配结果“加权平均”。

记“每对交配平均产生 N 个受精卵”:

  1. XX × X Y^S 这类交配:

    • 全部受精卵都能发育 (无双 Y 问题),后代雌雄 1 : 1;
    • 若共有 N 个受精卵,则可存活 N 个后代,雌雄各 N/2、N/2。
  2. X Y^- × X Y^S 这类交配:

    • 受精卵里有 1/4 出现双 Y 而不发育,故存活后代数 = 3/4 N;
    • 存活的后代里雌雄比 = 2 : 1;
    • 因此雌 = (2/3)×(3/4 N) = 1/2 N,雄 = (1/3)×(3/4 N) = 1/4 N。

两种交配各占一半时,总体来看(把两类加起来):

  • 雌性总数 = (来自第一类) N/2 + (来自第二类) 1/2 N = N/2 + N/2 = N。
  • 雄性总数 = (来自第一类) N/2 + (来自第二类) 1/4 N = N/2 + N/4 = 3N/4。
  • 总存活后代 = N + 3N/4 = 7N/4。

因此,F_2 存活后代的 雌 : 雄 = N : (3N/4) = 4 : 3


对应题目选项

题目中给出的四个比值选项分别是:

A. 4 : 3
B. 3 : 4
C. 8 : 3
D. 7 : 8

由以上分析可知,当父鼠是较“常规”的 X Y^S 型雄鼠时,最终 F_2 的雌雄比会是 4 : 3。故正确答案即为:

A. 4 : 3


补充说明

  • 如果父鼠具有其他带 S 基因的性染色体组合(如 X^S X、X^S Y^- 等),在 F_1、F_2 各代也可能出现不一样的雌雄统计结果;不过一般解答竞赛/考试类题目,找到其中“能够出现的某种比值”即可说明题目的正确选项。
  • 题目问的是“可能的”雌雄比例,而不是“唯一的”雌雄比例,因此只要证明 4 : 3 能出现,便足以匹配选项 A。

总结

根据典型且最简便的父本基因型 X Y^S 分析,F_2 最终能够出现的雌雄比例为 4 : 3。故答案选择 A

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